ACT Mathematics Tests 2024

$ \angle ACD = \angle BAC + \angle ABC \quad\dots\text{exterior angle of a triangle = sum of the two interior opposite angles.} \\[3ex] \angle ACD = 35 + 95 \\[3ex] = 130^\circ $

Cost for dog collars @ $4 per dog collar = $4 \cdot x = 4x$
Shipping and Handling Fee = $5
∴ $y = 4x + 5$

$ \text{width} = 6\;km \\[3ex] \text{length} = 4 + 2\cdot \text{width} \\[3ex] = 4 + 2(6) \\[3ex] = 4 + 12 \\[3ex] = 16\;km \\[5ex] \text{Area} = \text{length} \cdot \text{width} \\[3ex] = 16 \cdot 6 \\[3ex] = 96\;km^2 $

$ |4 - 3| - |2 - 5| \\[3ex] |1| - |-3| \\[3ex] 1 - 3 \\[3ex] -2 $

$ (8p - 3q) - (2q + 6p) \\[3ex] 8p - 3q - 2q - 6p \\[3ex] 2p - 5q $

400 miles per hour implies 400 miles in 1 hour
60 minutes = 1 hour

$ \underline{\text{Unity Fraction Method}} \\[3ex] \text{Set up units: } 300\;miles * \dfrac{...\;hour}{...\;miles} * \dfrac{...\;minutes}{...\;hour} \\[5ex] \text{Set up measurements and units: } 300\;miles * \dfrac{1\;hour}{400\;miles} * \dfrac{60\;minutes}{1\;hour} \\[5ex] = 45\;minutes $

$ \left(x^{10}\right)^4 \\[4ex] = x^{10 * 4} \quad\dots\text{Law 5 Exp} \\[4ex] = x^{40} $

$ y = f(x) \\[3ex] f(x) = y \\[3ex] f(4) = 9 \\[3ex] \implies \\[3ex] x = 4 \\[3ex] y = 9 \\[3ex] \underline{Test\;\;each\;\;option} \\[3ex] Option\;A \\[3ex] f(x) = x^2 - 7 \\[3ex] f(4) = 4^2 - 7 \\[3ex] = 16 - 7 \\[3ex] = 9 \\[3ex] $ Option A is the correct answer.

1st day: 9 new spelling words
2nd day: 6 new spelling words, total = 9 + 6 = 15 new spelling words
3rd day: 6 new spelling words, total = 9 + 6(2) = 9 + 12 = 21 new spelling words
4th day: 6 new spelling words, total = 9 + 6(3) = 9 + 18 = 27 new spelling words
This is an Arithmetic Sequence

$ a = 9 \\[3ex] d = 6 \\[3ex] n = 30 \\[3ex] AS_n = a + d(n - 1) \\[5ex] AS_{30} = 9 + 6(30 - 1) \\[5ex] = 9 + 6(29) \\[3ex] = 9 + 174 \\[3ex] = 183 $

$ \dfrac{3x - 6}{2} + 5 = 18 \\[5ex] \dfrac{3x - 6}{2} = 18 - 5 \\[5ex] \dfrac{3x - 6}{2} = 13 \\[5ex] 3x - 6 = 2(13) \\[3ex] 3x - 6 = 26 \\[3ex] 3x = 26 + 6 \\[3ex] 3x = 32 \\[3ex] x = \dfrac{32}{3} \\[5ex] $ Check

$ x = \dfrac{32}{3}$

LHS	RHS
$ \dfrac{3x - 6}{2} + 5 \\[5ex] \dfrac{3\left(\dfrac{32}{3}\right) - 6}{2} + 5 \\[8ex] \dfrac{32 - 6}{2} - 5 \\[5ex] \dfrac{26}{2} - 5 \\[5ex] 13 - 5 \\[3ex] 8 $	18

At noon, Mia sees mile marker 117
Then, after 20 minutes, she sees mile marker 97
This implies that her mother drove 117 − 97 = 20 miles in 20 minutes

$ distance = 20\;miles \\[3ex] time = 20\;minutes \\[3ex] speed = \dfrac{distance}{time} \\[5ex] = \dfrac{20}{20} \\[5ex] = 1\;mile/minute \\[3ex] $ On average, Mia's mother drove 1 mile in 1 minute.
In other words, her average speed is 1 mile per minute.
But the question wants us to calculate the speed in miles per hour

$ 60\;minutes = 1\;hour \\[3ex] 1\;mile\;\;per\;\;minute = ?\;mile\;\;per\;\;hour \\[3ex] \underline{\text{Unity Fraction Method}} \\[3ex] \text{Set up units: } \dfrac{1\;mile}{minute} * \dfrac{...\;minute}{...\;hour} \\[5ex] \text{Set up measurements and units: } \dfrac{1\;mile}{minute} * \dfrac{60\;minute}{1\;hour} \\[5ex] = 60\;miles/hour \\[3ex] $ On average, Mia's mother drove 60 miles in 1 hour.
Her average speed is 60 miles per hour. (60 mph)

$ \angle 5 \cong \angle 8 \quad\dots\text{alternate exterior angles are congruent} $

The coordinates of the road sign is found by the Midpoint Formula.

$ Endpoint\;1 = (30, 50) \\[3ex] x_1 = 30 \\[3ex] y_1 = 50 \\[3ex] Endpoint\;2 = (90, 100) \\[3ex] x_2 = 90 \\[3ex] y_2 = 100 \\[3ex] \text{Midpoint} = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right) \\[5ex] = \left(\dfrac{30 + 90}{2}, \dfrac{50 + 100}{2}\right) \\[5ex] = \left(\dfrac{120}{2}, \dfrac{150}{2}\right) \\[5ex] = (60, 75) $

Total Expenses = $150 + $400 + $50 = $600
Expected Profit = $300
Expected Income = $600 + $300 = $900
Expected Number of students = 500
Price per student to generate expected income = 900 ÷ 500 = $1.80

F. is a circle with radius, 1 unit
G. is a square with apothem, 1 unit
H. is a square with radius, 1 unit
J. is an octagon with radius, 1 unit
K. is an octagon with apothem, 1 unit

As the number of sides of the polygon increases, the shape becomes more circular, making the area approach that of a circle.
This typically implies that the area of the circle is greater than the area of the octagon.
So, options J. and K. are out.

For the same radius of 1 unit, a circle has a greater area than a square because a circle encloses the maximum area possible for a given perimeter due to its continuous curvature, while a square has corners that do not use the radius as efficiently.
So, option H, is out.

We are now left with the two options of F. and G.
For the Square: The apothem is a shorter distance within the square (from center to midpoint of a side) which allows for a longer side length.
For the Circle: The radius of the circle defines the circle's circumference, leading to a smaller enclosed area when compared to the larger side length of the square.
Hence, a square with an apothem of 1 unit has a greater area than a circle with a radius of 1 unit.
Option F. is out.

This implies that the correct option is G.

Student: SamDom For Peace, I want to see these areas worked out so I know the correct option.
Teacher: No problem. We shall work it out.
However, please note that each question on the ACT is limited to about 1 minute.
Working it out will take more than a minute.
So, keep that in mind when taking your ACT.

$ \underline{\text{Option F: Circle}} \\[3ex] r = 1\;unit \\[3ex] A = \pi r^2 \\[3ex] A = \pi \cdot (1)^2 \\[4ex] A = 3.141592654\;square\;\;units \\[5ex] \underline{\text{Option G: Square}} \\[3ex] a = 1\;unit \\[3ex] L = 2a = 2(1) = 2 \\[3ex] A = L^2 = 2^2 \\[3ex] A = 4\;square\;\;units \\[5ex] \underline{\text{Option H: Square}} \\[3ex] r = 1\;unit \\[3ex] \implies\;\;4\;\;Right\;\;\triangle s \\[3ex] Area\;\;of\;\;1\;\;Right\;\;\triangle \\[3ex] b = r = 1\;cm \\[3ex] h = 1\;cm \\[3ex] A = \dfrac{1}{2} \cdot b \cdot h \\[5ex] A = \dfrac{1}{2} \cdot 1 \cdot 1 \\[5ex] A = \dfrac{1}{2}\;units \\[5ex] Area\;\;of\;\;4\;\;Right\;\;\triangle s \\[3ex] A = \dfrac{1}{2} \cdot 4 \\[5ex] A = 2\;square\;\;units \\[5ex] \underline{\text{Option J: Octagon}} \\[3ex] r = 1\;unit \\[3ex] n = 8 \\[3ex] A = \dfrac{\pi r^2}{2}\sin\left(\dfrac{360^\circ}{n}\right) \\[7ex] A = \dfrac{\pi \cdot (1)^2}{2}\sin\left(\dfrac{360^\circ}{8}\right) \\[7ex] A = \dfrac{\pi}{2}\sin 45^\circ \\[5ex] A = 1.110720735\;square\;\;units \\[5ex] \underline{\text{Option K: Octagon}} \\[3ex] a = 1\;unit \\[3ex] A = \dfrac{\pi a^2}{\cos^2\left(\dfrac{180^\circ}{n}\right)} \sin\left(\dfrac{360^\circ}{n}\right) \\[7ex] A = \dfrac{\pi \cdot (1)^2}{\cos^2\left(\dfrac{180^\circ}{8}\right)} \sin\left(\dfrac{360^\circ}{8}\right) \\[7ex] A = \dfrac{\pi\sin(45^\circ)}{\cos^2(22.5^\circ)} \\[7ex] A = 2.602580569\;square\;\;units \\[3ex] $ The results further confirm that Option G. is the correct answer.

$ \underline{Ann} \\[3ex] Grade\;\;Point\;\;Average \\[3ex] = \dfrac{\Sigma \text{Number of Grade Points}}{\Sigma \text{Number of class credits}} \\[5ex] = \dfrac{1(5) + 3(4) + 2(3) + 0(2) + 0(0)}{6(1)} \\[5ex] = \dfrac{5 + 12 + 6}{6} \\[5ex] = \dfrac{23}{6} \\[5ex] = 3.833333333 \\[3ex] \approx 3.83 \quad\dots\text{to the nearest 0.01} \\[3ex] $ Student: We were given the grades for 5 classes
But you used the number of credits for 6 classes
May you please explain?
Teacher: Nice observation.
The question specified 6 classes of 1 credit each.
We have to use what the question gave us.
Besides, if you decide to divide by 5(1), the answer is not listed in the option.
Be it as it may, we have to use what we are given.

Based on the Standard Form of the Equation of a Circle, compare: $(x - h)^2 + (y - k)^2 = r^2$ with $x^2 + y^2 = 16$
This implies that:

$ x^2 + y^2 = 16 \\[4ex] (x - 0)^ + (y - 0)^2 = 16 \\[4ex] \implies \\[3ex] center = (h, k) = (0, 0) \\[3ex] r^2 = 16 \\[4ex] r = \pm \sqrt{16} \\[3ex] r = \pm 4 \\[3ex] $ The radius cannot be a negative number.
However, we shall consider the two values in finding the intersection on the x-axis.

This implies that:
Intersecion on the x-axis = (-4, 0) and (4, 0)

$ s...t...d \implies speed \cdot time = distance \\[3ex] \underline{\text{Entire Journey}} \\[3ex] Distance = 450\;miles \\[3ex] Time = 9\;hours \\[5ex] \underline{\text{Already Driven}} \\[3ex] Distance = 180\;miles \\[3ex] Speed = 60\;miles\;per\;hour \\[5ex] Time = \dfrac{Distance}{Speed} = \dfrac{180}{60} = 3\;hours \\[5ex] \underline{\text{Remaining Journey}} \\[3ex] Distance = 450 - 180 = 270\;miles \\[3ex] Time = 9 - 3 = 6\;hours \\[3ex] Speed = \dfrac{Distance}{Speed} \\[5ex] Speed = \dfrac{270}{6} \\[5ex] Speed = 45\text{miles per hour} \\[3ex] $ The minimum average speed that JoAnna can drive for the remainder of the route and have a driving time of 9 hours for the entire trip is 45 miles per hour.

Due to the fact that one minute is allocated to each question on the ACT (60 minutes for 60 questions), it is better to check the solution to this question by their answer options.
So, let us check and eliminate until we get the answer.

$ \underline{Option\;F} \\[3ex] 18 \div 5 = 3 \;R\; 3 \; \checkmark \\[3ex] 18 \div 6 = 3 \;R\;0 \;\text{remainder is 0, not 4} \\[3ex] NEXT \\[5ex] \underline{Option\;G} \\[3ex] 23 \div 5 = 4 \;R\; 3 \; \checkmark \\[3ex] 23 \div 6 = 3 \;R\; 5 \;\text{remainder is 3, not 4} \\[3ex] NEXT \\[5ex] \underline{Option\;H} \\[3ex] 28 \div 5 = 5 \;R\; 3 \; \checkmark \\[3ex] 28 \div 6 = 4 \;R\; 4 \; \checkmark \\[3ex] STOP \\[3ex] $ Option H is the correct answer.
This is the answer because the answer choices are in ascending order.

Student: Is there another way to do this question without checking by the answer options?
Teacher: Yes, we can do it: Modular Arithmetic
However, I think it takes more than a minute to do.

$ Let: \\[3ex] dividend = d \\[3ex] quotient = q \\[3ex] 1st:\;\; \text{Remainder of 3 when divided by 5} \\[3ex] d \equiv 3 \mod 5...cong.(1) \\[5ex] 2nd:\;\; \text{Remainder of 4 when divided by 6} \\[3ex] d \equiv 4 \mod 6 ...cong.(2) \\[5ex] \implies \\[3ex] d = 6q + 4 ...eqn.(1) \\[3ex] Substitute\;\;eqn.(1) \;\;for\;\;d\;\;in\;\;cong.(1) \\[3ex] 6q + 4 \equiv 3 \mod 5 \\[3ex] \text{Test positive integers for q beginning from the first positive integer} \\[3ex] 6(1) + 4 = 10 \equiv 0 \mod 5...Not\;\;3 \\[3ex] 6(2) + 4 = 16 \equiv 1 \mod 5...Not\;\;3 \\[3ex] 6(3) + 4 = 22 \equiv 2 \mod 5...Not\;\;3 \\[3ex] 6(4) + 4 = 28 \equiv 3 \mod 5 \;\checkmark \\[3ex] \implies \\[3ex] d = 28 $

Let the number of hours after 8:00 am be h
at a constant rate of $\dfrac{3}{4}$ inch per hour ⇒ $\dfrac{3}{4}h$

until there was 6 inches of snow on the ground.

$ \dfrac{3}{4}h = 6 \\[5ex] 3h = 6(4) \\[3ex] 3h = 24 \\[3ex] h = \dfrac{24}{3} \\[5ex] h = 8\text{ hours} $

Let the:
diameter of the larger snowball = D
circumference of the larger snowball = C
diameter of the smaller snowball = d

$ \underline{\text{Larger Snowball}} \\[3ex] C = \pi D ...formula \\[3ex] C = 36\pi ...given \\[3ex] C = C \implies \\[3ex] \pi D = 36\pi \\[3ex] D = 36\;inches \\[3ex] \underline{\text{Smaller Snowball}} \\[3ex] 3d = D \\[3ex] d = \dfrac{D}{3} \\[5ex] d = \dfrac{36}{3} \\[5ex] d = 12\;inches $

This is a case of Conditional Probability
It may rain.
It may not rain.
If it rains, the snow may melt.
If it does not rain, the snow may melt.
Let us discuss the conditions.

The probability that rain will occur = 50% = 0.5
The probability that rain will not occur = 100% − 50% = 50% = 0.5 ...Complementary Rule

The probability that the snow will melt if it rains = 60% = 0.6
The probability that the snow will not melt if it rains = 1 − 0.6 = 0.4 ...Complementary Rule

The probability that the snow will melt if it does not rain = 10% = 0.1
The probability that the snow will not melt if it does not rain = 1 − 0.1 = 0.9 ...Complementary Rule

The question asked for the probability that the snow will melt, so we are going to look at the conditions that the snow will melt.

$ P(Rain) = 0.5 \\[3ex] P(Rain') = 0.5 \\[3ex] P(Melts | Rain) = 0.6 \\[3ex] P(Melts | Rain') = 0.1 \\[3ex] P(Melts) = P(Rain) \cdot P(Melts | Rain) + P(Rain') \cdot P(Melts | Rain') \\[3ex] = 0.5(0.6) + 0.5(0.1) \\[3ex] = 0.3 + 0.05 \\[3ex] = 0.35 $

$ \underline{Sample} \\[3ex] n = 15 \\[3ex] x = 5 \\[3ex] \hat{p} = \dfrac{x}{n} \\[5ex] = \dfrac{5}{15} \\[5ex] = \dfrac{1}{3} $

Let us convert the minutes to hours because:
12 noon is the 12th hour and
1 hour apart is in hour

$ 2\dfrac{1}{2}\;minutes \\[5ex] = \dfrac{5}{2}\;minutes \\[5ex] = \dfrac{5}{2}\;minutes \cdot \dfrac{1\;hour}{60\;minutes} \\[5ex] = \dfrac{1}{24}\;hour \\[7ex] 1\dfrac{1}{4}\;minutes \\[5ex] = \dfrac{5}{4}\;minutes \\[5ex] = \dfrac{5}{4}\;minutes \cdot \dfrac{1\;hour}{60\;minutes} \\[5ex] = \dfrac{1}{48}\;hour \\[7ex] \text{Let the number of hours = x} \\[3ex] Gaining\;\;2\dfrac{1}{2}\;minutes \implies Gaining\;\;\dfrac{1}{24}\;hour \implies +\dfrac{1}{24}x\;hours \\[5ex] Losing\;\;1\dfrac{1}{4}\;minutes \implies Losing\;\;\dfrac{1}{48}\;hour \implies -\dfrac{1}{48}x\;hours \\[7ex] \text{both watches are set correctly at noon} \implies \\[3ex] 12 + \dfrac{1}{24}x \;\;and\;\; 12 - \dfrac{1}{48}x\;hours \\[7ex] \text{exactly 1 hour apart} \implies \\[3ex] \left(12 + \dfrac{1}{24}x\right) - \left(12 - \dfrac{1}{48}x\right) = 1 \\[5ex] 12 + \dfrac{1}{24}x - 12 + \dfrac{1}{48}x = 1 \\[5ex] \dfrac{x}{24} + \dfrac{x}{48} = 1 \\[5ex] \dfrac{2x + x}{48} = 1 \\[5ex] 3x = 1(48) \\[3ex] x = \dfrac{48}{3} \\[5ex] x = 16\;hours $

Prism	Length	Width	Height	Volume = Length × Width × Height
A	4	4	3	48
B	3	3	4	36
C	4	3	1	12

$ 48 - 36 = what \cdot 12 \\[3ex] 12 = what \cdot 12 \\[3ex] what = 1 $

The complex conjugate of 2 + 7i is 2 − 7i

$ g(x) = x^2 - 2 \\[3ex] g(-3) = (-3)^2 - 2 \\[3ex] = 9 - 2 \\[3ex] = 7 \\[5ex] f(x) = 3x + 4 \\[3ex] f(g(-3)) = f(7) \\[3ex] = 3(7) + 4 \\[3ex] = 21 + 4 \\[3ex] = 25 $

$ \dfrac{21A^2B + 6AB^2}{3AB} \\[5ex] = \dfrac{3AB(7A + 2B)}{3AB} \\[5ex] = 7A + 2B $

$ 0.0002^n \\[4ex] = \dfrac{0\color{darkblue}{.0002}^n}{\color{darkblue}{10000}^n} \\[6ex] = \dfrac{2^n}{(10^4)^n} \\[6ex] = \dfrac{2^n}{10^{4n}}...Law\;5...Exp \\[6ex] = 2^n \times 10^{-4n} ...Law\;6...Exp $

$ 0.6, \;\; 0.08, \;\; \dfrac{5}{8} \\[5ex] 0.6, \;\; 0.08, \;\; 0.625 \\[3ex] 0.08 \lt 0.6 \lt 0.625 \\[3ex] \implies \\[3ex] 0.08, \;\; 0.6, \;\; \dfrac{5}{8} $

Selecting each digit from the set {2, 3, 9}
It's better to use a Punnett Square

$2nd\:\:Set\:\:\rightarrow$ $1st\:\:Set\:\:\downarrow$	2	3	9
2	22	23	29
3	32	33	39
9	92	93	99

$ P(99) = \dfrac{n(99)}{n(S)} \\[5ex] = \dfrac{1}{9} $

$ \underline{SOHCAHTOA} \\[3ex] \tan \theta = \dfrac{5}{\sqrt{39}} = \dfrac{opp}{adj} \\[5ex] \underline{Pythagorean\;\;Theorem} \\[3ex] hyp^2 = opp^2 + adj^2 \\[3ex] hyp^2 = 5^2 + \left(\sqrt{39}\right)^2 \\[4ex] hyp^2 = 25 + 39 \\[3ex] hyp^2 = 64 \\[3ex] hyp = \sqrt{64} \\[3ex] hyp = 8\;units \\[3ex] \sin\theta = \dfrac{opp}{hyp} \quad\dots\text{SOHCAHTOA} \\[5ex] \sin\theta = \dfrac{5}{8} $

Teacher: The answer is Option F.
Student: How do you know?
Just by looking at it?
Teacher: Yes...all other options do not have any term in y²
Student: But I would like to know how to expand it.
Teacher: Sure. We have at least two ways to do it.
But I recommend the process of eliminating incorrect options on the ACT due to the time limit of 1 minute per question.
This question takes about 2 seconds to do.
The remaining 58 seconds should be used for questions that may take more than a minute to solve.

$ \underline{\text{1st Approach: Pascal's Triangle}} \\[3ex] ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~1 \\[3ex] ~~~~~~~~~~~~~~~~~~~~~~~1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~1 \\[3ex] ~~~~~~~~~~~~~~~~~~1~~~~~~~~~~~~~~~~~~~~~~~~~~~~2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~1 \\[3ex] ~~~~~~~~~~~~~1~~~~~~~~~~~~~~~~~~~~3~~~~~~~~~~~~~~~~~~~~~~~~~~~3~~~~~~~~~~~~~~~~~~~~~~~~~~~1 \\[5ex] \implies \\[3ex] (y + 8)^3 \\[4ex] = y^3 + 3y^2(8) + 3y(8)^2 + 8^3 \\[4ex] = y^3 + 24y^2 + 192y + 512 \\[5ex] \underline{\text{2nd Approach: Algebraic Expansion}} \\[3ex] (y + 8)^3 = (y + 8)^2 \cdot (y + 8) \\[3ex] (y + 8)^2 = (y + 8)(y + 8) \\[3ex] = y^2 + 8y + 8y + 64 \\[3ex] = y^2 + 16y + 64 \\[5ex] (y + 8)^2 \cdot (y + 8) \\[3ex] = (y^2 + 16y + 64)(y + 8) \\[3ex] = y^3 + 8y^2 + 16y^2 + 128y + 64y + 512 \\[3ex] = y^3 + 24y^2 + 192y + 512 $

$ \underline{Original\;\;List} \\[3ex] \bar{x} = 88 \\[3ex] n = 4 \\[3ex] \Sigma x = \bar{x} \cdot n \\[3ex] \Sigma x = 88(4) \\[3ex] \Sigma x = 352 \\[3ex] 4th\;\;number = 50 \\[3ex] \Sigma three = 352 - 50 \\[3ex] \Sigma three = 302 \\[5ex] \underline{New\;\;List} \\[3ex] \Sigma three = 302 \\[3ex] 4th\;\;number = 70 \\[3ex] \Sigma x = \Sigma four = 302 + 70 \\[3ex] \Sigma x = 372 \\[3ex] \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] \bar{x} = \dfrac{372}{4} \\[5ex] \bar{x} = 93 $

$ -2x + 7 \ge 19 \\[3ex] -2x \ge 19 - 7 \\[3ex] -2x \ge 12 \\[3ex] x \le \dfrac{12}{-2} \\[5ex] x \le -6 \\[3ex] $ Check

$x \le -6$

LHS	RHS
$ Let\;\; x = -7 \\[3ex] -2x + 7 \\[3ex] -2(-7) + 7 \\[3ex] 14 + 7 \\[3ex] 21 $	19
21 ≥ 19

Area of the shaded region = Area of the Square − Area of the Circle

$ \underline{Square} \\[3ex] apothem,\;a = 4\;cm \\[3ex] Length,\;L = 2a = 2(4) = 8\;cm \\[3ex] Area,\;A = L^2 \\[3ex] A = 8^2 \\[3ex] A = 64\;cm^2 \\[5ex] \underline{Circle} \\[3ex] radius,\;r = 4\;cm \\[3ex] Area,\;A = \pi r^2 \\[4ex] A = \pi(4)^2 \\[4ex] A = 16\pi\;cm^2 \\[5ex] \implies \\[3ex] \text{Area of the shaded region} = 64 - 16\pi $

Rectangle A	Rectangle B
$ Length = L_A \\[4ex] Width = W_A \\[4ex] Perimeter = P_A \\[5ex] L_A = L_B \\[4ex] W_A = W_B - 3 \\[4ex] P_A = 2(L_A + W_A) \\[4ex] = 2(L_B + W_B - 3) \\[4ex] = 2L_B + 2W_B - 6 $	$ Length = L_B \\[4ex] Width = W_B \\[4ex] Perimeter = P_B \\[5ex] P_B = 2(L_B + W_B) \\[4ex] = 2L_B + 2W_B $
How many inches less is the perimeter of Rectangle A than the perimeter of Rectangle B? $ P_B - P_A \\[4ex] = (2L_B + 2W_B) - (2L_B + 2W_B - 6) \\[4ex] = 2L_B + 2W_B - 2L_B - 2W_B + 6 \\[4ex] = 6 $

$ \left(x^4\right)^{\dfrac{5}{4}}\left(x^5\right)^{\dfrac{2}{5}} \\[7ex] x^{4 * \dfrac{5}{4}} \cdot x^{5 * \dfrac{2}{5}} \quad\dots\text{Law 5 Exp} \\[7ex] x^5 \cdot x^2 \\[4ex] x^{5 + 2} \quad\dots\text{Law 1 Exp} \\[4ex] x^7 $

To get the length of the 3rd piece, we need to consider:
1st: the length of the 1st piece
2nd: the length of the 2nd piece
3rd: the width of each cut, also known as the kerf. The kerf is the thickness (width) of the cut made by the blade. It is an important factor to consider when making accurate cuts.
We shall determine the sum of these cuts, and subtract the sum from the length of the lumber to get the length of the third piece.

$ \underline{Total\;\;Length\;\;of\;\;Cuts} \\[3ex] Length\;\;of\;\;1st\;\;Piece = 3\;ft\;6\;in \\[3ex] Length\;\;of\;\;2nd\;\;Piece = 2\;ft\;3\;in \\[3ex] Thickness\;\;of\;\;1st\;\;cut = \dfrac{1}{8} = 0.125\;in \\[5ex] Thickness\;\;of\;\;2nd\;\;cut = \dfrac{1}{8} = 0.125\;in \\[5ex] \begin{array}{c} 3ft\;\;\;6in \\ +\hspace{7em} \\ 2ft\;\;\;3in \\ +\hspace{7em} \\ ~~~~~~~0ft\;\;\;0.125in \\ +\hspace{7em} \\ ~~~~~~~0ft\;\;\;0.125in \\ \hline ~~~~~5ft\;\;\;9.25in \\ \hline \end{array} $

$ \underline{Length\;\;of\;\;3rd\;\;Piece} \\[3ex] \begin{array}{c} 8ft\;\;\;0in \\ -\hspace{7em} \\ ~~~~~5ft\;\;\;9.25in \\ \hline ~~~~~2ft\;\;\;2.75in \\ \hline \end{array} $

$ \text{Length of 3rd Piece} = 2ft\;\;\;2.75in \\[3ex] Convert\;\;decimal\;\;to\;\;fraction \\[3ex] 2.75 = 2 + 0.75 \\[3ex] 0.75 = \dfrac{75}{100} = \dfrac{3}{4} \\[5ex] \implies \\[3ex] \text{Length of 3rd Piece} = 2\;ft\;2\dfrac{3}{4}\;in $

Convert all inches to feet

$ \underline{\text{Unity Fraction Method}} \\[3ex] ...\;in * \dfrac{...\;ft}{...\;in} \\[5ex] 6\;in = \dfrac{6}{12}\;ft \\[5ex] 3\;in = \dfrac{3}{12}\;ft \\[5ex] \dfrac{1}{8}\;in = \dfrac{1}{8}\;in * \dfrac{1\;ft}{12\;in} = \dfrac{1}{96}\;in \\[5ex] $



$\text{Length of 3rd Piece} = 2\;ft\;2\dfrac{3}{4}\;in$

The question is asking for the Least Common Multiple (LCM) of 35 and 50 up until 1000

The colors besides red indicate the common factors that should be counted only one time.
Begin with them in the multiplication for the LCM.
Then, include the rest.

$ Numbers = 35, 50 \\[3ex] 35 = \color{black}{5} * 7 \\[3ex] 50 = 2 * \color{black}{5} * 5 \\[5ex] LCM = 2 * \color{black}{5} * 7 * 5 \\[3ex] LCM = 350 \\[3ex] $ The 350th person will receive both a T-shirt and a concert ticket
The 700th person will receive both a T-shirt and a concert ticket
The next person (700 + 350 = 1050th person) is more than 1000, hence only 2 people will receive both a T-shirt and a concert ticket.

You may also find the quotient of 1000 and 350 (do not include the remainder because human beings are whole number people).
1000 ÷ 350 = 2.857142857

$ \$1 = x\;currency \\[3ex] \$y = ? \\[3ex] \underline{Unity\;\;Fraction\;\;Method} \\[3ex] \$y * \dfrac{x\;currency}{\$1} \\[5ex] xy\;currency $

$ Let: \\[3ex] |AC| = y \\[3ex] |CD| = x \\[5ex] \underline{\triangle ACD} \\[3ex] \angle CAD + \angle ADC + \angle ACD = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;a\;\;\triangle \\[3ex] \angle CAD + 60 + 90 = 180 \\[3ex] \angle CAD = 180 - 60 - 90 \\[3ex] \angle CAD = 30^\circ \\[5ex] \dfrac{x}{\sin 30^\circ} = \dfrac{y}{\sin 60^\circ}...Sine\;\;Law \\[5ex] x = \dfrac{y\sin 30}{\sin 60} \\[5ex] \underline{\triangle ACB} \\[3ex] \angle CAB + \angle ABC + \angle ACB = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;a\;\;\triangle \\[3ex] \angle CAB + 30 + 90 = 180 \\[3ex] \angle CAB = 180 - 30 - 90 \\[3ex] \angle CAB = 60^\circ \\[5ex] \angle CAB = \angle CAD + \angle DAB ...diagram \\[3ex] 60 = 30 + \angle DAB \\[3ex] \angle DAB = 60 - 30 \\[3ex] \angle DAB = 30^\circ \\[5ex] \dfrac{x + 100}{\sin 60^\circ} = \dfrac{y}{\sin 30^\circ} ...Sine\;\;Law \\[5ex] x + 100 = \dfrac{y\sin 60}{\sin 30} \\[5ex] x = \dfrac{y\sin 60}{\sin 30} - 100 \\[5ex] $ This is the updated diagram


$ x = x \implies \\[3ex] \dfrac{y\sin 30}{\sin 60} = \dfrac{y\sin 60}{\sin 30} - 100 \\[5ex] y * \sin 30 * \dfrac{1}{\sin 60} = y * \sin 60 * \dfrac{1}{\sin 30} - 100 \\[5ex] \sin 30^\circ = \dfrac{1}{2}...Special\;\angle s \\[5ex] \dfrac{1}{\sin 30^\circ} = \dfrac{2}{1} = 2 \\[5ex] \sin 60^\circ = \dfrac{\sqrt{3}}{2}...Special\;\angle s \\[5ex] \dfrac{1}{\sin 60^\circ} = \dfrac{2}{\sqrt{3}} \\[5ex] \implies \\[3ex] y * \dfrac{1}{2} * \dfrac{2}{\sqrt{3}} = y * \dfrac{\sqrt{3}}{2} * \dfrac{2}{1} - 100 \\[5ex] \dfrac{y}{\sqrt{3}} = y\sqrt{3} - 100 \\[5ex] \sqrt{3}\left(\dfrac{y}{\sqrt{3}}\right) = \sqrt{3}\left(y\sqrt{3}\right) - \sqrt{3}(100) \\[5ex] y = 3y - 100\sqrt{3} \\[3ex] 100\sqrt{3} = 3y - y \\[3ex] 2y = 100\sqrt{3} \\[3ex] y = \dfrac{100\sqrt{3}}{2} \\[5ex] y = 50\sqrt{3} $

Two students can make two different grades on the same test... a Function
Two students can make the same grade on the same test... a Function
However, a student cannot make two different grades on the same test... Not a function
An input may not have more than one output in a function. In other words, a function is a relation in which each input has a unique output.
The input, x-value of 3 may not have output images of y-values of 6 and 4
Hence, the correct answer is Option H.

$ \underline{Algebra\;III\;\;Test} \\[3ex] Let: \\[3ex] \text{number of juniors} = x \\[3ex] \text{number of seniors} = y \\[3ex] Enrolled = x + y \\[3ex] \text{3 juniors for each senior} \implies x = 3y \\[3ex] \implies \\[3ex] Enrolled = 3y + y = 4y \\[5ex] 80\% = \dfrac{80}{100} = 0.8 \\[5ex] 70\% = \dfrac{70}{100} = 0.7 \\[5ex] \text{Passed: } 0.8x \\[3ex] \text{Did not pass: } x - 0.8x = 0.2x \\[5ex] \text{Passed: } 0.7y \\[3ex] \text{Did not pass: } y - 0.7y = 0.3y \\[5ex] \implies \\[3ex] \text{Did not pass: } = 0.2x + 0.3y \\[3ex] = 0.2(3y) + 0.3y \\[3ex] = 0.6y + 0.3y \\[3ex] = 0.9y \\[5ex] \text{Percent who did not pass} = \dfrac{\text{Number who did not pass}}{\text{Number of Enrolled Students}} * 100 \\[5ex] =\dfrac{0.9y}{4y} * 100 \\[5ex] = 22.5\% \\[3ex] = 22\dfrac{1}{2}\% $

Let the number of shoppers who own at least 1 cat = C
Let the number of shoppers who own at least 1 dog = D
Let us represent the information on a Venn Diagram



Those who own at least 1 dog but did NOT own at least 1 cat are those who own only dog = 31 − 7 = 24 shoppers.

Let the number of shoppers who own at least 1 cat = C
Let the number of shoppers who own at least 1 dog = D

$ n(S) = 50 \\[3ex] n(C) = 20 \\[3ex] n(D) = 31 \\[3ex] n(C \cap D) = 7 \\[3ex] n(C \cup D) = n(C) + n(D) - n(C \cap D) ...Addition\;\;Rule \\[3ex] = 20 + 31 - 7 \\[3ex] = 44 \\[5ex] P(C \cup D) = \dfrac{n(C \cup D)}{n(S)} \\[5ex] = \dfrac{44}{50} \\[5ex] = \dfrac{22}{25} $

Let the number of shoppers who own at least 1 cat = C
Let the number of shoppers who own at least 1 dog = D

$ n(D) = 31 \\[3ex] \text{monetary value} = \$4 \\[3ex] \text{total monetary value} = 31(4) = \$124 \\[5ex] n(C) = 20 \\[3ex] \text{monetary value} = \$3 \\[3ex] \text{total monetary value} = 20(3) = \$60 \\[5ex] n(neither) = 6 \\[3ex] \text{monetary value} = \$g \\[3ex] \text{total monetary value} = 6(g) = \$6g \\[5ex] \text{total dollar value},\;t = 124 + 60 + 6g \\[3ex] t = 184 + 6g \\[3ex] 6g = t - 184 \\[3ex] g = \dfrac{t - 184}{6} $

$ \dfrac{x}{7b} = \dfrac{3a}{5a}\quad\dots\text{ratio of corresponding sides of similar}\;\triangle s \\[5ex] x = \dfrac{3a \cdot 7b}{5a} \\[5ex] x = \dfrac{21b}{5} $

Let us use the Punnett Square to list the possibilities. Then, we can select the seating location in Row 5 but NOT in Seat 1
Please Note:
Beacuse the ACT is timed, you do not need to complete the entire table, even though I shall complete it here.
But you need to know that the cardinality of the sample space is: 4 × 7 = 28

1, 1 implies Row 1, Seat 1
5, 1 implies Row 5, Seat 1

$Row\:\:\rightarrow$ $Seat\:\:\downarrow$	1	2	3	4	5	6	7
1	1, 1	2, 1	3, 1	4, 1	5, 1 No	6, 1	7, 1
2	1, 2	2, 2	3, 2	4, 2	5, 2	6, 2	7, 2
3	1, 3	2, 3	3, 3	4, 3	5, 3	6, 3	7, 3
4	1, 4	2, 4	3, 4	4, 4	5, 4	6, 4	7, 4

$ P(\text{Row 5 but NOT in Seat 1}) = \dfrac{n(\text{Row 5 but NOT in Seat 1})}{n(S)} \\[5ex] = \dfrac{3}{28} $

$ \log_5{625} \\[4ex] = \log_5{5^4} \\[4ex] = 4 \log_5{5} \quad\dots\text{Law 5 Log} \\[4ex] = 4 \cdot 1 \quad\dots\text{Law 4 Log} \\[4ex] 4 $

$7x^2y \lt 0$
Less than 0 implies a negative number

For $7x^2y$:
7 is a positive number
$x^2$ is a positive number because the square of any number is positive
The square of a positive number is positive
The square of a negative number is also positive
So, $7x^2$ is positive
⇒
$y$ must be negative because the product of positive and negative is negative
Option J. is the correct answer.

$ \dfrac{bc - bd}{6c - 6d} \lt 0 \\[5ex] \dfrac{b(c - d)}{6(c - d)} \lt 0 \\[5ex] \dfrac{b}{6} \lt 0 \\[5ex] b \lt 0(6) \\[3ex] b \lt 0 \\[3ex] $ All negative real numbers are less than 0

Negative values of k implies that k is less than 0
Let us test some numbers for k

$ If\;\;k = -1 \\[3ex] 2^{-1} = \dfrac{1}{2^1} = \dfrac{1}{2} \quad\dots\text{Law 6 Exp} \\[5ex] \dfrac{1}{2} \text{ is a rational number} \\[5ex] \dfrac{1}{2} \text{ is less than 1} \\[7ex] If\;\;k = -2 \\[3ex] 2^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4} \quad\dots\text{Law 6 Exp} \\[5ex] \dfrac{1}{4} \text{ is a rational number} \\[5ex] \dfrac{1}{4} \text{ is less than 1} \\[7ex] If\;\;k = -\dfrac{1}{2} \\[5ex] 2^{-\dfrac{1}{2}} = \dfrac{1}{2^{\dfrac{1}{2}}} \quad\dots\text{Law 6 Exp} \\[7ex] \dfrac{1}{2^{\dfrac{1}{2}}} = \dfrac{1}{\sqrt{2}} \quad\dots\text{Law 7 Exp} \\[7ex] \dfrac{1}{\sqrt{2}} \text{ is an irrational number} \\[5ex] \dfrac{1}{\sqrt{2}} \text{ is less than 1} \\[5ex] $ This implies that for all negative values of k, the range of values of $2^k$ is all positive numbers less than 1.

$ c^2 = a^2 + b^2 - 2ab(\cos C) \quad\dots\text{Cosine Law} $

Compare the events to sets
Let us write examples of the events (sets) based on the question

$ A = \{1, 2, 3, 4, 5, 6\} \\[3ex] |A| = n(A) = 6 \\[5ex] B = \{7, 8, 9\} \\[3ex] |B| = n(B) = 3 \\[5ex] C = A \cup B \\[3ex] C = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \\[3ex] |C| = n(C) = 9 \\[5ex] D = A \cap B \\[3ex] D = \phi \\[3ex] |D| = n(D) = 0 \\[5ex] \implies \\[3ex] |D| \lt |B| \lt |A| \lt |C| $

The base of a right square pyramid is a square
The base has a side length of 20 ft.
This implies that:
The square has a side length of 20 ft.
This implies that the base area is the area of the square
Base Area, BA = Area of the square = side² = (20)² = 400 square feet

The slant height of the pyramid = 15 ft.
Lateral Surface Area, LSA = 2 × base × slant height
LSA = 2 × 20 × 15 = 600 square feet

Total Surface Area, TSA = BA + LSA
TSA = 400 + 600
TSA = 1000 square feet.

Let us test some integers for x to see what we can get.
We shall skip all positive numbers less than or equal to 4 because we need to get a positive prime number.


$ (x - 1)(x - 4) \\[3ex] Test\;\;x = 5 \\[3ex] (5 - 1)(5 - 4) = 4(1) = 4 \quad\dots\text{Not prime} \\[5ex] Test\;\;x = 7 \\[3ex] (7 - 1)(7 - 4) = 6(3) = 18 \quad\dots\text{Not prime} \\[5ex] Test\;\;x = -1 \\[3ex] (-1 - 1)(-1 - 4) = (-2)(-5) = 10 \quad\dots\text{Not prime} \\[5ex] Test\;\;x = -3 \\[3ex] (-3 - 1)(-3 - 4) = (-4)(-7) = 28 \quad\dots\text{Not prime} \\[5ex] \text{This applies to all odd numbers: 9, 11, 13, ... and −1, −3, −5, ...} \\[3ex] \text{The odd numbers will give a positive composite (not positive prime)} \\[5ex] Test\;\;x = 6 \\[3ex] (6 - 1)(6 - 4) = 5(2) = 10 \quad\dots\text{Not prime} \\[5ex] Test\;\;x = 8 \\[3ex] (8 - 1)(8 - 4) = 7(4) = 28 \quad\dots\text{Not prime} \\[5ex] Test\;\;x = -2 \\[3ex] (-2 - 1)(-2 - 4) = (-3)(-6) = 18 \quad\dots\text{Not prime} \\[5ex] Test\;\;x = -4 \\[3ex] (-4 - 1)(-4 - 4) = (-5)(-8) = 40 \quad\dots\text{Not prime} \\[5ex] \text{This applies to all even numbers: 10, 12, 14, ... and −2, −4, −6, ...} \\[3ex] \text{The even numbers will give a positive composite (not positive prime)} \\[3ex] $ As you can see, there is no number that we substutute for x which will give a positive prime number.
Hence, the correct answer is zero: Option F.

The denominator should not be zero
So, any value that would make the denominator, zero is NOT in the domain
Let us set the denominator to zero to find that value

$ 1 + \sec x = 0 \\[3ex] \sec x = 0 - 1 \\[3ex] \sec x = -1 \\[3ex] \implies \\[3ex] \dfrac{1}{\cos x} = -1 \\[5ex] \cos x * -1 = 1 \\[3ex] \cos x = \dfrac{1}{-1} \\[5ex] \cos x = -1 \\[3ex] x = \cos^{-1}(-1) \\[4ex] x = 180^\circ \\[3ex] $ 180° is NOT in the domain of the function.

$ r(n) = ab^n \\[4ex] r(1) = ab^1 = \dfrac{1}{2} \\[5ex] ab = \dfrac{1}{2}...eqn.(1) \\[7ex] r(2) = ab^2 = \dfrac{3}{2} \\[5ex] ab^2 = \dfrac{3}{2}...eqn.(2) \\[7ex] eqn.(2) \div eqn.(1) \implies \\[3ex] \dfrac{ab^2}{ab} = \dfrac{3}{2} \div \dfrac{1}{2} \\[6ex] b = \dfrac{3}{2} \cdot \dfrac{2}{1} \\[5ex] b = 3 \\[3ex] Substitute\;\;b = 3\;\;into\;\; eqn.(1) \\[3ex] a \cdot 3 = \dfrac{1}{2}...eqn.(1) \\[5ex] a = \dfrac{1}{2} \cdot \dfrac{1}{3} \\[5ex] a = \dfrac{1}{6} \\[5ex] \implies \\[3ex] r(n) = \dfrac{1}{6} \cdot 3^n \\[6ex] = \dfrac{1}{2 \cdot 3} \cdot 3^n \\[6ex] = \dfrac{1}{2} \cdot \dfrac{1}{3} \cdot 3^n \\[6ex] = \dfrac{1}{2}\left(\dfrac{3^n}{3^1}\right) ...Law\;2...Exp \\[6ex] = \dfrac{1}{2}(3)^{n - 1} $

$ \text{Old room rate} = \$80 \\[3ex] 20\%\text{ Increase} \\[3ex] = 20\% \text{ of } \$80.00 \\[3ex] = 0.2(80) \\[3ex] = \$16 \\[5ex] \text{New room rate} \\[3ex] = \text{Old room rate} + \text{Increase} \\[3ex] = \$80 + \$16 \\[3ex] = \$96.00 $

$ 5(1) = 5 \\[3ex] 5(2) = 10 \\[3ex] STOP \\[3ex] $ The correct answer is Option E.

$ 5\begin{bmatrix} 1 & 2 \\[2ex] 3 & 4 \end{bmatrix} = \begin{bmatrix} 5(1) & 5(2) \\[2ex] 5(3) & 5(4) \end{bmatrix} = \begin{bmatrix} 5 & 10 \\[2ex] 15 & 20 \end{bmatrix} $

Average implies Arithmetic Mean
$30 on Monday and Tuesday (2 days)
$60 on Wednesday, Thursday, and Friday (3 days)

$ Average = \bar{x} = \dfrac{\Sigma x}{n} \\[5ex] = \dfrac{(30 \cdot 2) + (60 \cdot 3)}{5} \\[5ex] = \dfrac{60 + 180}{5} \\[5ex] = \dfrac{240}{5} \\[5ex] = \$48 $

$ -x + 6 = 3x - 10 \\[3ex] 6 + 10 = 3x + x \\[3ex] 16 = 4x \\[3ex] 4x = 16 \\[3ex] x = \dfrac{16}{4} \\[5ex] x = 4 \\[3ex] $ Check

$x = 4$

LHS	RHS
$ -x + 6 \\[3ex] -4 + 6 \\[3ex] 2 $	$ 3x - 10 \\[3ex] 3(4) - 10 \\[3ex] 12 - 10 \\[3ex] 2 $

Let the number of minutes Fred could use in 1 month without exceeding his budget be x

$ \underline{Monthly} \\[3ex] \text{cell phone service cost} = \$25 \\[3ex] \text{over 100 minutes}, cost = \$0.1 \;\;per\;\;minute \implies 0.1(x - 100) \\[3ex] \text{budget cost} = \$33 \\[3ex] \implies \\[3ex] 25 + 0.1(x - 100) \le 33 \\[3ex] 25 + 0.1x - 10 \le 33 \\[3ex] 0.1x + 15 \le 33 \\[3ex] 0.1x \le 33 - 15 \\[3ex] 0.1x \le 18 \\[3ex] x \le \dfrac{18}{0.1} \\[5ex] x \le 180\;minutes \\[3ex] $ Fred can use up to 180 minutes in 1 month without exceeding his budget.

$ f(c) = \dfrac{9}{5}c + 32 \\[5ex] c = 13^\circ C \\[4ex] f(13) = \dfrac{9}{5}\cdot 13 + 32 \\[5ex] = \dfrac{277}{5} \\[5ex] = 55.4^\circ F $

Let that number be p

$ \dfrac{1}{3} + p = \dfrac{1}{4} + \dfrac{1}{6} \\[5ex] p = \dfrac{1}{4} + \dfrac{1}{6} - \dfrac{1}{3} \\[5ex] \text{LCD of 4, 6, 3} = 12 \\[3ex] \implies \\[3ex] p = \dfrac{3 + 2 - 4}{12} \\[5ex] p = \dfrac{1}{12} $

$ \text{Midpoint of}\; \overline{RS} = \dfrac{1}{2} \cdot 6 = 3 \\[5ex] \text{Distance between T and the midpoint of}\; \overline{RS} \\[3ex] = \text{Distance between the midpoint of}\; \overline{RS} + \text{length of}\; \overline{ST} \\[3ex] = 3 + 20 \\[3ex] = 23 $

As seen from the diagram:



The steepest decline is indiated in red color.
It is the time during which the average daily consumption decreased the most.
It is the most negative slope.
It is from 2004 to 2005

$ \underline{\text{Rectangular Field}} \\[3ex] Length,\;L = 300\;m \\[3ex] Width,\;W = 175\;m \\[3ex] Perimeter = P \\[3ex] P = 2(L + W) \\[3ex] = 2(300 + 175) \\[3ex] = 2(475) \\[3ex] = 950\;m \\[5ex] \underline{\text{Fencing Cost}} \\[3ex] 950\meters\;\;@\;\;\$2.05\;\;per\;\;meter \\[3ex] = 950(2.05) \\[3ex] = \$1947.50 $

$ h = 300 - 30t - 16t^2 \\[4ex] h(t) = 300 - 30t - 16t^2 \\[4ex] t = 3\text{ seconds} \\[3ex] h(3) = 300 - 30(3) - 16(3)^2 \\[4ex] = 66\;feet $

Sorting the dataset in ascending order gives:

$ \text{4, 4, 8, 14, 20} \\[3ex] \text{median = middle number} = 8 $

For this specific question, it is not recommended to use a calcuator.
Be it as it may, let us see how it can be done on the calculator.

Proportional Reasoning Method

inch	feet
4	90
5	what

$ \dfrac{what}{90} = \dfrac{5}{4} \\[5ex] what \cdot 4 = 90 \cdot 5 \\[3ex] what = \dfrac{90 \cdot 5}{4} \\[5ex] what = 112.5\;feet $

$ \angle BCD = 65^\circ + 30^\circ \quad\dots\text{the exterior angle of a triangle is the sum of the two interior opposite angles.} \\[3ex] \angle BCD = 95^\circ $

Let us test this statement with some values of a and b

$ Let: \\[3ex] \underline{Example\;1} \\[3ex] a = 6 \quad\dots\text{positive and even} \\[3ex] b = -3 \quad\dots\text{negative and odd} \\[3ex] a - b \\[3ex] 6 - (-3)\quad\dots\text{subtraction operation} \\[3ex] 6 + 3 \\[3ex] 9\quad\dots\text{positve and odd} \\[5ex] \underline{Example\;2} \\[3ex] a = 2 \quad\dots\text{positive and even} \\[3ex] b = -5 \quad\dots\text{negative and odd} \\[3ex] a - b \\[3ex] 2 - (-5)\quad\dots\text{subtraction operation} \\[3ex] 2 + 5 \\[3ex] 7\quad\dots\text{positve and odd} $

Let the perpendicular altitude = h



$ \underline{\text{2 Right } \triangle s} \\[3ex] Perimeter = 32\;cm \\[3ex] \text{Length of base} = 32 - (10 + 10) \\[3ex] = 32 - 20 \\[3ex] = 12\;cm \\[3ex] \text{Base of each Right } \triangle = \dfrac{12}{2} = 6\;cm \\[5ex] \quad\dots\text{altitude splits the triangle into 2 congruent right triangles} \\[3ex] \text{Using any of the Right Triangles} \\[3ex] 10^2 = h^2 + 6^2 \quad\dots\text{Pythagorean Theorem} \\[4ex] h^2 = 10^2 - 6^2 \\[4ex] h^2 = 100 - 36 \\[3ex] h^2 = 64 \\[3ex] h = \sqrt{64} \\[3ex] h = 8\;cm $

$ \dfrac{a^2a^4}{a^6} \\[6ex] a^{2 + 4 - 6} \quad\dots\text{Laws 1 and 2 Exp} \\[4ex] a^0 \quad\dots\text{Law 3 Exp} \\[3ex] 1 $

Let the angle that the wire makes with the level ground = θ

$ \sin\theta = \dfrac{opp}{hyp} ...SOHCAHTOA \\[5ex] \sin\theta = \dfrac{8}{12} \\[5ex] \theta = \sin^{-1}\left(\dfrac{8}{12}\right) $

Let the number of toys = n
Cost Price for n toys = $2.25n
Selling Price for n toys = $4.05n
Profit = Selling Price − Cost Price
At least $81.00 implies ≥$81.00
This implies that:

$ 4.05n - 2.25n \ge 81 \\[3ex] 1.8n \ge 81 \\[3ex] n \ge \dfrac{81}{1.8} \\[5ex] n \ge 45 \\[3ex] $ Marcy needs to make and sell at least 45 toys to earn a profit of at least $81.00

$ g(x) = x^2 - 2 \\[4ex] g(-3) = (-3)^2 - 2 \\[4ex] g(-3) = 7 \\[5ex] f(x) = 4x + 3 \\[3ex] f(7) = 4(7) + 3 \\[3ex] f(7) = 31 $

$ z = 4 + 5i \\[3ex] w = 3 + 7i \\[3ex] 2z + w \\[3ex] = 2(4 + 5i) + (3 + 7i) \\[3ex] = 8 + 10i + 3 + 7i \\[3ex] = 11 + 17i $

OR

We have at least two approaches to solve this question.
Use any approach you like.

$ \underline{Factoring\;\;Method} \\[3ex] x^2 - 5x - 6 = 0 \\[4ex] (x + 1)(x - 6) = 0 \\[3ex] x + 1 = 0 \;\;OR\;\; x - 6 = 0 \\[3ex] x = -1 \;\;OR\;\; x = 6 \\[3ex] \text{Sum of roots} = -1 + 6 \\[3ex] = 5 \\[5ex] \underline{Formula\;\;Method} \\[3ex] \text{Compare the equation to the standard form} \\[3ex] a^2 + bx + c = 0 \\[4ex] a = 1 \\[3ex] b = -5 \\[3ex] \text{Sum of roots} = \dfrac{-b}{a} \\[5ex] = \dfrac{-(-5)}{1} \\[5ex] = 5 $

Let:
the amount of chicken = C
the amount of beef = B

Amount of beef @ $3.00 per pound = B × 3 = 3B
Amount of chicken @ $2.00 per pound = C × 2 = 2C

Kate has a budget of $90.00 to spend on meat for the party.
This implies that: 3B + 2C ≤ 90

$3B + 2C \le 90$

Beef, B	Chicken, C	Point (B, C)
$0$	$ 3B + 2C = 90 \\[3ex] 3(0) + 2C = 90 \\[3ex] 2C = 90 \\[3ex] C = \dfrac{90}{2} \\[5ex] C = 45 $	$(0, 45)$
$ 3B + 2C = 90 \\[3ex] 3B + 2(0) = 90 \\[3ex] 3B = 90 \\[3ex] B = \dfrac{90}{3} \\[5ex] B = 30 $	$0$	$(30, 0)$

Indicate the two points on the graph
Join the points with a solid line because of the equality sign in ≤
Shade the region below the solid line because of the less than sign in ≤
This implies that the correct answer is Option G

72 inches would be cut into equal pieces (piecese of equal length)
Let us analyze each option to determine the correct answer.

Option A
1 ÷ 16 = 0.0625 inches
Can we have equal lengths of 0.0625 inch from a string length of 72 inches?
72 ÷ 0.0625 = 1152
Yes, we can have 1152 pieces of equal lengths of 0.0625 inch each

Option B
1 ÷ 2 = 0.5 inches
Can we have equal lengths of 0.5 inch from a string length of 72 inches?
72 ÷ 0.5 = 144
Yes, we can have 144 pieces of equal lengths of 0.5 inch each

Option C
1 inch
Can we have equal lengths of 1 inch from a string length of 72 inches?
72 ÷ 1 = 72
Yes, we can have 72 pieces of equal lengths of 1 inch each

Option D
16 inches
Can we have equal lengths of 16 inches from a string length of 72 inches?
72 ÷ 16 = 4.5
This is not an integer. There is a remainder.
We can have 4 equal lengths of 16 inches each; however the remainder will not be an equal length of 16 inches.
So, this option is the correct answer but let us go ahead and analyze the final option.

Option E
36 inches
Can we have equal lengths of 36 inches from a string length of 72 inches?
72 ÷ 36 = 22
Yes, we can have 2 pieces of equal lengths of 36 inches each

Option D is the correct answer.

The axis of a vertical parabola is the vertical line through the vertex of the parabola.
It is the x-coordinate of the vertex.
It is the line of symmetry that shows two same halves of the parabola when the parabola is folded across that axis.
This implies that the axis passes through the midpoint of the x-intercepts
This implies that the axis is the midpoint of the x-intercepts because the midpoint of the x-intercepts and the x-coordinate of the vertex both have the same x-value
In that regard:

$ Vertex = (-1, 81) \\[3ex] x-coordinate\;\;of\;\;vertex = -1 \\[3ex] x-intercepts = (-4, 0)\;\;and\;\;(a, 0) \\[5ex] Midpoint\;\;of\;\;x-intercept = \left(\dfrac{-4 + a}{2}, \dfrac{0 + 0}{2}\right) \\[5ex] = \left(\dfrac{-4 + a}{2}, 0\right) \\[5ex] x-value\;\;of\;\;Midpoint = \dfrac{-4 + a}{2} \\[5ex] \implies \\[3ex] \dfrac{-4 + a}{2} = -1 \\[5ex] -4 + a = -1(2) \\[3ex] -4 + a = -2 \\[3ex] a = -2 + 4 \\[3ex] a = 2 $

Let the fraction = p

$ p \text{ of } 4\dfrac{1}{4} \text{ is } 2\dfrac{1}{8} \\[5ex] p * \dfrac{17}{4} = \dfrac{17}{8} \\[5ex] p = \dfrac{17}{8} \div \dfrac{17}{4} \\[5ex] p = \dfrac{17}{8} * \dfrac{4}{17} \\[5ex] p = \dfrac{1}{2} $

Due to the fact that about one minute is allocated to each question on the ACT, it is better to check the solution to this question by their answer options.
So, let us check and eliminate until we get the answer.

$ \underline{Option\;F} \\[3ex] 22 \div 6 = 3 \;R\; 4 \; \checkmark \\[3ex] 22 \div 7 = 3 \;R\; 1 \;\text{remainder is 1, not 5} \\[3ex] NEXT \\[5ex] \underline{Option\;G} \\[3ex] 33 \div 6 = 5 \;R\; 3 \;\;\text{remainder is 3, not 4} \\[3ex] NEXT \\[5ex] \underline{Option\;H} \\[3ex] 40 \div 6 = 6 \;R\; 4 \; \checkmark \\[3ex] 40 \div 7 = 5 \;R\; 5 \; \checkmark \\[3ex] STOP \\[3ex] $ Option H is the correct answer.
The answer choices in the options are arranged is ascending order.
So, 40 is the least positive number among the remaining choices that we did not test.

Student: Is there another way to do this question without checking by the answer options?
Teacher: Yes, we can do it: Modular Arithmetic
However, I think it takes more than a minute to do.

$ Let: \\[3ex] dividend = d \\[3ex] quotient = q \\[3ex] 1st:\;\; \text{Remainder of 4 when divided by 6} \\[3ex] d \equiv 4 \mod 6...cong.(1) \\[5ex] 2nd:\;\; \text{Remainder of 5 when divided by 7} \\[3ex] d \equiv 5 \mod 7 ...cong.(2) \\[5ex] \implies \\[3ex] d = 7q + 5 ...eqn.(1) \\[3ex] Substitute\;\;eqn.(1) \;\;for\;\;d\;\;in\;\;cong.(1) \\[3ex] 7q + 5 \equiv 4 \mod 6 \\[3ex] \text{Test positive integers for q beginning from the first positive integer} \\[3ex] 7(1) + 5 = 12 \equiv 0 \mod 6...Not\;\;4 \\[3ex] 7(2) + 5 = 19 \equiv 1 \mod 6...Not\;\;4 \\[3ex] 7(3) + 5 = 26 \equiv 2 \mod 6...Not\;\;4 \\[3ex] 7(4) + 5 = 33 \equiv 3 \mod 6...Not\;\;4 \\[3ex] 7(5) + 5 = 40 \equiv 4 \mod 6 \;\checkmark \\[3ex] \implies \\[3ex] d = 40 $

$ s...t...d \\[3ex] speed * time = distance \\[3ex] time = \dfrac{distance}{speed} \\[5ex] \underline{Joe} \\[3ex] d = 3\;miles \\[3ex] s = 4\;mph \\[3ex] t = \dfrac{3}{4}\;hour \\[5ex] \underline{Tom} \\[3ex] d = 3\;miles \\[3ex] s = 5\;mph \\[3ex] t = \dfrac{3}{5}\;hour \\[5ex] \underline{Alexis} \\[3ex] d = 3\;miles \\[3ex] s = 6\;mph \\[3ex] t = \dfrac{3}{6} = \dfrac{1}{2}\;hour \\[5ex] \underline{3-Person\;\;Team} \\[3ex] time = \dfrac{3}{4} + \dfrac{3}{5} + \dfrac{1}{2} \\[5ex] = \dfrac{15 + 12 + 10}{20} \\[5ex] = \dfrac{37}{20} \\[5ex] = 1.85\;hours \\[3ex] = 1\;hour + 0.85\;hour \\[3ex] = 1\;hour + \left(0.85\;hour * \dfrac{60\;minutes}{1\;hour}\right) \\[5ex] = 1\;hour + 51\;minutes $

$ \tan\theta = \dfrac{\sin\theta}{\cos\theta}...Quotient\;\;Identities \\[5ex] \cos\theta \cdot \tan\theta = \sin\theta \\[3ex] \implies \\[3ex] \cos\theta = \dfrac{\sin\theta}{\tan\theta} \\[5ex] = \sin\theta \div \tan\theta \\[3ex] = \dfrac{a}{b} \div \dfrac{a}{c} \\[5ex] = \dfrac{a}{b} \cdot \dfrac{c}{a} \\[5ex] = \dfrac{c}{b} $

$ \text{Distance between the fractions} = \dfrac{5}{4} - \dfrac{1}{4} \\[5ex] = \dfrac{4}{4} \\[5ex] = 1\quad\dots\text{represented by 8 lines} \\[5ex] 8\;lines\;\;from\;\; \dfrac{1}{4} \;\;to\;\; \dfrac{5}{4}\quad\dots\text{not including the line on}\;\;\dfrac{1}{4} \\[5ex] \dfrac{1}{4}\;\;\text{will be included when finding the value} \\[5ex] P\;\;\text{is on the 7th line} \\[3ex] 8\;\;lines \rightarrow 1 \\[3ex] 7th\;\;line \rightarrow \dfrac{7(1)}{8} = \dfrac{7}{8} \\[5ex] \text{Value on the 7th line} = \dfrac{1}{4} + \dfrac{7}{8} \\[5ex] = \dfrac{2}{8} + \dfrac{7}{8} \\[5ex] = \dfrac{9}{8} $

5 lines of symmetry intersecting at the origin implies that the graph is symmetrical at regular intervals of each rotation.
A full rotation = 360°
This is equivalent to 5 rotations corresponding to the 5 lines of symmetry

$ 5\;rotations \rightarrow 360^\circ \\[4ex] 1\;rotation \rightarrow \dfrac{360}{5} = 72^\circ \\[3ex] $ The smallest positive clockwise rotation about the origin = 72°

Using the process of elimination to save time:
Option A. is out because it has a zero slope
Option C. is out because it has a negative slope: the y-values decrease as the x-values increase.

Option B. is the most likely answer because the slope is 8
Options D. and E. has positive slopes, however, the values are less than 8

Student: May you please explain each option?
Teacher: Sure, let's do it.

$ \text{Slope,}\;\; m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] \text{Slope-Intercept Form:}\;\; y = mx + b \\[5ex] \underline{Option\;\;A.} \\[3ex] f(x) = 15 \\[3ex] f(x) = 0x + 15 \\[3ex] m = 0 \\[5ex] \underline{Option\;\;B.} \\[3ex] g(x) = 8x + 7 \\[3ex] m = 8 \\[5ex] \underline{Option\;\;C.} \\[3ex] (0, 6) \;\;\;and\;\;\; (11, -16) \\[3ex] x_1 = 0 ~~~~~~~~~~~~~~ x_2 = 11 \\[4ex] y_1 = 6 ~~~~~~~~~~~~~~ y_2 = -16 \\[4ex] m = \dfrac{-16 - 6}{11 - 0} \\[5ex] = -\dfrac{22}{11} \\[5ex] = -2 \\[5ex] \underline{Option\;\;D.} \\[3ex] (-6, 0) \;\;\;and\;\;\; (14, 10) \\[3ex] x_1 = -6 ~~~~~~~~~~~~~~ x_2 = 14 \\[4ex] y_1 = 0 ~~~~~~~~~~~~~~ y_2 = 10 \\[4ex] m = \dfrac{10 - 0}{14 - (-6)} \\[5ex] = \dfrac{10}{14 + 6} \\[5ex] = \dfrac{10}{20} \\[5ex] = \dfrac{1}{2} \\[5ex] \underline{Option\;\;E.} \\[3ex] (-1, 1) \;\;\;and\;\;\; (0, 5) \\[3ex] x_1 = -1 ~~~~~~~~~~~~~~ x_2 = 0 \\[4ex] y_1 = 1 ~~~~~~~~~~~~~~ y_2 = 5 \\[4ex] m = \dfrac{5 - 1}{0 - (-1)} \\[5ex] = \dfrac{5}{0 + 1} \\[5ex] = \dfrac{5}{1} \\[5ex] = 5 \\[3ex] $ Option B. is the correct answer.

$ P:\;Vertex\;1:\;\;(x_1, y_1) = (8, 12) \\[4ex] Q:\;Vertex\;2:\;\;(x_2, y_2) = (10, 8) \\[4ex] R:\;Vertex\;3:\;\;(x_3, y_3) = (8, 2) \\[4ex] x_1 = 8 ~~~~~~~~~~~~~~~~ y_1 = 12 \\[4ex] x_2 = 10 ~~~~~~~~~~~~~~~ y_2 = 8 \\[4ex] x_3 = 8 ~~~~~~~~~~~~~~~~~ y_3 = 2 \\[4ex] Area = \dfrac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \\[4ex] = \dfrac{1}{2}|8(8 - 2) + 10(2 - 12) + 8(12 - 8)| \\[5ex] = \dfrac{1}{2}|8(6) + 10(-10) + 8(4)| \\[5ex] = \dfrac{1}{2}|48 - 100 + 32| \\[5ex] = \dfrac{1}{2}|-20| \\[5ex] = \dfrac{1}{2} \cdot 20 \\[5ex] = 10\;square\;\;units $

The graph has a vertical asymptote at x = 0
This implies that it never touches or crosses the x-axis at x = 0
This implies that there is no x-intercept at (0, 0)
Also, we can see from the diagram that the graph did not cross the y-axis at y = 0
This implies that this is false: The graph of f(x) has an intercept at (0,0).
Option E. is the correct answer.

The other options are true.

$ P(11th | No) = \dfrac{75}{267} ...Given \\[7ex] P(11th | No) = \dfrac{n(11th \cap No)}{n(No)}\quad\dots\text{Conditional Probability} \\[5ex] = \dfrac{y}{65 + x + y + 56} \\[5ex] = \dfrac{y}{x + y + 121} \\[5ex] \implies \\[3ex] \dfrac{y}{x + y + 121} = \dfrac{75}{267} \\[5ex] y = 75...eqn.(1) \quad\dots\text{Equality of Numerators} \\[3ex] x + y + 121 = 267 ...eqn.(2) \quad\dots\text{Equality of Denominators} \\[3ex] \text{Substituting eqn.(1) into eqn.(2)} \implies \\[3ex] x + 75 + 121 = 267 \\[3ex] x = 267 - 75 - 121 \\[3ex] x = 71 $

Let:
r be the radius of the circle modeled by the top edge of the lampshade
R be teh radius of the circle modeled by the bottom edge of the lampshade

$ radius = \dfrac{diameter}{2} \\[5ex] Area = \pi radius^2 \\[4ex] r = \dfrac{6}{2} = 3\;in \\[5ex] R = \dfrac{8}{2} = 4\;in \\[5ex] \text{Ratio of the areas} \\[3ex] = \dfrac{\pi (3)^2}{\pi (4)^2} \\[5ex] = \dfrac{9}{16} \\[5ex] = 9 : 16 \\[3ex] $

Given:
4 different print designs
4 different color schemes
1 fabric: burlap, linen, or silk: 3 materials

To Select:
1 print design
1 color scheme
1 fabric

The number of different possible lampshade designs does she have to choose from is:

$ = 4 \cdot 4 \cdot 3 \quad\dots\text{Fundamental Counting Principle} \\[3ex] = 48\text{ designs} $

The shape of Josie's lampshade is a truncated cone (also known as a frustrum).
When rotated about its vertical symmetry line, the isosceles trapezoid will form the shape of a truncated cone.

Let the coordinate of K = −n
Let the coordinate of M = p

Let us represent the information diagrammatically:


Let us use a table to solve the question for easier comprehension.

$|KL|$	$|KL|$	$|KM|$	$|KM|$
$ 0 - (-n) \\[3ex] 0 + n \\[3ex] n $	$ 32 - 26 \\[3ex] 6 $	$ p - (-n) \\[3ex] p + n \\[3ex] $	15
$n = 6$		$ p + n = 15 \\[3ex] p + 6 = 15 \\[3ex] p = 15 - 6 \\[3ex] p = 9 $

The coordinate of M is 9.

$ \text{Area of a rectangle} = Length \cdot Width \\[3ex] \text{Area of the 1st (inner) rectangle},A_1 = f \cdot e = ef \\[4ex] \text{Area of the 2nd rectangle}, A_2 = d \cdot c = cd \\[4ex] \text{Area of the shaded region} = A_2 - A_1 \\[4ex] = cd - ef $

$ \underline{Rectangular\;\;Pyramid} \\[3ex] Volume = \dfrac{1}{3} \cdot Length \cdot Width \cdot \perp Height \\[5ex] a = \dfrac{1}{3} \cdot \dfrac{12}{x} \cdot x \cdot 12 \\[5ex] a = 48\;cubic\;feet \\[5ex] b = \dfrac{1}{3} \cdot 12 \cdot x \cdot \dfrac{12}{x} \\[5ex] b = 48\;cubic\;feet \\[5ex] a = b $

Let:
the number of remixes = R
the number of hip-hop somgs = H

Let us represent the information on a Venn Diagram



$ 39 + 60 + 25 + n(Neither) = 249 \\[3ex] n(Neither) = 249 - 39 - 60 - 25 \\[3ex] n(Neither) = 125 $

$ x(x - 2) + (2 - x) \\[3ex] x(x - 2) + 1(2 - x) \\[3ex] x(x - 2) + -1(x - 2) \\[3ex] x(x - 2) - 1(x - 2) \\[3ex] (x - 2) (x - 1) \\[3ex] (x - 1)(x - 2) $

$ \dfrac{\dfrac{2}{a} + \dfrac{2}{b}}{\dfrac{3}{ab}} \\[10ex] = \left(\dfrac{2}{a} + \dfrac{2}{b}\right) \div \dfrac{3}{ab} \\[6ex] = \left(\dfrac{2b}{ab} + \dfrac{2a}{ab}\right) \cdot \dfrac{ab}{3} \\[6ex] = \dfrac{2b + 2a}{ab} \cdot \dfrac{ab}{3} \\[5ex] = \dfrac{2a + 2b}{3} \\[5ex] = \dfrac{2(a + b)}{3} $

$ s.......t.......d \\[3ex] s \cdot t = d \\[3ex] s = \dfrac{d}{t} \\[5ex] $

(time, distance)

A	B	C	D	E
$\dfrac{1}{0.25} = 4$	$\dfrac{0.5}{0.2} = \color{darkblue}{2.5}$	$\dfrac{7}{2} = \color{darkblue}{3.5}$	$\dfrac{8}{0.5} = 16$	$\dfrac{9}{0.75} = 12$

The ordered pairs that represents speeds less than 4 miles per hour (in darkblue color) are: B and C

$ \text{Area of Red Square} = 4^2 = 16\text{ square inches} \\[4ex] \text{Area of Dartboard} = (3 + 4 + 3)^2 = 10^2 = 100\text{ square inches} \\[4ex] P(\text{Red Square}) = \dfrac{\text{Area of Red Square}}{\text{Area of Dartboard}} \\[5ex] = \dfrac{16}{100} \\[5ex] = \dfrac{4}{25} $

$ \angle Q + 40^\circ + 80^\circ = 180^\circ \quad\dots\text{sum of interior angles of a } \triangle \\[4ex] \angle Q = 180 - 40 - 80 \\[3ex] \angle Q = 60^\circ \\[5ex] \dfrac{QR}{\sin \angle P} = \dfrac{PR}{\sin \angle Q} \quad\dots\text{Sine Law} \\[5ex] \dfrac{QR}{\sin 40^\circ} = \dfrac{110}{\sin 60^\circ} \\[5ex] QR = \dfrac{110\sin 40^\circ}{\sin 60^\circ} $

$ \log_5{25} \\[4ex] = \log_5{5^2} \\[5ex] = 2\log_5{5} \quad\dots\text{Law 5 Log} \\[4ex] = 2 \cdot 1 \quad\dots\text{Law 4 Log} \\[3ex] = 2 $

$ \sqrt{420} = 20.49390153 \\[3ex] A = 20 \\[5ex] \sqrt{56} = 7.483314774 \\[3ex] B = 8 \\[5ex] A - B \\[3ex] = 20 - 8 \\[3ex] = 12 $

Let us test each option.
In other words, let us determine any two real numbers whose product would not be a nonzero rational number.

$ \underline{\text{Option A}} \\[3ex] \sqrt{2} \;\;\;and\;\;\; \sqrt{2} \\[3ex] \sqrt{2} \cdot \sqrt{2} = \sqrt{2 \cdot 2} = \sqrt{4} = 2...True \\[3ex] NEXT \\[5ex] \underline{\text{Option B}} \\[3ex] \dfrac{5}{2} \;\;\;and\;\;\; \dfrac{2}{5} \\[5ex] \dfrac{5}{2} \cdot \dfrac{2}{5} = 1...True \\[5ex] NEXT \\[5ex] \underline{\text{Option C}} \\[3ex] 2 \;\;\;and\;\;\; 3 \\[3ex] 2 \cdot 3 = 6...True \\[3ex] NEXT \\[5ex] \underline{\text{Option D}} \\[3ex] 3 \;\;\;and\;\;\; -4 \\[3ex] 3 \cdot -4 = -12...True \\[3ex] NEXT \\[5ex] \underline{\text{Option E}} \\[3ex] 2 \;\;\;and\;\;\; \sqrt{2} \\[3ex] 2 \cdot \sqrt{2} = 2\sqrt{2}...Not\;\;True \\[3ex] $ Option E. is the correct answer.

V(t) is a periodic function with a period of 9 seconds.
This implies: 9 seconds in 1 period
So, how many periods is 1 hour?

$ \underline{\text{Unity Fraction Method}} \\[3ex] \text{Set up units: } 1\;hour * \dfrac{...minutes}{...hour} * \dfrac{...seconds}{...minutes} * \dfrac{...periods}{...seconds} \\[5ex] \text{Set up measurements and units: } 1\;hour * \dfrac{60\;minutes}{1\;hour} * \dfrac{60\;seconds}{1\;minutes} * \dfrac{1\;periods}{9\;seconds} \\[5ex] 400\text{ periods} $

From our observations of the graph:
The graph follows a periodic function...the same cycle repeats each period
The same cycle repeats every 9 seconds
So, in 51 seconds:

$ 51 = 9(5) + 6 \\[3ex] $ At the 6th second, the volume is 50 gallons.
This implies that after 51 seconds, the volume is 50 gallons.

The pump is turned on at t = 0 seconds when there are 20 gallons in the tank.
The function V(t) represents the volume, in gallons, of water in the tank t seconds after the pump is turned on.
This implies that: $V(t) = 20$

A 2nd tank has 30 gallons in it when its pump is turned on at t = 0 seconds.

$ 30 = 20 + 10 \\[3ex] 30 = V(t) + 10 \\[3ex] $ This represents the volume, in gallons, of water in the 2nd tank t seconds after the pump is turned on.

1st Sample:
Number of tagged bass in the 1st sample = 32
Let the entire population of the bass in the lake = n

2nd Sample:
Number of tagged bass in the 2nd sample = 9
Sample size of 2nd sample = 60

$ \underline{\text{Capture-Recapture Method}} \\[3ex] \dfrac{32}{n} = \dfrac{9}{60} \\[5ex] 9n = 60 \cdot 32 \\[3ex] n = \dfrac{60 \cdot 32}{9} \\[5ex] n = 213.3\bar{3} \\[3ex] $ The correct answer is Option C.

$ \underline{\text{1st Graph}} \\[3ex] y^2 = x + 8 \\[4ex] y = \pm\sqrt{x + 8} \\[3ex] \text{For Point A:}\; y = \sqrt{x + 8} \quad\dots\text{because it is a positive}\;y-value \\[5ex] \underline{\text{2nd Graph}} \\[3ex] y = 3x - 3 \\[5ex] \underline{\text{Intersection of 1st Graph and 2nd Graph}} \\[3ex] y = y \\[3ex] \implies \\[3ex] 3x - 3 = \sqrt{x + 8} $

	1st Element	2nd Element	3rd Element	4th Element
Length (mm)	$8 + 8 = 16$	$16 + 8 = 24$	$16 + 8(2) = 32$	$16 + 8(3) = 40$
Width (mm)	$8 + 8 = 16$	$16 + 8 = 24$	$16 + 8(2) = 32$	$16 + 8(3) = 40$
Area (mm²)	$16 \cdot 16$	$24 \cdot 24$	$32 \cdot 32$	$40 \cdot 40 = 1600$

Convert 1600 mm² to cm²

$ \underline{\text{Unity Fraction Method}} \\[3ex] \text{Set up units: } 1600\;mm * mm * \dfrac{...m}{...mm} * \dfrac{...m}{...mm} * \dfrac{...cm}{...m} * \dfrac{...cm}{...m} \\[5ex] \text{Set up measurements and units: } 16 \cdot 10^2\;mm * mm * \dfrac{10^{-3}\;m}{1\;mm} * \dfrac{10^{-3}\;m}{1\;mm} * \dfrac{1\;cm}{10^{-2}\;m} * \dfrac{1\;cm}{10^{-2}\;m} \\[6ex] 16 \cdot 10^{2 - 3 - 3 + 2 + 2} \\[4ex] 16 \cdot 10^0 \\[4ex] 16 \cdot 1 \\[3ex] 16\;cm^2 $

Let:
a long-sleeved shirt = L
a white shirt = W

$ P(L) = \dfrac{13}{23} \\[5ex] P(W) = \dfrac{8}{23} \\[5ex] P(L \;\;and\;\; W) = \dfrac{3}{23} \\[5ex] P(L \;\;or\;\; W) = ? \\[5ex] P(L \;\;or\;\; W) = P(L) + P(W) - P(L \;\;and\;\; W) ...Addition\;\;Rule \\[3ex] = \dfrac{13}{23} + \dfrac{8}{23} - \dfrac{3}{23} \\[5ex] = \dfrac{13 + 8 - 3}{23} \\[5ex] = \dfrac{18}{23} $

This is a Horizontal Ellipse

The easiest approach to solve these kinds of questions on ellipses because ACT is a timed test.

$ a = 8 \\[3ex] b = 5 \\[3ex] (h, k) = (0, 0) \\[3ex] \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1 \\[6ex] \dfrac{(x - 0)^2}{8^2} + \dfrac{(y - 0)^2}{5^2} = 1 \\[6ex] \dfrac{x^2}{64} + \dfrac{y^2}{25} = 1 \\[5ex] $ Student: I would like to learn to solve it without considering the 1-minute per question time of the ACT.
May you please explain the solution?
Show all work.
Teacher: Sure, let's do it.

$ \underline{Center} \\[3ex] Center = (h, k) = origin = (0, 0) \\[3ex] $ We can calculate the length of both axis using at least two approaches: the major axis and the minor axis
Use any approach you prefer.

$ \underline{First\;\;Approach:\;\;Distance\;\;Formula} \\[3ex] distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[5ex] \text{Major Axis = x-axis} \\[3ex] Endpoints = (-8, 0)\;\;and\;\;(8, 0) \\[3ex] Point\;1 = (-8, 0) \\[3ex] x_1 = -8 \\[4ex] y_1 = 0 \\[4ex] Point\;2 = (8, 0) \\[3ex] x_2 = 8 \\[4ex] y_2 = 0 \\[4ex] distance = \sqrt{(8 - -8)^2 + (0 - 0)^2} \\[4ex] = \sqrt{(8 + 8)^2 + (0)^2} \\[4ex] = \sqrt{16^2 + 0^2} \\[4ex] = \sqrt{256 + 0} \\[4ex] = \sqrt{256} \\[3ex] = 16 \\[3ex] a = \dfrac{16}{2} \\[5ex] a = 8 \\[5ex] \text{Minor Axis = y-axis} \\[3ex] Endpoints = (0, 5)\;\;and\;\;(0, -5) \\[3ex] Point\;1 = (0, 5) \\[3ex] x_1 = 0 \\[4ex] y_1 = 5 \\[4ex] Point\;2 = (0, -5) \\[3ex] x_2 = 0 \\[4ex] y_2 = -5 \\[4ex] distance = \sqrt{(0 - 0)^2 + (-5 - 5)^2} \\[4ex] = \sqrt{0^2 + (-10)^2} \\[4ex] = \sqrt{0 + 100} \\[4ex] = \sqrt{100} \\[3ex] = 10 \\[3ex] b = \dfrac{10}{2} \\[5ex] b = 5 \\[5ex] \underline{Second\;\;Approach:\;\;Coordinates} \\[3ex] \text{Major Axis = x-axis} \\[3ex] Half-length = a \\[3ex] Endpoints = (-8, 0)\;\;and\;\;(8, 0) \\[3ex] x-coordinates\;\;of\;\;Endpoints = -8\;\;and\;\;8 \\[3ex] x-coordinate\;\;of\;\;Center = 0 \\[3ex] \implies \\[3ex] -8 + a = 0 \;\;\;OR\;\;\; 8 - a = 0 \\[3ex] a = 0 + 8 \;\;\;OR\;\;\; 8 - 0 = a \\[3ex] a = 8 \\[5ex] \text{Minor Axis = y-axis} \\[3ex] Half-length = b \\[3ex] Endpoints = (0, 5)\;\;and\;\;(0, -5) \\[3ex] y-coordinates\;\;of\;\;Endpoints = 5\;\;and\;\;-5 \\[3ex] y-coordinate\;\;of\;\;Center = 0 \\[3ex] \implies \\[3ex] 5 - b = 0 \;\;\;OR\;\;\; -5 + b = 0 \\[3ex] 5 + 0 = b \;\;\;OR\;\;\; b = 0 + 5 \\[3ex] b = 5 \\[5ex] \underline{Standard\;\;Form} \\[3ex] \dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1 \\[5ex] \dfrac{(x - 0)^2}{8^2} + \dfrac{(y - 0)^2}{5^2} = 1 \\[5ex] \dfrac{x^2}{64} + \dfrac{y^2}{25} = 1 \\[5ex] LCD = 64 \cdot 25 = 1600 \\[3ex] 1600 \cdot \dfrac{x^2}{64} + 1600 \cdot \dfrac{y^2}{25} = 1600(1) \\[5ex] 25x^2 + 64y^2 = 1600 $

$ For\;\; a - b \gt a + b \\[3ex] a - a \gt b + b \\[3ex] 0 \gt 2b \\[3ex] 2b \lt 0 \\[3ex] b \lt \dfrac{0}{2} \\[5ex] b \lt 0 \\[3ex] $ This is a necessary condition.
The correct answer is Option G.

$ \text{P(Card A)} + \text{P(Not Card A)} = 1 \quad\dots\text{Complementary Rule} \\[3ex] \dfrac{1}{8} + \text{P(Not Card A)} = 1 \\[3ex] \text{P(Not Card A)} = 1 - \dfrac{1}{8} \\[5ex] = \dfrac{7}{8} $

$ \text{Average} = \dfrac{40 + 60 + 10 + 80 + 30 + 40}{6} \\[5ex] = \dfrac{260}{6} \\[5ex] = 43.\bar{3} \\[3ex] \approx 43 \quad\dots\text{to the nearest whole number.} $

Sample Space, S = {153 rap songs, 72 country songs, 14 jazz songs, and 121 rock songs.}
n(S) = 153 + 72 + 14 + 121 = 360

$ P(\text{country song}) = \dfrac{n(\text{country songs})}{n(S)} \\[5ex] = \dfrac{72}{360} \\[5ex] = 0.2 $

Let the number of passengers picked up at Stop 1 = x

$ \text{Stop 1: Pick up } x \text{ passengers} \\[3ex] \text{Remaining: } x \\[5ex] \text{Stop 2: Picks up 3 passengers} \implies x + 3 \\[3ex] \text{Remaining: } x + 3 \\[5ex] \text{Stop 3: Drops off } \dfrac{1}{2} \text{ of all passengers} \implies \dfrac{1}{2} * (x + 3) = \dfrac{x + 3}{2} \\[5ex] \text{Remaining: } (x + 3) - \dfrac{x + 3}{2} \\[5ex] = \dfrac{2(x + 3) - (x + 3)}{2} \\[5ex] = \dfrac{2x + 6 - x - 3}{2} \\[5ex] = \dfrac{x + 3}{2} \\[5ex] \text{Stop 4: Picks up 11 passengers} \implies \dfrac{x + 3}{2} + 11 \\[5ex] \text{Remaining: } \dfrac{x + 3}{2} + 11 \\[5ex] = \dfrac{x + 3 + 22}{2} \\[5ex] = \dfrac{x + 25}{2} \\[5ex] \text{Stop 5: Drops off } \dfrac{1}{2} \text{ of all passengers} \implies \dfrac{1}{2} * \dfrac{x + 25}{2} \\[5ex] = \dfrac{x + 25}{4} \\[5ex] \text{Remaining: } \dfrac{x + 25}{2} - \dfrac{x + 25}{4} \\[5ex] = \dfrac{2(x + 25) - (x + 25)}{4} \\[5ex] = \dfrac{2x + 50 - x - 25}{4} \\[5ex] = \dfrac{x + 25}{4} \\[5ex] \text{Remaining: } = 13 \quad\dots\text{Given} \\[3ex] \implies \\[3ex] \dfrac{x + 25}{4} = 13 \\[5ex] x + 25 = 4(13) \\[3ex] x = 52 - 25 \\[3ex] x = 27 $

$ \text{hypotenuse}^2 = \text{leg}^2 + \text{leg}^2 \quad\dots\text{Pythagorean Theorem} \\[3ex] = 8^2 + 3^2 \\[3ex] = 64 + 9 \\[3ex] = 73 \\[3ex] \text{hypotenuse} = \sqrt{73}\text{ inches.} $

$ x = 100 \\[3ex] y = -20 \\[3ex] x + y = 100 + (-20) \\[3ex] = 100 - 20 \\[3ex] = 80 \\[5ex] 80 = -5 * \text{what} \\[3ex] \text{what} = \dfrac{80}{-5} \\[5ex] \text{what} = -16 $

$ a = 5 \\[3ex] a \otimes b = a^2b = 100 \\[3ex] \implies \\[3ex] 5^2 * b = 100 \\[3ex] b = \dfrac{100}{5^2} \\[5ex] = \dfrac{100}{25} \\[5ex] = 4 $

$ \text{Midpoint of } \overline{RS} = \dfrac{1}{2} * 6 = 3 \\[5ex] \text{Distance between } T \text{ and the midpoint of } \overline{RS} \\[3ex] = 3 + 16 \\[3ex] = 19\text{cm} $

$ A - B \\[3ex] = \begin{bmatrix} -2 & 6 \\ 3 & -8 \end{bmatrix} - \begin{bmatrix} 8 & 5 \\ 1 & -4 \end{bmatrix} \\[7ex] = \begin{bmatrix} -2 - 8 & 6 - 5 \\ 3 - 1 & -8 - (-4) \end{bmatrix} \\[7ex] = \begin{bmatrix} -10 & 1 \\ 2 & -4 \end{bmatrix} $

Let us identify the nature of the slope and the y-intercept for each line.

Line r
Negative slope
Negative y-intercept
Option E. is the correct answer.

Line q
Negative slope
Positive y-intercept

Line m
Positive slope
Positive y-intercept

Line n
Positive slope
Negative y-intercept

Line p
Zero slope
Positive y-intercept

$ \text{base angle} + \text{base angle} + \text{vertex angle} = 180^\circ \quad\dots\text{Sum of angles in a triangle} \\[3ex] 2\text{base angle} + 34 = 180 \\[3ex] \text{base angle} = \dfrac{180 - 34}{2} \\[5ex] = \dfrac{146}{2} \\[5ex] = 73^\circ $

Let the:
number of large figurines = p
number of small figurines = k

This month, Reed sold 40 figurines in 2 sizes.
$p + k = 40$

Revenue from the sale of large figurines:p
The large figurines sold for $12 each
p figurines @ $12 per figurine = $12p$

Revenue from the sale of small figurines:k
and the small figurines sold for $4 each.
k figurines @ $4 per figurine = $4k$

The amount of money he received from the sales of the large figurines was equal to the amount of money he received from the sales of the small figurines.
$12p = 4k$

$ p + k = 40 \quad\dots eqn.(1) \\[3ex] 12p = 4k \quad\dots eqn.(2) \\[5ex] \text{From } eqn.(2) \\[3ex] k = \dfrac{12p}{4} \\[5ex] k = 3p \\[5ex] \text{Substitute for } k \text{ in } eqn.(1) \\[3ex] p + 3p = 40 \\[3ex] 4p = 40 \\[3ex] p = \dfrac{40}{4} \\[5ex] p = 10 $

$ p + k = 40 \quad\dots eqn.(1) \\[3ex] 12p = 4k \\[3ex] 12p - 4k = 0 \quad\dots eqn.(2) \\[3ex] $

$ \text{first term} = a \\[3ex] \text{common difference} = d \\[3ex] \text{number of terms} = n \\[3ex] \text{nth term} = AS_{n} = a + d(n - 1) \\[5ex] AS_7 = a + 6d = 4 \quad\dots eqn.(1) \\[3ex] AS_8 = a + 7d = 12 \quad\dots eqn.(2) \\[3ex] d = AS_8 - AS_7 \quad\dots\text{common difference} \\[3ex] d = 12 - 4 = 8 \\[5ex] \text{Substitute for } d \text{ in } eqn.(1) \\[3ex] a + 6(8) = 4 \\[3ex] a + 48 = 4 \\[3ex] a = 4 - 48 \\[3ex] a = -44 \\[5ex] AS_2 = a + d \\[3ex] = -44 + 8 \\[3ex] = -36 $

$ \underline{\text{Cross High School Girls' Softball Team}} \\[3ex] 14\text{ wins} \\[3ex] 11\text{ losses} \\[3ex] 0\text{ ties} \\[3ex] 12\text{ remaining games} \\[3ex] \text{All teams games} = 14 + 11 + 0 + 12 = 37 \\[5ex] \text{50% of 37} = \dfrac{50}{100} * 37 = 18.5 \\[5ex] \text{Remaining must wins} = 18.5 - 14 = 4.5 \quad\dots\text{for 50% of all teams games} \\[3ex] \approx 5 \\[3ex] $ The least number of its remaining 12 games the team must win to finish the season winning more than 50% of all the team’s games is 5 games.

$ h(t) = -4.9t^2 + 49t + 980 \\[3ex] t = 6\text{ seconds} \\[3ex] h(6) = -4.9(6)^2 + 49(6) + 980 \\[3ex] = -176.4 + 294 + 980 \\[3ex] = 1097.6\text{ meters.} $

$ \underline{\text{Pattern A}} \\[3ex] \text{1st term: } 2 \\[3ex] \text{2nd term: } 2(2) - 1 = 4 - 1 = 3 \\[3ex] \text{3rd term: } 2(3) - 1 = 6 - 1 = 5 \\[3ex] \text{4th term: } 2(5) - 1 = 10 - 1 = 9 \\[5ex] \underline{\text{Pattern B}} \\[3ex] \text{1st term: } 2 \\[3ex] \text{2nd term: } (2 - 1) * 2 = 1 * 2 = 2 \\[3ex] \text{3rd term: } (2 - 1) * 2 = 1 * 2 = 2 \\[3ex] \text{4th term: } (2 - 1) * 2 = 1 * 2 = 2 \\[5ex] \text{4th term of Pattern A} + \text{4th term of Pattern B} \\[3ex] = 9 + 2 \\[3ex] = 11 $

$ x^2 - x - 156 \\[3ex] \text{Factors are: } 12 \text{ and } -13 \\[3ex] (x + 12)(x - 13) $

$ 4x - 2 \le 10 \\[3ex] 4x \le 10 + 2 \\[3ex] 4x \le 12 \\[3ex] x \le \dfrac{12}{4} \\[5ex] x \le 3 \\[3ex] $ The correct answer is:

Multiplying an irrational number by a rational number besides zero gives an irrational number.
If y is an irrational number, then 15y must be an irrational number.

$ \underline{\text{Equation of a Straight Line}} \\[3ex] y = mx + b \\[3ex] \text{where } \\[3ex] m = \text{slope} \\[3ex] b = y-\text{intercept} \\[3ex] $ The scatterplot has a negative slope and a positive y-intercept
This corresponds to option B.

$ \text{Points: } G(2, 6) \text{ and } H(10, 14) \\[3ex] \text{Midpoint of } \overline{GH} = \left(\dfrac{2 + 10}{2}, \dfrac{6 + 14}{2}\right) \\[5ex] = \left(\dfrac{12}{2}, \dfrac{20}{2}\right) \\[5ex] = (6, 10) $

When a two-dimensional shape is rotated about one of its sides and if the rotation causes the shape’s area to sweep through space, it generates a three-dimensional solid.
In this case, since the blade (a right triangle) rotates about its longer leg, the solid formed is a cone.

Let the 2 numbers be x, y respectively.

The sum of 2 numbers is 22.
$x + y = 22$

The sum of the squares of the 2 numbers is 260.
$x^2 + y^2 = 260$

$ x + y = 22 \quad\dots eqn.(1) \\[3ex] x^2 + y^2 = 260 \quad\dots eqn.(2) \\[5ex] \text{From } eqn.(2) \\[3ex] (x + y)^2 = (x + y)(x + y) \\[3ex] (x + y)^2 = x^2 + xy + xy + y^2 \\[3ex] (x + y)^2 = x^2 + y^2 + 2xy \\[3ex] x^2 + y^2 + 2xy = (x + y)^2 \\[3ex] \text{Substitute for the LHS expressions accordingly in both equations} \\[3ex] 260 + 2xy = 22^2 \\[3ex] 2xy = 484 - 260 \\[3ex] xy = \dfrac{224}{2} \\[5ex] xy = 112 $

There are several transformations that can map Trapezoid A onto Trapezoid B.
We are not given the points of the vertices of both trapezoids.
So, we shall visually check the answer choices to determine the correct option.



Reflection of image A over the x-axis leads to image A'
Then, translation (sliding) of image A' to the left leads to image B

The highest point is the topmost point of the circle, found by adding the radius to the y-coordinate of the center.
It is found by moving vertically upward from the center by the length of the radius.
So, for a circle center (5, 4) and a radius of 3 units:
the y-coordinate = 4 + 3 = 7.

The graph is that of a quadratic (squaring) function.
The correct answer is Option D.

$ \text{Car's fuel tank} = 16 \text{ gallons} \\[3ex] \dfrac{3}{4} \text{ full of fuel} = \dfrac{3}{4} * 16 = 12\text{ gallons} \\[5ex] 231\text{ miles away from home} \\[3ex] 22\text{ miles per gallon} \\[3ex] \text{How many gallons are needed to get home?} \\[3ex] 231\text{ miles} * \dfrac{1\text{ gallon}}{22\text{ miles}} \\[5ex] = 10.5\text{ gallons} \\[5ex] \text{Remaining gallons} = 12 - 10.5 \\[3ex] = 1.5\text{ gallons} $

The point on each triangle that is farthest from O is labeled A, B, C, D, or E.
The length of each of these points from the center are the hypotenuses of the right triangles.
They are also the radii of the circle.

A circle centered at O with a radius of 10 feet will be drawn.
So, any hypotenuse greater than 10 feet will be outside the circle.
The hypotenuse is the longest side of a right traingle.
For the point E, the hypotenuse will be greater than 10 feet because one of the legs is 10 feet.
In other words, $\overline{OE}$ will be outside of the circle because it is greater than 10 feet.

$ n = 365\text{ days} \\[3ex] t_1 = \$ 1 \\[3ex] t_n = \$ 365 \\[3ex] S_n = \dfrac{n(t_1 + t_n)}{2} \\[5ex] S_{365} = \dfrac{365(1 + 365)}{2} \\[5ex] = \dfrac{365(366)}{2} \\[5ex] = \$ 66795 $

$ \text{For } g(f(5)) \\[3ex] \text{1st: } f(5) = 4 \\[3ex] \text{2nd: } g(4) = -5 \\[3ex] \therefore g(f(5)) = -5 $

$ (3x + 5)^\circ + (x + 15)^\circ = 180^\circ \quad\dots\text{consecutive interior angles are supplementary} \\[3ex] 3x + 5 + x + 15 = 180 \\[3ex] 4x = 180 - 5 - 15 \\[3ex] 4x = 160 \\[3ex] x = \dfrac{160}{4} \\[5ex] x = 40^\circ $

$ (2\sqrt{7} + 1)(\sqrt{7} - 1) - (\sqrt{7} - 1) \\[3ex] 2(7) - 2\sqrt{7} + \sqrt{7} - 1 - \sqrt{7} + 1 \\[3ex] 14 - 2\sqrt{7} $

$ -2i(3 + 4i) - 3(2 + 5i) \\[3ex] -6i - 8i^2 - 6 - 15i \\[3ex] -8(-1) - 6 - 6i - 15i \\[3ex] 8 - 6 - 21i \\[3ex] 2 - 21i $

$ n(\text{Vanilla with Sprinkles}) = 77 \\[3ex] n(\text{Total}) = 645 \\[5ex] P(\text{Vanilla with Sprinkles}) = \dfrac{n(\text{Vanilla with Sprinkles})}{n(\text{Total})} \\[5ex] = \dfrac{77}{645} \\[5ex] = 0.119379845 \\[3ex] \approx 0.12 \quad\dots\text{to the nearest 0.01} $

Looking at the graph, we notice that:

$ f(x) = 5 \text{ when } x \lt -2 \text{ open circle} \\[3ex] f(x) = -3 \text{ when } x = -2 \text{ closed circle} \\[3ex] f(x) = 2 \text{ when } x \gt -2 \text{ open circle} \\[5ex] f(x) = \begin{cases} 5 & \text{for} & x \lt -2 \\ 3 & \text{for} & x = -2 \\ 2 & \text{for} & x \gt -2 \end{cases} $

$ n(\text{Like corn bread but dislikes banana bread}) = 82 \\[3ex] n(\text{Like corn bread}) = 160 \\[5ex] P(\text{Dislikes banana bread \underline{Given that} Likes corn bread}) \\[3ex] = \dfrac{n(\text{Like corn bread but dislikes banana bread})}{n(\text{Like corn bread})} \\[5ex] = \dfrac{82}{160} \\[5ex] = \dfrac{41}{80} $

For all positive values of a and b:

$ 8^{\dfrac{a}{b}} = \sqrt[b]{8^a} \quad\dots\text{Law 7 Exp} $

6-digit positive integers containing the digits:
1, 3, 4, 6, 7, and 8
Notice that these digits are not repeated. Once a digit is used, it cannot be used again.

$ \begin{array}{*{6}{c}} \text{1st Position} & \text{2nd Position} & \text{3rd Position} & \text{4th Position} & \text{5th Position} & \text{6th Position} \\[1ex] \underline{\quad\quad} & \underline{\quad\quad} & \underline{\quad\quad} & \underline{\quad\quad} & \underline{\quad\quad} & \underline{\quad\quad} \\[1ex] 6 & 5 & 4 & 3 & 2 & 1 \end{array} $

1st position: Any of the 6 available digits (1, 3, 4, 6, 7, or 8) can be chosen $\rightarrow$ 6 ways
2nd position: Any of the remaining 5 digits can be chosen $\rightarrow$ 5 ways
3rd position: Any of the remaining 4 digits can be chosen $\rightarrow$ 4 ways
4th position: Any of the remaining 3 digits can be chosen $\rightarrow$ 3 ways
5th position: Any of the remaining 2 digits can be chosen $\rightarrow$ 2 ways
6th position: The last remaining digit must be placed here $\rightarrow$ 1 way

Number of 6-digit positive integers containing those digits
= 6 × 5 × 4 × 3 × 2 × 1
= 720 integers.

$ \underline{\text{Denominators}} \\[3ex] x^2 - 1 = x^2 - 1^2 \\[3ex] = (x + 1)(x - 1) \quad\dots\text{Difference of Two Squares} \\[5ex] 7x - 7 = 7(x - 1) \\[5ex] \underline{\text{Least Common Denominator}} \\[3ex] 7(x + 1)(x - 1) $

The Law of Large Numbers states that as the number of independent random trials increases, the sample average (experimental probability) becomes closer to the theoretical expected value (true probability).

According to the Law of Large Numbers:
When a fair coin is flipped a large number of times, the number of heads should be approximately equal to the number of tails.
This implies that as the number of flips increases, the percentage of heads should get closer and closer to 50%.

$ \underline{\text{Experimental Probability}} \\[3ex] n(\text{heads}) = 4 \\[3ex] n(\text{flips}) = 6 \\[5ex] P(\text{heads}) = \dfrac{n(\text{heads})}{n(\text{flips})} \\[5ex] = \dfrac{4}{6} \\[5ex] = 0.6\bar{6} \\[3ex] \approx 67\% \quad\dots\text{to the nearest percent} \\[3ex] $ While 67% is greater than 50%, a sample size of only 6 flips is far too small to draw any reliable conclusion.
To accurately predict if a coin is unfair, Thomas would need to flip at least a hundred times at the minimum.
Therefore, not enough flips were performed to make a definitive claim.

$ \underline{\text{Unit Circle Trigonometry}} \\[3ex] \tan 60^\circ = \sqrt{3} \\[5ex] \underline{\triangle ABD} \\[3ex] \tan 60^\circ = \dfrac{|BD|}{|AD|} \quad\dots\text{SOHCAHTOA} \\[5ex] \sqrt{3} = \dfrac{|BD|}{10} \\[5ex] |BD| = 10\sqrt{3}\;m \\[5ex] \underline{\triangle ABC} \\[3ex] \text{base} = |AC| = 10 + 2 = 12\;m \\[3ex] \perp\text{height} = |BD| = 10\sqrt{3}\;m \\[5ex] \text{Area} = \dfrac{1}{2} * \text{base} * \perp\text{height} \\[5ex] = \dfrac{1}{2} * 12 * 10\sqrt{3} \\[5ex] = 60\sqrt{3}\;m $

Side Length - Angle Measure Theorem:
If any two side lengths of a triangle are unequal; the angles of the triangle are also unequal, and the side length facing an angle is opposite the angle measure as regards size.
In other words:
The small side faces small angle, middle side faces middle angle, big side faces big angle
This also implies that:
The middle side and the long side has the small angle
The small side and the middle side has the big angle
The small side and the big side has the middle angle



$ \text{Area} = \dfrac{1}{2} * 10 * 14 * \sin 23^\circ \\[5ex] \approx 70 * 0.39 \\[3ex] \approx 27.3 \\[3ex] \approx 27\;ft^2 \quad\dots\text{to the nearest whole number} $

$ \underline{\text{Points}} \\[3ex] \text{Point 1: } (0, 8) \hspace{3em} \text{Point 2: } \left(\dfrac{4}{3}, 0\right) \\[5ex] x_1 = 0 \hspace{5em} x_2 = \dfrac{4}{3} \\[5ex] y_1 = 8 \hspace{5em} y_2 = 0 \\[5ex] y-\text{intercept} = (0, 8) \\[5ex] \text{Slope} = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{0 - 8}{\dfrac{4}{3} - 0} \\[7ex] = -8 \div \dfrac{4}{3} \\[5ex] = -8 * \dfrac{3}{4} \\[5ex] = -6 $

$ (x + 1)^2 + 1 = 0 \\[3ex] (x + 1)(x + 1) + 1 = 0 \\[3ex] x^2 + x + x + 1 + 1 = 0 \\[3ex] x^2 + 2x + 2 = 0 \\[3ex] \underline{\text{Completing the Square Method}} \\[3ex] x^2 + 2x = -2 \\[3ex] \text{Coefficient of }x = 2 \\[3ex] \text{Half of it} = \dfrac{1}{2} * 2 = 1 \\[5ex] \text{Square it} = 1^2 \\[3ex] \implies \\[3ex] x^2 + 2x + 1^2 = -2 + 1^2 \\[3ex] (x + 1)^2 = -2 + 1 \\[3ex] (x + 1)^2 = -1 \\[3ex] x + 1 = \pm\sqrt{-1} \\[3ex] x + 1 = \pm i \\[3ex] x = -1 \pm i $

We need to put the quadratic equation in standard form before we can use the calculator.

$ \text{Number of employees with monthly salary greater than }\$5000 = 1 + 5 + 3 = 9 \\[3ex] \text{Total number of employees with monthly salaries} = 1 + 5 + 3 + 2 + 1 = 12 \\[3ex] \text{Percent of employees with monthly salary greater than }\$5000 \\[3ex] = \dfrac{9}{12} * 100\% \\[5ex] = 75\% $

$ \text{Rodrigo's monthly salary as a Hardware Engineer} = \$5,500 \\[3ex] |5500 - x| \gt 800 \\[3ex] \implies \\[3ex] 5500 - x \gt 800 \hspace{2em}\text{or}\hspace{2em} -(5500 - x) \gt 800 \\[3ex] ................................................. \\[3ex] 5500 - x \gt 800 \\[3ex] -x \gt 800 - 5500 \\[3ex] -x \gt -4700 \\[3ex] x \lt \dfrac{-4700}{-1} \\[5ex] x \lt 4700 \\[5ex] -(5500 - x) \gt 800 \\[3ex] -5500 + x \gt 800 \\[3ex] x \gt 800 + 5500 \\[3ex] x \gt 6300 \\[3ex] ................................................. \\[3ex] x \lt 4,700 \text{ or } x \gt 6,300 $

$ y = 5x^2 \\[3ex] \text{Reflection over the x-axis: Vertical Reflection} \\[3ex] -y = 5x^2 \\[3ex] y = -5x^2 \\[3ex] \text{Translated to the right 3 coordinate units: Horizontal Shift} \\[3ex] y = -5(x - 3)^2 $

$ x^4 - 9 = 0 \\[3ex] (x^2)^2 - 3^2 = 0 \\[3ex] (x^2 + 3)(x^2 - 3) = 0 \quad\dots\text{ Difference of Two Squares} \\[3ex] \underline{\text{Real Solutions}} \\[3ex] x^2 - 3 = 0 \\[3ex] x^2 = 3 \\[3ex] x = \pm\sqrt{3} $

$ \text{100 feet given 45 miles per hour} \\[3ex] 100\text{ feet} \\[3ex] 45\text{ miles} \\[3ex] 1\text{ hour} \\[5ex] \underline{\text{Unity Fraction Method}} \\[3ex] \text{Set up units:} \\[3ex] \quad\dots ft * \dfrac{\quad\dots mi}{\quad\dots ft} * \dfrac{\quad\dots hr}{\quad\dots mi} * \dfrac{\quad\dots min}{\quad\dots hr} * \dfrac{\quad\dots sec}{\quad\dots min} \\[5ex] \text{Set up measurements and units:} \\[3ex] 100\;ft * \dfrac{1\;mi}{5280\;ft} * \dfrac{1\;hr}{45\;mi} * \dfrac{60\;min}{1\;hr} * \dfrac{60\;sec}{1\;min} \\[5ex] = \dfrac{50}{33} \\[5ex] = 1.51\overline{51} \\[3ex] \approx 1.5\text{ seconds} \quad\dots\text{to the nearest 0.1 second} $

$ \underline{\text{Cylindrical Container}} \\[3ex] \text{diameter, } d = 16\text{ inches} \\[3ex] \text{radius, } r = \dfrac{d}{2} = \dfrac{16}{2} = 8\text{ inches} \\[5ex] \text{height, } h = 10\text{ inches} \\[3ex] \text{volume, } v = \pi r^2 h \\[3ex] v = \pi * 8^2 * 10 \\[3ex] v = 640\pi\text{ cubic inches} \\[5ex] \text{rate} = 40\pi \text{ cubic inches per minute} \\[3ex] 40\pi \text{ cubic inches} \\[3ex] 1\text{ minute} \\[3ex] \text{how many minutes will the container be completely filled?} \\[5ex] \underline{\text{Unity Fraction Method}} \\[3ex] \text{Set up units:} \\[3ex] \quad\dots\text{ cubic inches} * \dfrac{\quad\dots \text{minutes}}{\quad\dots \text{cubic inches}} \\[5ex] \text{Set up measurements and units:} \\[3ex] 640\pi\text{ cubic inches} * \dfrac{1\text{ minutes}}{40\pi \text{ cubic inches}} \\[5ex] = 16\text{ minutes} $

$ \dfrac{3^x 3^4}{(3^4)^7} = \dfrac{1}{9} \\[5ex] \dfrac{3^{x + 4}}{3^{4(7)}} = \dfrac{1}{9} \quad\dots\text{Laws 1 and 5 Exp} \\[5ex] 3^{x + 4} = \dfrac{3^{4(7)} * 1}{9} \\[5ex] 3^{x + 4} = \dfrac{3^{28}}{3^2} \\[5ex] 3^{x + 4} = 3^{28 - 2} \\[4ex] 3^{x + 4} = 3^{26} \\[4ex] \text{same base; equate exponents} \\[3ex] x + 4 = 26 \\[3ex] x = 26 - 4 \\[3ex] x = 22 $

Before the mistake
Let the balance in the check register = x

Craig writes a check for $18.
What should have been written in the check register (What it should show):
Balance in the check register (correct) = $x - 18$

When he records the check in his check register, he accidentally adds $18 to his balance instead of subtracting $18
After the mistake (What it shows):
Balance in the check register (incorrect) = $x + 18$

So, we know that what is showing in the check register is greater than what it should show
By how much is What is showing greater than What it should show?

$ (x + 18) - (x - 18) \\[3ex] x + 18 - x + 18 \\[3ex] 36 \\[3ex] $ Because of his mistake, Craig’s check register shows $36 more than it should.

Let $\overline{CE} = p$
Area of pentagon ABCDE = Area of rectangle ABCE + Area of $\triangle CDE$
Area of pentagon ABCDE = 200 square centimeters
Area of rectangle ABCE = $15 * p$
Area of $\triangle CDE = \dfrac{1}{2} * p * 4$
⇒

$ 15 \cdot p + \dfrac{1}{2} \cdot p \cdot 4 = 200 \\[5ex] 15p + 2p = 200 \\[3ex] 17p = 200 \\[3ex] p = \dfrac{200}{17} \\[5ex] p = 11.76470588 \\[3ex] |CE| \approx 11.8\;cm \quad\dots\text{to one decimal place.} $

Construction:
Label point B on the circumference of the circle.

Not drawn to scale.

$ \angle AOB + 150^\circ = 180^\circ \quad\dots\text{sum of angles on a straight line} \\[3ex] \angle AOB = 180 - 150 \\[3ex] \angle AOB = 30^\circ \\[5ex] \underline{\text{Unit Circle Trigonometry}} \\[3ex] \sin 30^\circ = \dfrac{1}{2} \\[5ex] \cos 30^\circ = \dfrac{\sqrt{3}}{2} \\[5ex] \underline{\triangle AOB} \\[3ex] \cos 30^\circ = \dfrac{-x}{2} \quad\dots\text{SOHCAHTOA} \\[5ex] -x = 2\cos 30 \\[3ex] -x = 2 * \dfrac{\sqrt{3}}{2} \\[5ex] x = -\sqrt{3} \\[3ex] \sin 30^\circ = \dfrac{y}{2} \quad\dots\text{SOHCAHTOA} \\[5ex] y = 2\sin 30 \\[3ex] y = 2 * \dfrac{1}{2} \\[5ex] y = 1 \\[5ex] A(x, y) = A(-\sqrt{3}, 1) $

$ \text{Total number of miles} \\[3ex] = 1\left(\dfrac{1}{8}\right) + 4\left(\dfrac{1}{4}\right) + 2\left(\dfrac{1}{2}\right) + 2\left(\dfrac{5}{8}\right) + 1\left(\dfrac{3}{4}\right) + 1\left(\dfrac{7}{8}\right) \\[5ex] = \dfrac{1}{8} + 1 + 1 + \dfrac{5}{4} + \dfrac{3}{4} + \dfrac{7}{8} \\[5ex] = 2 + \dfrac{8}{4} + \dfrac{8}{8} \\[5ex] = 2 + 2 + 1 \\[3ex] = 5\text{ miles} \\[5ex] \text{At } \dfrac{1}{2}\text{ mile per day:} \\[5ex] \text{Total number of days} = 5\text{ miles} * \dfrac{1\text{ day}}{\dfrac{1}{2}\text{ mile}} \\[7ex] = 5 \div \dfrac{1}{2} \\[5ex] = 5 * 2 \\[3ex] = 10\text{ days} $

Bookcases sold, $x$	Probability, $P(x)$	$x \cdot P(x)$
0 1 2 3 4 5	0.06 0.12 0.18 0.34 0.22 0.08	0 0.12 0.36 1.02 0.88 0.4
$\Sigma [x \cdot P(x)] = 2.78$

$ \text{Expected Value for 1 day} = \Sigma [x \cdot P(x)] = 2.78\text{ bookcases} \\[3ex] \text{Expected Value for 100 days} = 100(2.78) = 278\text{ bookcases} $

$ (2\cos x + 1)(\cos x - 1) = 0 \\[3ex] 2\cos x + 1 = 0 \hspace{2em}\text{or}\hspace{2em} \cos x - 1 = 0 \quad\dots\text{Zero Product Property} \\[3ex] ..................................................... \\[3ex] 2\cos x + 1 = 0 \\[3ex] 2\cos x = -1 \\[3ex] \cos x = -\dfrac{1}{2} \\[5ex] x = \cos^{-1}\left(-\dfrac{1}{2}\right) \\[5ex] x = 120^\circ \\[5ex] ..................................................... \\[3ex] \cos x - 1 = 0 \\[3ex] \cos x = 1 \\[3ex] x = \cos^{-1}(1) \\[3ex] x = 0^\circ \\[3ex] ..................................................... \\[3ex] $ The real solutions are 0° and 120°

$ \underline{\text{Circle ABCD}} \\[3ex] \text{circumference} = C = 16\pi \text{ feet} \\[3ex] \text{diameter} = d \\[3ex] C = \pi d \\[3ex] 16\pi = \pi * d \\[3ex] d = \dfrac{16\pi}{\pi} \\[5ex] d = 16\text{ feet} \\[3ex] $ Let the side of the square = x feet
Let us represent this information diagrammatically.



$ \underline{\text{Right }\triangle BCD} \\[3ex] \text{hyp}^2 = \text{leg}^2 + \text{leg}^2 \quad\dots\text{Pythagorean Thereom} \\[3ex] 16^2 = x^2 + x^2 \\[3ex] 2x^2 = 256 \\[3ex] x^2 = \dfrac{256}{2} \\[5ex] x^2 = 128 \\[3ex] x = \sqrt{128} \\[3ex] x = \sqrt{64 \cdot 2} \\[3ex] x = \sqrt{64} \cdot \sqrt{2} \\[3ex] x = 8\sqrt{2} \\[5ex] \underline{\text{Square ABCD}} \\[3ex] \text{Perimeter} = x + x + x + x \\[3ex] = 4x \\[3ex] = 4(8\sqrt{2}) \\[3ex] = 32\sqrt{2}\text{ feet} $

Let the amount of each of his monthly payments = p

$ \text{Initial payment} = \$900 \\[3ex] \text{36 equal payments }@\;\;\$p\;\;\text{per payment} = 36p \\[3ex] \text{Total payment} = \$9000 \\[3ex] \implies \\[3ex] 900 + 36p = 9000 \\[3ex] 36p = 9000 - 900 \\[3ex] 36p = 8100 \\[3ex] p = \dfrac{8100}{36} \\[5ex] p = \$225 $

$ (a, b) ♦ (c, d) = \dfrac{ac + bd}{ab - cd} \\[5ex] For\;\; (4,3) ♦ (1,2) \\[3ex] a = 4 \\[3ex] b = 3 \\[3ex] c = 1 \\[3ex] d = 2 \\[5ex] (4, 3) ♦ (1, 2) = \dfrac{4(1) + 3(2)}{4(3) - 1(2)} \\[5ex] = \dfrac{4 + 6}{12 - 2} \\[5ex] = \dfrac{10}{10} \\[5ex] = 1 $

$ \text{test scores are: } 71, 76, 81, 84, 95 \\[3ex] \text{lowest test score} = 71 \quad\dots\text{omit it} \\[3ex] \text{remaining test scores: } 76, 81, 84, 95 \\[3ex] \text{overall test score} \\[3ex] = \dfrac{76 + 81 + 84 + 95}{4} \\[5ex] = \dfrac{336}{4} \\[5ex] = 84 $

Area of the paved rectangular portion + Area of the unpaved field = Area of the rectangular field

$ \text{Area of the rectangular field} = 300 \cdot 200 = 60000\;ft^2 \\[3ex] \text{Area of the paved rectangular portion} = 225 \cdot 175 = 39375\;ft^2 \\[3ex] \text{Area of the unpaved field} \\[3ex] = 60000 - 39375 \\[3ex] = 20625\;ft^2 $

n(Universal set) = 26
Let the students enrolled in:
Choir = C
Band = B
both Choir and Band = p



$ 14 - p + p + 18 - p = 26 \\[3ex] 32 - 26 = p \\[3ex] p = 6 \\[3ex] $ 6 students are enrolled in both choir and band.

Given:
4 shirts
8 pairs of jeans
4 coats

To Select:
1 shirt
1 pair of jeans
1 coat

The number of different outfits available for Jesse to wear is:

$ = 4 \cdot 8 \cdot 4 \quad\dots\text{Fundamental Counting Principle} \\[3ex] = 128\text{ designs} $

$ (3x + 2)(3x - 2) \\[3ex] = (3x)^2 - 2^2 \quad\dots\text{Difference of Two Squares} \\[3ex] = 9x^2 - 4 $

1 hand broom @ $3.45 per hand broom = $3.45
1 rubber mallet @ $6.00 per rubber mallet = $6.00
2 telescoping forks @ $3.25 per telescoping fork = 2(3.25) = $6.5
Total money spent so far = 3.45 + 6 + 6.5 = $15.95

Let the greatest number of tent stakes she can buy with her remaining money = p
p tent stakes @ $1.20 per tent stake = $1.2p$

$25 to spend.


$ 15.95 + 1.2p = 25 \\[3ex] 1.2p = 25 - 15.95 \\[3ex] 1.2p = 9.05 \\[3ex] p = \dfrac{9.05}{1.2} \\[5ex] p = 7.541666667 \\[3ex] p \approx 7 \quad\dots\text{rounded down.} $

$ \text{Number of 10th graders at Hayes County High School} = 425\text{ students} \\[3ex] \text{Enrolled in Geometry} = \dfrac{3}{5} \cdot 425 = 255\text{ students} \\[5ex] \text{Earned a grade of A in Geometry} = \dfrac{1}{5} \cdot 255 = 51\text{ students} $

$ \angle EFA + \angle AFB = 180^\circ \quad\dots\text{sum of angles on a straight line} \\[3ex] \angle EFA + 90 = 180 \\[3ex] \angle EFA = 180 - 90 \\[3ex] \angle EFA = 90^\circ \\[5ex] \angle AFB + \angle BFC + \angle CFD + \angle DFE + \angle EFA = 360^\circ \quad\dots\text{sum of angles in a circle} \\[3ex] 90 + 18 + 108 + \angle DFE + 90 = 360 \\[3ex] \angle DFE + 306 = 360 \\[3ex] \angle DFE = 360 - 306 \\[3ex] \angle DFE = 54^\circ \\[5ex] m\overset{\huge\frown}{DEA} = \angle DFE + \angle EFA \\[3ex] \quad\dots\text{The measure of the intercepted arc is equal to the measure of the central angle} \\[3ex] = 54 + 90 \\[3ex] = 144^\circ $

A tire is circular in shape.
1 revolution (rev) = circumference of the circle = π d feet
To find the number of revolutions a tire with a diameter of d feet will make in a trip m miles long;

$ \text{Set up units: } \dfrac{1\;rev}{\pi d\;feet} * \dfrac{...feet}{...mile} * ...miles \\[5ex] \text{Set up measurements and units: } \dfrac{1\;rev}{\pi d\;feet} * \dfrac{5280\;feet}{1\;mile} * m\;miles \\[5ex] = \dfrac{5280m}{\pi d}\text{ revolutions} $

$ 3\begin{bmatrix} -2 & 1 \\ 0 & -5 \end{bmatrix} = \begin{bmatrix} 3(-2) & 3(1) \\ 3(0) & 3(-5) \end{bmatrix} \\[5ex] = \begin{bmatrix} -6 & 3 \\ 0 & -15 \end{bmatrix} $

Let the height of the flagpole = h
Let us represent the information diagrammatically.



$ \angle ABC \cong \angle DEF \quad\dots\text{right angles} \\[3ex] \angle ACB \cong \angle DFE \\[3ex] $ This is because the angle of elevation from the tip of the shadow of the rod to the top of the rod is congruent to the angle of elevation from the tip of the shadow of the flagpole to the top of the flagpole.

$ \therefore \triangle ABC \sim \triangle DEF \quad\dots\text{Angle–Angle Similarity Postulate} \\[3ex] \dfrac{h}{42} = \dfrac{42}{14} \\[5ex] h = 42 \cdot 3 \\[3ex] h = 126\text{ feet} $

Check whether LHS = RHS

LHS	RHS
$ (x, y) = (-4, 6) \\[3ex] x + 2y \\[3ex] -4 + 2(6) \\[3ex] -4 + 12 \\[3ex] 8 $ $ -2x + y \\[3ex] -2(-4) + 6 \\[3ex] 8 + 6 \\[3ex] 14 $	$8$ $14$

Luckily, Option A. checks out.

Student: SamDom For Peace, is it not better to just solve it rather than check it?
I mean...considering the time limit.
Teacher: Good point.
Let's solve it.

$ \underline{\text{Substitution Method}} \\[3ex] x + 2y = 8 \quad\dots eqn. (1) \\[3ex] -2x + y = 14 \quad\dots eqn.(2) \\[5ex] \text{From } eqn.(1) \\[3ex] x = 8 - 2y \quad\dots eqn.(3) \\[5ex] \text{Substitute for } x \text{ in } eqn.(2) \\[3ex] -2(8 - 2y) + y = 14 \\[3ex] -16 + 4y + y = 14 \\[3ex] 5y = 14 + 16 \\[3ex] y = \dfrac{30}{5} \\[5ex] y = 6 \\[5ex] \text{Substitute for } y \text{ in } eqn.(3) \\[3ex] x = 8 - 2(6) \\[3ex] x = 8 - 12 \\[3ex] x = -4 \\[5ex] (x, y) = (-4, 6) $

$ x^2 - 13x + 30 \\[3ex] \text{Factors are: } -3x \text{ and } -10x \\[3ex] \implies \text{using shortcut approach} \\[3ex] (x - 3)(x - 10) $

$ x^2 + y^2 = 36 \\[3ex] \text{At } x-axis, \;\; y = 0 \\[3ex] x^2 + 0^2 = 36 \\[3ex] x^2 = 36 \\[3ex] x = \pm\sqrt{36} \\[3ex] x = \pm 6 \\[3ex] \implies \\[3ex] \text{Points} = (-6, 0) \text{ and } (6, 0) $

Number of vacuums manufactured = 1000
Number of vacuums sold 3 months after they were manufactured = 301
Number of vacuums sold 4 months after they were manufactured = 297
Number of vacuums sold 3 or 4 months after they were manufactured = 301 + 297 = 598 (Addition Rule for Independent Events)
Probability that 1 of the 1,000 vacuums was sold 3 or 4 months after it was manufactured
= $\dfrac{598}{1000} = 0.598$

$ \angle BFE = \angle CBF = 34^\circ \quad\dots\text{alternate interior angles are equal} \\[3ex] \angle BEF = \angle BFE = 34^\circ \quad\dots\text{base angles of isosceles } \triangle EBF \\[3ex] \angle BED + \angle BEF = 180^\circ \quad\dots\text{sum of angles on a straight line} \\[3ex] \angle BED + 34 = 180 \\[3ex] \angle BED = 180 - 34 \\[3ex] \angle BED = 146^\circ $

Multiplying an irrational number by a rational number besides zero gives an irrational number.
If y is an irrational number, then 5y must be an irrational number.

$ \text{45 miles per hour to feet per second} \\[3ex] \text{Set up units: } \\[3ex] \dfrac{...miles}{...hour} * \dfrac{...feet}{...mile} * \dfrac{...hour}{...minute} * \dfrac{...minute}{...second} \\[5ex] \text{Set up measurements and units: } \\[3ex] \dfrac{45\;miles}{1\;hour} * \dfrac{5280\;feet}{1\;mile} * \dfrac{1\;hour}{60\;minutes} * \dfrac{1\;minute}{60\;seconds} \\[5ex] = 66\text{ feet per second} $

$ h(z) = (-5z) + 8 \\[3ex] g(z) = -2z^2 + 2z + 8 \\[3ex] (h + g)(z) \\[3ex] = (-5z + 8) + (-2z^2 + 2z + 8) \\[3ex] = -5z + 8 - 2z^2 + 2z + 8 \\[3ex] = -2z^2 - 3z + 16 $

$ n\text{Gateway respondents} = 169 \\[3ex] n(\text{Gateway respondents who responded "yes"}) = 98 \\[3ex] P(\text{selected guest who was asked about the Gateway responded “yes”}) \\[3ex] = \dfrac{98}{169} \\[5ex] = 0.5798816568 \\[3ex] \approx 0.58 \text{ to the nearest 0.01} $

$ \text{volume, } V = 616\;m^3 \\[3ex] \text{height, } h = 5\;m \\[3ex] \text{base radius, } r = ? \\[3ex] V = \dfrac{\pi r^2 h}{3} \\[5ex] 3V = \pi r^2 h \\[3ex] r^2 = \dfrac{3V}{\pi h} \\[5ex] r = \sqrt{\dfrac{3V}{\pi h}} \\[5ex] r = \sqrt{\dfrac{3 \cdot 616}{\pi \cdot 5}} \\[5ex] r = \sqrt{\dfrac{616(3)}{5\pi}} $

$ 3[\sin^2(2\theta) + \cos^2(2\theta)] \\[3ex] .................................................. \\[3ex] \sin^2\alpha + \cos^2\alpha = 1 \quad\dots\text{Pythagorean Identity} \\[3ex] \text{Let } \alpha = 2\theta \\[3ex] \sin^2(2\theta) + \cos^2(2\theta) \quad\dots\text{Pythagorean Identity} \\[3ex] .................................................. \\[3ex] = 3(1) \\[3ex] = 3 $

Assume $\theta = 45^\circ$

$w = 6m + 9$
9 is the w-intercept, found by setting m to 0 and solving for w
m represents the number of cookies in the carton.
So, when there are no cookies in the carton, (when $m = 0$), we have an empty carton.
The weight of that empty carton is the w-intercept.
9 is the weight, in ounces, of the empty carton.

$ \sin\alpha = \dfrac{144}{145} \\[5ex] \tan\alpha = \dfrac{144}{17} \\[5ex] \tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} \quad\dots\text{Quotient Identity} \\[5ex] \implies \\[3ex] \cos\alpha = \dfrac{\sin\alpha}{\tan\alpha} \\[5ex] = \dfrac{144}{145} \div \dfrac{144}{17} \\[5ex] = \dfrac{144}{145} \cdot \dfrac{17}{144} \\[5ex] = \dfrac{17}{145} $

$ \underline{\text{Number of } 1 \times 1 \text{ squares}} \\[3ex] \text{1st element: } 4 \\[3ex] \text{2nd element: } 9 \\[3ex] \text{3rd element: } 16 \\[3ex] \text{These represents the squares of natural numbers } \gt 1 \\[3ex] \text{Sequence: } 2^2, 3^2, 4^2, \\[3ex] \text{Sequence of first 6 elements: } 2^2, 3^2, 4^2, 5^2, 6^2, 7^2 \\[3ex] 4, 9, 16, 25, 36, 49 \\[3ex] \text{Sum of the first 6 elements} = 4 + 9 + 16 + 25 + 36 + 49 \\[3ex] = 139 $

$ \underline{\text{Rectangle B}} \\[3ex] \text{length} = L \\[3ex] \text{width} = W \\[3ex] \text{Perimeter} = 2(L + W) \\[3ex] = 2L + 2W \\[5ex] \underline{\text{Rectangle A}} \\[3ex] \text{length} = L \\[3ex] \text{width} = W - 6 \\[3ex] \text{Perimeter} = 2[L + (W - 6)] \\[3ex] = 2[L + W - 6] \\[3ex] = 2L + 2W - 12 \\[5ex] \text{Perimeter of Rectangle B} - \text{Perimeter of Rectangle A} \\[3ex] = 2L + 2W - (2L + 2W - 12) \\[3ex] = 2L + 2W - 2L - 2W + 12 \\[3ex] = 12 \\[3ex] $ Perimeter of Rectangle A is 12 meters less than the perimeter of Rectangle B

$ n(\text{Sample Space, S}) = 6 * 8 = 48 \\[5ex] \text{Event Space, E} \\[3ex] \text{1st fair die} + \text{2nd fair die} = 12 \\[3ex] 4 + 8 = 12 \\[3ex] 5 + 7 = 12 \\[3ex] 6 + 6 = 12 \\[3ex] n(E) = 3 \\[5ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{3}{48} $

$ \text{shaded sector, S} = \{1, 3, 5, 7, 9\} \\[3ex] n(S) = 5 \\[5ex] \text{multiple of 3 that is a shaded sector, E} = \{3, 9\} \\[3ex] n(E) = 2 \\[5ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] = \dfrac{2}{5} $

$ f(x) = x + 2 \\[3ex] g(x) = \dfrac{1}{x^2} \\[5ex] g(f(x)) \\[3ex] = g(x + 2) \\[3ex] = \dfrac{1}{(x + 2)^2} $

$ \underline{\text{Rectangle } EFGH} \\[3ex] \text{Length} = |EF| = |2 - (-1)| \\[3ex] = |2 + 1| \\[3ex] = |3| \\[3ex] = 3\text{ units} \\[5ex] \text{Width} = |FG| = |-1 - 3| \\[3ex] = |-4| \\[3ex] = 4\text{ units} \\[5ex] \text{Area} = \text{Length} \cdot \text{Width} \\[3ex] = 3 \cdot 4 \\[3ex] = 12\text{ square units} \\[5ex] \underline{\text{Quadrant IV}} \\[3ex] \text{Length} = |2 - 0| \\[3ex] = |2| \\[3ex] = 2\text{ units} \\[5ex] \text{Width} = |-1 - 0| \\[3ex] = |-1| \\[3ex] = 1\text{ unit} \\[5ex] \text{Area} = \text{Length} \cdot \text{Width} \\[3ex] = 2 \cdot 1 \\[3ex] = 2\text{ square units} \\[5ex] \text{Percent of rectangle's area that lies in Quadrant IV} \\[3ex] = \dfrac{2}{12} * 100\% \\[5ex] = \dfrac{50}{3}\% \\[5ex] = 16\dfrac{2}{3}\% $

$ \dfrac{4}{c^2}\left(\dfrac{c + 2}{c}\right) \\[5ex] = \dfrac{4(c + 2)}{c^2 \cdot c} \\[5ex] = \dfrac{4c + 8}{c^3} $

When drawing a line of best fit (with the least sum of squared residuals), we: (1.) Balance the points so that there is roughly an equal number of points above and below the line.
(2.) Ensure that the points above and below are distributed evenly along the length of the line rather than clustered at one end.
In that regard, if we apply a linear model to this data, the regression line is a straight line with a positive slope.
Because the data points are widely scattered and follow a non-linear trend, a linear slope is only an approximation. Estimating by eye, the slope of this line is about $\dfrac{1}{3}$

$ (9i - 2)(3 - 5i) \\[3ex] 27i - 45i^2 - 6 + 10i \\[3ex] - 45(-1) + 37i - 6 \\[3ex] 45 + 37i - 6 \\[3ex] 39 + 37i $

The denominator of a rational function must not be zero because divisibility by zero is undefined.
Even though the function can be simplified to −1, however, before we simplify the function, is there any value of x that will make the function undefined?

$ f(x) = \dfrac{(-1)(x + 2)}{(x + 2)} \\[5ex] \text{Set the denominator to } 0 \text{ and solve for } x \\[3ex] x + 2 = 0 \\[3ex] x = -2 \\[3ex] $ The function is undefined at $x = -2$

Let the:
cost of each gourmet chocolate = g
cost of each premiere chocolate = p

A gift box with 5 gourmet chocolates and 10 premiere chocolates costs $30.35
$5g + 10p = 30.35$
Dividing both sides by 5, we have:
$g + 2g = 6.07$

6 gourmet chocolates and 6 premiere chocolates costs $22.68.
$6g + 6p = 22.68$
Dividing both sides by 5, we have:
$g + p = 3.78$

$ g + 2p = 6.07 \quad\dots eqn.(1) \\[3ex] g + p = 3.78 \quad\dots eqn.(2) \\[3ex] \underline{\text{Elimination by Subtraction Method}} \\[3ex] eqn.(1) - eqn.(2) \implies \\[3ex] (g + 2p) - (g + p) = 6.07 - 3.78 \\[3ex] g + 2p - g - p = 2.29 \\[3ex] p = \$2.29 \\[5ex] 2 * eqn.(2) \implies \\[3ex] 2(g + p) = 2(3.78) \\[3ex] 2g + 2p = 7.56 \quad\dots eqn.(3) \\[3ex] eqn.(3) - eqn.(1) \implies \\[3ex] (2g + 2p) - (g + 2p) = 7.56 - 6.07 \\[3ex] 2g + 2p - g - 2p = 1.49 \\[3ex] g = \$1.49 \\[3ex] $ What is the cost of a gift box with 2 gourmet chocolates and 1 premiere chocolate?

$ 2g + p \\[3ex] = 2(1.49) + 2.29 \\[3ex] = \$5.27 $

$ \left(\dfrac{4^9}{4^{15}}\right)^{\dfrac{1}{3}} \\[7ex] = \dfrac{4^{9 \cdot \dfrac{1}{3}}}{4^{15 \cdot \dfrac{1}{3}}} \\[9ex] = \dfrac{4^3}{4^5} \\[5ex] = 4^{3 - 5} \\[5ex] = 4^{-2} \quad\dots\text{Option B.} \\[3ex] = \dfrac{1}{4^2} \quad\dots\text{Option C.} \\[5ex] (4^{-6})^{\dfrac{1}{3}} \quad\dots\text{Option A.} \\[5ex] = 4^{-6 \cdot \dfrac{1}{3}} \\[5ex] = 4^{-2} \quad\dots\text{Option B.} \\[5ex] \dfrac{4^3}{4^5} \quad\dots\text{Option D.} \\[5ex] = 4^{3 - 5} \quad\dots\text{Law 2 Exp} \\[3ex] = 4^{-2} \quad\dots\text{Option B.} \\[5ex] (4^{24})^{\dfrac{1}{3}} \quad\dots\text{Option E.} \\[5ex] = 4^{24 \cdot \dfrac{1}{3}} \\[5ex] = 4^8 \quad\dots\text{Not equivalent} $

Let us try the variables with relatively prime positive integers.

$ a \gt b \gt c \gt d \\[3ex] 11 \gt 7 \gt 5 \gt 3 \\[3ex] a = 11 \\[3ex] b = 7 \\[3ex] c = 5 \\[3ex] d = 3 \\[5ex] \dfrac{a}{b} = \dfrac{11}{7} \approx 1.571 \\[5ex] \dfrac{a}{c} = \dfrac{11}{5} = 2.2 \\[5ex] \dfrac{b}{c} = \dfrac{7}{5} = 1.4 \\[5ex] \dfrac{c}{a} = \dfrac{5}{11} = 0.45\overline{45} \\[5ex] \dfrac{d}{a} = \dfrac{3}{11} = 0.27\overline{27} \\[5ex] \text{Greatest Value} = \dfrac{a}{c} $

Construction:
Extend Point X to Point Y
Extend $|WZ|$ to $|WY|$
Label Point V



$ \angle YWV + 147^\circ = 180^\circ \quad\dots\text{sum of angles on a straight line} \\[3ex] \angle YWV = 180 - 147 \\[3ex] \angle YWV = 33^\circ \\[5ex] \angle WYX = \angle YWV = 33^\circ \quad\dots\text{alternate angles are congruent} \\[5ex] \angle XZW = \angle WYX + 37^\circ \quad\dots\text{exterior angle of a triangle is the sum of the interior opposite angles} \\[3ex] \angle XZW = 33 + 37 \\[3ex] \angle XZW = 70^\circ $

$ z^2 + z + 3 = 0 \\[3ex] \underline{\text{Completing the Square Method}} \\[3ex] z^2 + z = -3 \\[3ex] ................................................. \\[3ex] \text{coefficient of } z = 1 \\[3ex] \text{half of it} = \dfrac{1}{2} \cdot 1 = \dfrac{1}{2} \\[5ex] \text{square it} = \left(\dfrac{1}{2}\right)^2 \\[5ex] ................................................. \\[3ex] z^2 + z + \left(\dfrac{1}{2}\right)^2 = -3 + \left(\dfrac{1}{2}\right)^2 \\[5ex] \left(z + \dfrac{1}{2}\right)^2 = -3 + \dfrac{1}{4} \\[5ex] = -\dfrac{12}{4} + \dfrac{1}{4} \\[5ex] = \dfrac{-12 + 1}{4} \\[5ex] = -\dfrac{11}{4} \\[5ex] z + \dfrac{1}{2} = \pm \sqrt{-\dfrac{11}{4}} \\[5ex] = \pm \dfrac{\sqrt{-11}}{\sqrt{4}} \\[5ex] = \pm \dfrac{\sqrt{-1 \cdot 11}}2 \\[5ex] = \pm \dfrac{i\sqrt{11}}{2} \\[5ex] z = -\dfrac{1}{2} \pm \dfrac{i\sqrt{11}}{2} \\[5ex] z = \dfrac{-1 \pm i\sqrt{11}}{2} $

Label:
Congruent Triangles: I, II, III, and IV
Square: V



$ \underline{\text{Square, } V} \\[3ex] \text{side} = 12 - 2 = 10\text{ units} \\[3ex] \text{Area of the Square} = side^2 = 10^2 = 100\text{ square units} \\[5ex] \underline{\text{Congruent Triangles}} \\[3ex] \text{Example II} \\[3ex] \text{base} = 10\text{ units} \\[3ex] \perp\text{height} = 24 - 12 = 12\text{ units} \\[3ex] \text{Area} = \dfrac{1}{2} \cdot \text{base} \cdot \perp\text{height} \\[5ex] = \dfrac{1}{2} \cdot 10 \cdot 12 \\[5ex] = 60\text{ square units} \\[5ex] \text{Area of the 4 Congruent Triangles} = 4(60) = 240\text{ square units} \\[5ex] \text{Total Surface Area of the Pyramid} \\[3ex] = \text{Area of the Square} + \text{Area of the 4 Congruent Triangles} \\[3ex] = 100 + 240 \\[3ex] = 340\text{ square units} $

$ f(x) = 4x + 3 \\[3ex] g(x) = -4x^2 + 10x + 1 \\[3ex] f(x) = g(x) \implies \\[3ex] 4x + 3 = -4x^2 + 10x + 1 \\[3ex] 4x + 3 + 4x^2 - 10x - 1 = 0 \\[3ex] 4x^2 - 6x + 2 = 0 \\[3ex] 2(2x^2 - 3x + 1) = 0 \\[3ex] 2x^2 - 3x + 1 = \dfrac{0}{2} \\[5ex] 2x^2 - 3x + 1 = 0 \\[3ex] ........................................... \\[3ex] 2x^2 \cdot 1 = 2x^2 \\[3ex] \text{Factors are: } -2x \text{ and } -x \\[3ex] ........................................... \\[3ex] 2x^2 - 2x - x + 1 = 0 \\[3ex] 2x(x - 1) - 1(x - 1) = 0 \\[3ex] (x - 1)(2x - 1) = 0 \\[3ex] x - 1 = 0 \hspace{1em}\text{or}\hspace{1em} 2x - 1 = 0 \\[3ex] x = 1 \hspace{1em}\text{or}\hspace{1em} 2x = 1 \\[3ex] \hspace{2em}\text{or}\hspace{2em} x = \dfrac{1}{2} \\[5ex] \text{When } x = 1 \\[3ex] f(1) = 4(1) + 3 \\[3ex] = 4 + 3 \\[3ex] = 7 \\[3ex] \text{Point 1} = (1, 7) \quad\dots\text{correct answer in the option} \\[5ex] \text{When } x = \dfrac{1}{2} \\[5ex] f\left(\dfrac{1}{2}\right) = 4\left(\dfrac{1}{2}\right) + 3 \\[5ex] = 2 + 3 \\[3ex] = 5 \\[3ex] \text{Point 2} = \left(\dfrac{1}{2}, 5\right) \\[5ex] $

The Pythagorean Triples are Options:
B. 3 — 4 — 5
C. 5 — 12 — 13
D. 6 — 8 — 10
E. 8 — 15 — 17

So, the exception is Option A.

Student: How do you know?
Teacher: Pythagorean Triples
3 — 4 — 5
They have the same denominator
$5^2 = 3^2 + 4^2$
Student: Is there a way to know the next Pythagorean Triple?
Teacher: Yes, see one of the patterns
6 — 8 — 10
9 — 12 — 15
Student: 12 — 16 — 20
Teacher: Then, if we identify another set, we follow the same pattern
Student: So, for 5 — 12 — 13
10 — 24 — 26
15 — 36 — 39
Teacher: That is correct.
Student: Got it.
But, may you solve it? Teacher: Sure, let's do it.

The hypotenuse (hyp) is the longest side of a right triangle.
We shall use the Pythagorean Theorem to determine which of the side lengths form a right triangle.

$hyp^2 = leg^2 + leg^2 \quad\dots\text{Pythagorean Theorem}$

Options	LHS ($hyp^2$)	RHS ($leg^2 + leg^2$)
$ A.\;\; \left\{\dfrac{1}{2}, \dfrac{2}{3}, \dfrac{3}{4}\right\} \\[5ex] = \left\{\dfrac{6}{12}, \dfrac{8}{12}, \dfrac{9}{12}\right\} \\[5ex] hyp = \dfrac{9}{12} \\[5ex] leg = \dfrac{6}{12} \\[5ex] leg = \dfrac{8}{12} $	$ \left(\dfrac{9}{12}\right)^2 \\[5ex] \dfrac{81}{144} $	$ \left(\dfrac{6}{12}\right)^2 + \left(\dfrac{8}{12}\right)^2 \\[5ex] \dfrac{36}{144} + \dfrac{64}{144} \\[5ex] \dfrac{36 + 64}{144} \\[5ex] \dfrac{100}{144} $
$ B.\;\; \left\{\dfrac{3}{7}, \dfrac{4}{7}, \dfrac{5}{7}\right\} \\[5ex] hyp = \dfrac{5}{7} \\[5ex] leg = \dfrac{4}{7} \\[5ex] leg = \dfrac{3}{7} $	$ \left(\dfrac{5}{7}\right)^2 \\[5ex] \dfrac{25}{49} $	$ \left(\dfrac{4}{7}\right)^2 + \left(\dfrac{3}{7}\right)^2 \\[5ex] \dfrac{16}{49} + \dfrac{9}{49} \\[5ex] \dfrac{16 + 9}{49} \\[5ex] \dfrac{25}{49} $
$ C.\;\; \left\{\dfrac{5}{12}, 1, \dfrac{13}{12}\right\} \\[5ex] = \left\{\dfrac{5}{12}, \dfrac{12}{12}, \dfrac{13}{12}\right\} \\[5ex] hyp = \dfrac{13}{12} \\[5ex] leg = \dfrac{5}{12} \\[5ex] leg = \dfrac{12}{12} $	$ \left(\dfrac{13}{12}\right)^2 \\[5ex] \dfrac{169}{144} $	$ \left(\dfrac{5}{12}\right)^2 + \left(\dfrac{12}{12}\right)^2 \\[5ex] \dfrac{25}{144} + \dfrac{144}{144} \\[5ex] \dfrac{25 + 144}{144} \\[5ex] \dfrac{169}{144} $
$ D.\;\; \left\{\dfrac{6}{11}, \dfrac{8}{11}, \dfrac{10}{11}\right\} \\[5ex] hyp = \dfrac{10}{11} \\[5ex] leg = \dfrac{6}{11} \\[5ex] leg = \dfrac{8}{11} $	$ \left(\dfrac{10}{11}\right)^2 \\[5ex] \dfrac{100}{121} $	$ \left(\dfrac{6}{11}\right)^2 + \left(\dfrac{8}{11}\right)^2 \\[5ex] \dfrac{36}{121} + \dfrac{64}{121} \\[5ex] \dfrac{36 + 64}{121} \\[5ex] \dfrac{100}{121} $
$ E.\;\; \left\{2, \dfrac{15}{4}, \dfrac{17}{4}\right\} \\[5ex] = \left\{\dfrac{8}{4}, \dfrac{15}{4}, \dfrac{17}{4}\right\} \\[5ex] hyp = \dfrac{17}{4} \\[5ex] leg = \dfrac{8}{4} \\[5ex] leg = \dfrac{15}{4} $	$ \left(\dfrac{17}{4}\right)^2 \\[5ex] \dfrac{289}{16} $	$ \left(\dfrac{8}{4}\right)^2 + \left(\dfrac{15}{4}\right)^2 \\[5ex] \dfrac{64}{16} + \dfrac{225}{16} \\[5ex] \dfrac{64 + 225}{16} \\[5ex] \dfrac{289}{16} $

$ f(x) = -\dfrac{1}{3}x - 5 \\[3ex] \text{Domian: } -6 \le x \le 9 \\[5ex] \text{Considering the endpoints of the domain:} \\[3ex] \text{When } x = -6 \\[3ex] f(-6) = -\dfrac{1}{3} \cdot -6 - 5 \\[3ex] = 2 - 5 \\[3ex] = -3 \\[5ex] \text{When } x = -9 \\[3ex] f(-6) = -\dfrac{1}{3} \cdot 9 - 5 \\[3ex] = -3 - 5 \\[3ex] = -8 \\[5ex] \text{Range: } -8 \le f(x) \le -3 $

$ (10 - i) + (8 + 6i) = a + bi \\[3ex] 10 - i + 8 + 6i = a + bi \\[3ex] 18 + 5i = a + bi \\[5ex] a = 18 \\[3ex] b = 5 \\[3ex] ab = 18 \cdot 5 = 90 $

We were not given $m\angle ADB$
The information given is insufficient to determine the measure of $\angle ADC$.

$ \text{Dividend: } f(x) = x^7 - x^6 - 2x^4 + 2x^3 + x - 1 \\[3ex] \text{Divisor: } x - 1 \\[5ex] \text{Set divisor to } 0 \text{ and solve for } x \\[3ex] x - 1 = 0 \\[3ex] x = 1 \\[5ex] \text{When } x^7 - x^6 - 2x^4 + 2x^3 + x - 1 \text{ is divided by } x - 1 \\[3ex] \text{The remainder is } f(1) \\[3ex] f(1) = 1^7 - 1^6 - 2(1)^4 + 2(1)^3 + 1 - 1 \\[3ex] = 1 - 1 - 2(1) + 2(1) + 1 - 1 \\[3ex] = 0 - 2 + 2 + 1 - 1 \\[3ex] = 0 $

mean = 100
standard deviation = 20

68% of the data are within 1 standard deviation of the mean
within means below and above

$ 1\text{ standard deviation below the mean}
= \text{mean} - 1\text{ standard deviation} \\[3ex] = 100 - 1(20) \\[3ex] = 80 \\[5ex] 1\text{ standard deviation above the mean}
= \text{mean} + 1\text{standard deviation} \\[3ex] = 100 + 1(20) \\[3ex] = 120 \\[3ex] $ 68% of the data are within $(80, 120)$

Option E. is a hyperbola.
A hyperbola is a conic section formed by the intersection of a plane with both halves of a double cone.
In the standard (x, y) coordinate plane, the graph of a hyperbola always consists of two symmetric, disconnected curves (called branches) that open away from each other.

$f(x)$ is an exponential growth function.
$g(x)$ is a linear function.
$h(x)$ is a quadratic function.
$k(x)$ is a cubic function.

The exponential growth function will exceed the values of all the other listed functions as x approaches infinity.

$ \text{1st day of class typing speed} = \text{initial} = 35\;wpm \\[3ex] \text{99th day of class typing speed} = \text{new} = 55\;wpm \\[3ex] \text{Percent Increase} = \dfrac{new - initial}{initial} * 100\% \\[5ex] = \dfrac{35}{55} * 100\% \\[5ex] = 63.\overline{63}\% \\[3ex] \approx 64\% \quad\dots\text{to the nearest 1%} $

None of the information given in I, II, and III can be verified from the given table.

We shall assume a linear function.

$ \underline{\text{For Shakir}} \\[3ex] \text{1st day, } x_1 = 1 \\[3ex] \text{Typing speed, } y_1 = 25 \\[5ex] \text{99th day, } x_2 = 99 \\[3ex] \text{Typing speed, } y_2 = 55 \\[5ex] \text{Average rate of change} = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{55 - 25}{99 - 1} \\[5ex] = \dfrac{30}{98} \\[5ex] = 0.306122449 \\[3ex] \approx 0.306\;\text{ wpm per day to 3 decimal places} $

$ \text{center of dilation} = A(-2, 2) \\[3ex] \text{scale factor} = \dfrac{3}{2} \\[5ex] $ distance from center to image = scale factor * distance from center to object

$ \text{Image of A, } A' \\[3ex] A(-2, 2) \\[3ex] A' = (-2, 2) \quad\dots\text{Because A is the center of dilation} \\[5ex] \text{Image of B, } B' \\[3ex] A(-2, 2) \text{ to } B(0, 2) \\[5ex] \text{horizontal change} = 0 - (-2) = 0 + 2 = 2 \\[3ex] \text{new horizontal change} = \dfrac{3}{2} \cdot 2 = 3 \\[5ex] \text{new x-coordinate} = -2 + 3 = 1 \\[5ex] \text{vertical change} = 2 - 2 = 0 \\[3ex] \text{new vertical change} = \dfrac{3}{2} \cdot 0 = 0 \\[5ex] \text{new y-coordinate} = 2 + 0 = 2 \\[5ex] B' = (1, 2) \\[5ex] \text{Image of C, } C' \\[3ex] A(-2, 2) \text{ to } C(0, -2) \\[5ex] \text{horizontal change} = 0 - (-2) = 0 + 2 = 2 \\[3ex] \text{new horizontal change} = \dfrac{3}{2} \cdot 2 = 3 \\[5ex] \text{new x-coordinate} = -2 + 3 = 1 \\[5ex] \text{vertical change} = -2 - 2 = -4 \\[3ex] \text{new vertical change} = \dfrac{3}{2} \cdot -4 = -6 \\[5ex] \text{new y-coordinate} = 2 + (-6) = 2 - 6 = -4 \\[5ex] C' = (1, -4) \\[5ex] \text{Image of D, } D' \\[3ex] A(-2, 2) \text{ to } D(-2, -2) \\[5ex] \text{horizontal change} = -2 - (-2) = -2 + 2 = 0 \\[3ex] \text{new horizontal change} = \dfrac{3}{2} \cdot 0 = 0 \\[5ex] \text{new x-coordinate} = -2 + 0 = -2 \\[5ex] \text{vertical change} = -2 - 2 = -4 \\[3ex] \text{new vertical change} = \dfrac{3}{2} \cdot -4 = -6 \\[5ex] \text{new y-coordinate} = 2 + (-6) = 2 - 6 = -4 \\[5ex] D' = (-2, -4) \\[3ex] $ The correct answer is Option H.

$ \text{Point 1: } (-1, 1) \\[3ex] x_1 = -1 \\[3ex] y_1 = 1 \\[5ex] \text{Point 2: } (3, d) \\[3ex] x_2 = 3 \\[3ex] y_2 = d \\[5ex] \text{slope} = 2 \\[3ex] \text{slope} = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] 2 = \dfrac{d - 1}{3 - (-1)} \\[5ex] \dfrac{d - 1}{3 + 1} = 2 \\[5ex] d - 1 = 2(4) \\[3ex] d = 8 + 1 \\[3ex] d = 9 $

$ y = ax^2 + bx + c \quad\dots\text{standard form of a quadratic function} \\[3ex] \text{For } (0, -3) \\[3ex] x = 0 \\[3ex] y = -3 \\[5ex] -3 = a(0)^2 + b(0) + c \\[3ex] -3 = 0 + 0 + c \\[3ex] c = -3 \\[5ex] \text{For } (-2, 0) \\[3ex] x = -2 \\[3ex] y = 0 \\[5ex] 0 = a(-2)^2 + b(-2) + -3 \\[3ex] 0 = 4a - 2b - 3 \\[3ex] 0 + 3 = 4a - 2b \\[3ex] 4a - 2b = 3 \quad\dots eqn.(1) \\[5ex] \text{For } (3, 0) \\[3ex] x = 3 \\[3ex] y = 0 \\[5ex] 0 = a(3)^2 + b(3) + -3 \\[3ex] 0 = 9a + 3b - 3 \\[3ex] 0 + 3 = 9a + 3b \\[3ex] 9a + 3b = 3 \\[3ex] 3(3a + b) = 3 \\[3ex] 3a + b = \dfrac{3}{3} \\[5ex] 3a + b = 1 \quad\dots eqn.(2) \\[5ex] \underline{\text{Substitution Method}} \\[3ex] \text{From } eqn.(2) \\[3ex] b = 1 - 3a \quad\dots eqn.(3) \\[5ex] \text{Substitute for } b \text{ in } eqn.(1) \\[3ex] 4a - 2(1 - 3a) = 3 \\[3ex] 4a - 2 + 6a = 3 \\[3ex] 10a = 3 + 2 \\[3ex] a = \dfrac{5}{10} \\[5ex] a = \dfrac{1}{2} \\[5ex] \text{Substitute for } a \text{ in } eqn.(3) \\[3ex] b = 1 - 3\left(\dfrac{1}{2}\right) \\[5ex] b = \dfrac{2}{2} - \dfrac{3}{2} \\[5ex] b = -\dfrac{1}{2} \\[5ex] \implies \\[3ex] y = \dfrac{1}{2}x^2 - \dfrac{1}{2}x - 3 \\[5ex] \text{Horizontal Shift: Translation of 3 units to the right} \\[3ex] y' = \dfrac{1}{2} (x - 3)^2 - \dfrac{1}{2}(x - 3) - 3 \\[5ex] = \dfrac{1}{2}[(x - 3)^2 - (x - 3) - 6] \\[5ex] = \dfrac{1}{2}[(x - 3)(x - 3) - x + 3 - 6] \\[5ex] = \dfrac{1}{2}[x^2 - 3x - 3x + 9 - x + 3 - 6] \\[5ex] = \dfrac{1}{2}[x^2 - 7x + 6] \\[5ex] = \dfrac{1}{2}[(x - 6)(x - 1)] $

Let us represent the information diagrammatically



$ \underline{\triangle ABO} \\[3ex] \angle AOB + \alpha = 180^\circ \quad\dots\text{ sum of angles on a straight line} \\[3ex] \angle AOB = 180 - \alpha \\[5ex] |AO|^2 = |AB|^2 + |BO|^2 \quad\dots\text{Pythagorean Theorem} \\[3ex] = 7^2 + 3^2 \\[3ex] = 49 + 9 \\[3ex] = 58 \\[3ex] |AO| = \sqrt{58} \\[5ex] \cos\angle AOB = \dfrac{|BO|}{|AO|} \quad\dots\text{SOHCAHTOA} \\[5ex] \cos(180 - \alpha) = \dfrac{3}{\sqrt{58}} \\[5ex] \text{But }\cos(180 - \alpha) = -\cos\alpha \quad\dots\text{2nd Quadrant Identity} \\[3ex] \implies \\[3ex] -\cos\alpha = \dfrac{3}{\sqrt{58}} \\[5ex] \cos\alpha = -\dfrac{3}{\sqrt{58}} $

$ a \lt 0 \text{ implies that } a \text{ is negative} \\[3ex] b \lt 0 \text{ implies that } b \text{ is negative} \\[3ex] \dfrac{a}{b} \lt 1 \\[5ex] a \lt 1(b) \\[3ex] a \lt b \\[5ex] \text{Say } \\[3ex] a = -\dfrac{1}{4} \\[5ex] b = -\dfrac{1}{2} \\[5ex] a - b \\[3ex] = -\dfrac{1}{4} - \left(-\dfrac{1}{2}\right) \\[5ex] = -\dfrac{1}{4} + \dfrac{1}{2} \\[5ex] = -\dfrac{1}{4} + \dfrac{2}{4} \\[5ex] = \dfrac{1}{4} \\[5ex] \gt 0 $