Enhanced ACT Mathematics Tests

$ \left(\dfrac{1}{64}\right)^{-\dfrac{1}{2}} \\[5ex] = \dfrac{1}{\left(\dfrac{1}{64}\right)^{\dfrac{1}{2}}}\quad\dots\text{Law 6 Exp} \\[9ex] ................................................. \\[3ex] \left(\dfrac{1}{64}\right)^{\dfrac{1}{2}} \\[7ex] = \sqrt{\dfrac{1}{64}} \quad\dots\text{Law 7 Exp} \\[5ex] = \dfrac{1}{8} \\[5ex] ................................................. \\[3ex] = \dfrac{1}{\dfrac{1}{8}} \\[7ex] = 1 \div \dfrac{1}{8} \\[5ex] = 1 * \dfrac{8}{1} \\[5ex] = 8 $

The decimal expansion of a fraction is a:
(a.) terminating decimal if the prime factors of the denominator is 2 or 5.
(b.) repeating decimal if the prime factors of the denominator is not 2 or 5.

$ \underline{\text{Denominators of the Fractions}} \\[3ex] 4 = 2 \cdot 2 \quad\dots\text{gives a terminating decimal} \\[3ex] 5 = 5 \quad\dots\text{gives a terminating decimal} \\[3ex] 8 = 2 \cdot 2 \cdot 2 \quad\dots\text{gives a terminating decimal} \\[3ex] 3 \quad\dots\text{gives a repeating decimal} \\[3ex] $

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%.
To find the strongest evidence that a coin is not fair, we look for the result that deviates the furthest from the expected 50%.

$ \underline{\text{Experimental Probability}} \\[3ex] \text{Option F: } P(\text{heads}) = \dfrac{2}{3} = 0.6\bar{6} \approx 67\% \quad\dots\text{to the nearest percent} \\[5ex] \text{Option G: } P(\text{heads}) = \dfrac{3}{4} = 0.75 = 75\% \\[5ex] \text{Option H: } P(\text{heads}) = \dfrac{4}{5} = 0.8 = 80\% \\[5ex] \text{Option J: } P(\text{heads}) = \dfrac{5}{6} = 0.8\bar{3} \approx 83\% \quad\dots\text{to the nearest percent} \\[5ex] $ Option J. has both the largest number of trials (6 flips) and the highest deviation from the expected 50% mark.
Landing approximately 83% heads over 6 flips is the strongest evidence that the coin is not fair.

Let the number of weeks = w
$15 per week = $15w$

$ 45 + 15w \ge 200 \\[3ex] 15w \ge 200 - 45 \\[3ex] 15w \ge 155 \\[3ex] w \ge \dfrac{155}{15} \\[5ex] w \ge 10.3\bar{3} \\[3ex] $ The number of weeks must be an integer.
So, we need to round up.
In that regard, the minimum number of weeks is 11 weeks.

Similar triangles are triangles that have the same shape, with equal corresponding angles and proportional corresponding sides.
Angle—Angle (AA) Rule
If two angles of one triangle are equal to two angles of another triangle, the triangles are similar.

Sum of Angles of a Triangle Theorem
If two angles of one triangle are equal to two angles of another triangle, this implies that both triangles have all equal angles because the third angles must be the same to make the sum of the angles of each triangle to be equal to 180°

This implies that similar triangles must have the same angle measures.

$ (3x - 9)^2 \\[3ex] (3x - 9)(3x - 9) \\[3ex] 9x^2 - 27x - 27x + 81 \\[3ex] 9x^2 - 54x + 81 $

High Schools:     East High School and West High School
Number of Students:     3432
Ratio of Students:     11 : 13
Sum of Ratios: 11 + 13 = 24

$ \underline{\text{Attending East High School}} \\[3ex] \text{Number of Students} = \dfrac{11}{24} * 3432 \\[5ex] = 1573\text{ students} $

$ \underline{\text{1 week}} \\[3ex] \text{Worked Hours} = 54\text{ hours} \\[3ex] \text{Standard Hours} = 40\text{ hours} \\[3ex] \text{Overtime Hours} = 54 - 40 = 14\text{ hours} \\[5ex] \underline{\text{Standard Work}} \\[3ex] 40\text{ hours @ } \$20.10\text{ per hour} \\[3ex] = \$40(20.10) \\[5ex] \underline{\text{Overtime Work}} \\[3ex] 14\text{ hours @ } 1\dfrac{1}{2} * \$20.10\text{ per hour} \\[3ex] = \$14\left(\dfrac{1}{2}\right)(20.10) \\[5ex] \underline{\text{Total Pay in 1 Week}} \\[3ex] = \text{Standard Hours} + \text{Overtime Work} \\[3ex] = 40(20.10) + 14\left(\dfrac{1}{2}\right)(20.10) $

$ z^2 = -9 \\[3ex] z = \pm\sqrt{-9} \\[3ex] z = \pm 3i \\[3ex] $ Check

$z = \pm 3i; \;\; i^2 = -1$

LHS	RHS
$ z^2 \\[3ex] (3i)^2 \\[3ex] 3^2 \cdot i^2 \\[3ex] 9i^2 \\[3ex] 9(-1) \\[3ex] -9 $ $ (-3i)^2 \\[3ex] (-3)^2 \cdot i^2 \\[3ex] 9i^2 \\[3ex] 9(-1) \\[3ex] -9 $	$-9$

There are 12 points on the scatterplot.
Right now, the line touches only one point, and all the other points lie above it.
This shows that the line is not a good fit.
The line of best fit should be drawn so that the points are evenly spread on both sides of the line.
Since the line is too low, we need to shift it upward, which means increasing the y‑intercept.

n(Juniors) = 11
n(Seniors) = 79
n(Sample Space) = 11 + 79 = 90

Given that the Ist member selected is a junior:
1 person (a junior) is selected
Number of remaining students = 90 – 1 = 89

So, we have: 79 seniors out of 89 students (juniors and seniors)
Probability that the 2nd member selected will be a senior = $\dfrac{79}{89}$

$ \text{Smallest Engine Size, } x = 2.0\;L \\[3ex] \text{Actual Weight} = 1041\;kg \\[5ex] \text{Predicted Weight} = 375x + 370 \\[3ex] = 375(2) + 370 \\[3ex] = 1120\;kg \\[5ex] \text{Difference} = 1120 - 1041 \\[3ex] = 79\;kg $

The domain is the set of all the input values, t for which the function, d has an output.
The domain in interval notation is: [0, 13]

$ \dfrac{10i + 2}{2i} \\[5ex] = \dfrac{2(5i + 1)}{2i} \\[5ex] = \dfrac{5i + 1}{i} \\[5ex] = \dfrac{5i + 1}{i} * \dfrac{i}{i} \\[5ex] = \dfrac{5i^2 + i}{i^2} \\[5ex] = \dfrac{5(-1) + i}{-1} \\[5ex] = \dfrac{-5 + i}{-1} \\[5ex] = 5 - i $

$ 8x + y = 6...eqn.(1) \\[3ex] 9x - 6y = 11...eqn.(2) \\[5ex] \underline{\text{Substitution Method}} \\[3ex] \text{From } eqn.(1) \\[3ex] y = 6 - 8x \\[5ex] \text{Substitute for } y \text{ in } eqn.(2) \\[3ex] 9x - 6(6 - 8x) = 11 \\[3ex] 9x - 36 + 48x = 11 \\[3ex] 57x = 11 + 36 \\[3ex] 57x = 47 \\[3ex] x = \dfrac{47}{57} $

The Amplitude of a sinusoidal function is the height/distance from the maximum value of the sinusoidal function to the horizontal axis.
This means that the $\text{Amplitude} = 2$

The Period of a trigonometric function is the length of one cycle of the function (the length before it begins to repeat).
From the graph, there are three cycles of the function which stops at $x = 1$
⇒
$ 3 * \text{Period} = 1 \\[3ex] \text{Period} = \dfrac{1}{3} $

Let us solve for x

$ c(x - h) = q \\[3ex] \text{Divide both sides by } c \\[3ex] x - h = \dfrac{q}{c} \\[5ex] \text{Add } h \text{ to both sides} \\[3ex] x = \dfrac{q}{c} + h $

Let us represent the information diagrammatically as shown:


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 measure of an angle is opposite the length of the side facing that angle as regards size.
The measure of the smallest angle is opposite the shortest side length.
The measure of the greatest angle is opposite the longest side length.
The measure of the middle angle is opposite the middle side length.

The measure of $\angle R$ is the least.

$ f(x) = 2x - 5 \\[3ex] y = 2x - 5 \\[3ex] \text{Interchange } x \text{ and } y \\[3ex] x = 2y - 5 \\[3ex] \text{Solve for } y \\[3ex] 2y - 5 = x \\[3ex] 2y = x + 5 \\[3ex] y = \dfrac{x + 5}{2} \\[3ex] \implies \\[3ex] f^{-1}(x) = \dfrac{x + 5}{2} $

Let us cut out the unshaded region at the bottom right corner and place it in the corresponding shaded region at the top left corner


So, now we have an unshaded region that is a triangle.
Let the:
center of the circle = O
unlabeled vertices of the square be labeled as: E and F



For a Square circumscribed about a Circle:
The diameter of the circle is the length of the square
The radius of the circle is half the length of the square

$ \angle OAC = 90^\circ \quad\dots\text{radius, |OA| is } \perp \text{ to tangent at point of contact, A} \\[3ex] $ This gives a Right Triangle.
So, the unshaded region is a right triangle.
Hence, the area of the unshaded region is the area of the right triangle.

$ \text{Right Triangle COA} \\[3ex] \text{base} = 8\;inch \\[3ex] \perp \text{height} = 8\;inch \\[3ex] \text{Area} = \dfrac{1}{2} * \text{base} * \perp\text{height} \\[5ex] = \dfrac{1}{2} * 8 * 8 \\[5ex] = 32\;inch^2 \\[5ex] \text{Square CEDF} \\[3ex] \text{length} = 16\;inch \\[3ex] \text{Area} = \text{length}^2 \\[3ex] = 16^2 \\[3ex] = 256\;inch^2 \\[5ex] \text{Area of the shaded region} = \text{Area of Square CEDF} - \text{Area of Right Triangle COA} \\[3ex] = 256 - 32 \\[3ex] = 224\;inch^2 $

$ (x, y) \rightarrow (x', y') \\[3ex] (4, -16) \rightarrow (1, -4) \\[5ex] x * \text{Scale Factor} = x' \\[3ex] 4 * \text{Scale Factor} = 1 \\[3ex] \text{Scale Factor} = \dfrac{1}{4} \\[5ex] \text{To confirm:} \\[3ex] y * \text{Scale Factor} = y' \\[3ex] -16 * \dfrac{1}{4} = -4 \\[5ex] \text{Similarly, for } (12, -24) \\[3ex] 12 * \dfrac{1}{4} = 3 \\[3ex] -24 * \dfrac{1}{4} = -6 \\[3ex] (12, -24) \rightarrow (3, -6) $

Revenue, $x$	$P(x)$	$x \cdot P(x)$
$400	0.02	8
$100	0.02	2
$60	0.04	2.4
$0	0.92	0
$\Sigma [x \cdot P(x)] = 12.4$

The expected value of the of the revenue generated from an advertisement sent to a potential customer = $\Sigma [x \cdot P(x)] = \$12.40$

(1.) The normal distribution curve is symmetric about the mean.
The mean is the center of the normal distribution = 1,000.
Being symmetric about the mean implies that the area under the curve to the left of the mean is equal to the area under the curve to the right of the mean.

(2.) The total area under the normal curve is 100%.

This implies that 50% of the area is to the left of the mean and 50% is to the right of the mean.



$ P(1,100 \lt X \lt 1,200) = 14\% $

$ 0.\overline{7412} \\[3ex] \text{There are 4 digits before it begins to repeat.} \\[3ex] \text{After the decimal point:} \\[3ex] \text{1st digit} = 7 \\[3ex] \text{2nd digit} = 4 \\[3ex] \text{3rd digit} = 1 \\[3ex] \text{4th digit} = 2 \\[5ex] \dfrac{323}{4} \\[3ex] = 80.75 \\[3ex] = 80 + 0.75 \\[3ex] = 80 + \dfrac{0.75(4)}{4} \\[5ex] = 80 + \dfrac{3}{4} \\[5ex] = 80 \text{ remainder } 3 \\[5ex] \text{Remainder} = \text{3rd digit} = 1 \\[3ex] \therefore \text{323rd digit = 3rd digit} = 1 $

Let us represent the information diagrammatically:



$ \tan 25^\circ = \dfrac{\text{Height}}{45}\quad\dots\text{SOHCAHTOA} \\[5ex] \text{Height} = 45\tan 25^\circ $

$ \underline{\text{Cone}} \\[3ex] \text{radius, } r = 3\;in \\[3ex] \perp\text{height, } h = 2\;in \\[3ex] \text{Volume, } V = \dfrac{\pi}{3}r^2h \\[5ex] = \dfrac{\pi}{3} * 3^2 * 2 \\[5ex] = 6\pi\;in^3 \\[5ex] \underline{\text{Weight of Granulated Sugar (Leaked)}} \\[3ex] = 0.025\;lbs/in^3 \\[3ex] = \dfrac{0.025\;lbs}{1\;in^3} * 6\pi\;in^3 \\[5ex] = 0.15\pi\;lb \\[5ex] \underline{\text{Weight of Granulated Sugar (Unopened)}} \\[3ex] = 1.5\;lb \\[5ex] \underline{\text{Weight of Remaining Granulated Sugar}} \\[3ex] = \text{Weight of Granulated Sugar (Unopened)} - \text{Weight of Granulated Sugar (Leaked)} \\[3ex] = 1.5 - 0.15\pi \\[3ex] = 1.028761102 \\[3ex] \approx 1.029\;lbs $

The 2nd term in the expansion will be $8x^3y = 8x^3y^1$.
This implies that the degree = 3 + 1 = 4

For a nonnegative integer exponent n, the binomial expansion of $(x + y)^n$ has exactly $n + 1$ terms.
Hence, when $n = 4$, the expansion has 4 + 1 = 5 terms.

Let us represent the information on a table.

Time, t (hours)	Approximate Concentration $(mg/L)$
0	6
6	151
12	27

Quick observation of the output trend (approximate concentration) shows a decrease by approximately equal ratios rather than by equal differences.
This points to an Exponential Sequence.
Let's see.

$ \dfrac{151}{850} = 0.1776470588 \\[5ex] \dfrac{27}{151} = 0.178807947 \\[5ex] 0.1776470588 \approx 0.178807947 \\[3ex] $ The model is Exponential, because over equal time intervals, the concentration decreases by equal factors.

Because $-4\sqrt{14}$ is an irrational number, any rational number subtracted from it will make the difference an irrational number.
An irrational number divided by 2 still gives an irrational number.
Let us demonstrate with an example, say the rational number, $n = 1$

$ \dfrac{-4\sqrt{14} - 1}{2} \\[5ex] = -7.983314774\quad\dots\text{an irrational number} $

To find the x-intercept, set y to be zero, and solve for x

$ \dfrac{x^2}{9} - \dfrac{y^2}{49} = 1 \\[5ex] y = 0 \\[3ex] \dfrac{x^2}{9} - \dfrac{0^2}{49} = 1 \\[5ex] \dfrac{x^2}{9} = 1 \\[5ex] x^2 = 1(9) \\[3ex] x = \pm\sqrt{9} \\[3ex] x = \pm 3 \\[3ex] x-\text{intercepts} = (-3, 0) \text{ and } (3, 0) $

This is the graph of a transformed logarithmic function.
The vertical asymptote is $x = -2$
The general form of a transformed logarithmic function is: $y = a\log_p{c(x - h)} + k$ where h is the vertical asymptote

$ h = -2 \\[3ex] (x - h) \\[3ex] = x - (-2) \\[3ex] = (x + 2) \\[3ex] $ The correct answer is Option C.

Student: SamDom For Peace
Teacher: That's me. What's good?
Student: Let's assume that we were not given the multiple choice options.
Could you solve the question?
I understand that we are working with time, but let's also assume we have the time.
Teacher: Sure, let's do it.
What are the transformations that are observable?
Student: From the parent function, I think the transformations are:
Horizontal Shift to the Left and Vertical Shift.
I would also include a Vertical Stretch based on the multiple choice options.

$ \underline{\text{General Form of a Transformed Logarithmic Function}} \\[3ex] y = a\log_p{c(x - h)} + k \\[3ex] \text{where} \\[3ex] y = \text{Dependent Variable} \\[3ex] a = \text{Vertical Stretch/Compression/Reflection} \\[3ex] p = \text{base} \\[3ex] c = \text{Horizontal Stretch/Compression/Reflection} \\[3ex] x = \text{Independent Variable} \\[3ex] h = \text{Horizontal Shift (which is the location of the Vertical Asymptote)} \\[3ex] k = \text{Vertical Shift} \\[5ex] h = -2\quad\dots\text{From the graph} \\[3ex] c = 1 \quad\dots\text{Assume no horizontal stretch/compression/reflection} \\[3ex] \text{Noticeable Points on the Graph are:} (x, y) \\[3ex] (-1, 4) \\[3ex] (2, 8) \\[3ex] (6, 10) \\[5ex] \text{For } (x, y) = (-1, 4) \\[3ex] 4 = a\log_p{1[-1 - (-2)]} + k \\[3ex] 4 = a\log_p{1(-1 + 2)} + k \\[3ex] 4 = a\log_p{1} + k \\[3ex] 4 = a(0) + k \quad\dots\text{Law 3 Log} \\[3ex] 4 = 0 + k \\[3ex] k = 4 \\[5ex] \text{For } (x, y) = (2, 8) \\[3ex] 8 = a\log_p{1[2 - (-2)]} + 4 \\[3ex] 8 = a\log_p{1(2 + 2)} + 4 \\[3ex] 8 = a\log_p{4} + 4 \\[3ex] a\log_p{4} = 8 - 4 \\[3ex] a\log_p{4} = 4...eqn.(1) \\[5ex] \text{For } (x, y) = (6, 10) \\[3ex] 10 = a\log_p{1[6 - (-2)]} + 4 \\[3ex] 10 = a\log_p{1(6 + 2)} + 4 \\[3ex] 10 = a\log_p{8} + 4 \\[3ex] a\log_p{8} = 10 - 4 \\[3ex] a\log_p{8} = 6...eqn.(2) \\[5ex] eqn.(2) \div eqn.(1) \implies \\[3ex] \dfrac{a\log_p{8}}{a\log_p{4}} = \dfrac{6}{4} \\[5ex] \dfrac{\log_p{8}}{\log_p{4}} = \dfrac{3}{2} \\[5ex] \dfrac{\log_p{2^3}}{\log_p{2^2}} = \dfrac{3}{2} \\[5ex] \dfrac{3\log_p{2}}{2\log_p{2}} = \dfrac{3}{2} \\[5ex] \dfrac{3}{2} = \dfrac{3}{2} \\[5ex] $ Because the base cancels out, any valid base would work in this case.
In other words, p can be any valid base: $p \gt 0; \; p \ne 1$
Since any valid base works, let us choose a convenient base by setting the denominator equal to 2 (or the numerator equal to 3).

$ \text{Begin with the Numerator} \\[3ex] \log_p{8} = 3 \\[3ex] p^3 = 8 \quad\dots\text{Logarithm–Exponent Relationship} \\[3ex] p = \sqrt[3]{8} \\[3ex] p = 2 \\[5ex] \text{Check with the Denominator} \\[3ex] \log_p{4} = 2 \\[3ex] p^2 = 4 \quad\dots\text{Logarithm–Exponent Relationship} \\[3ex] p = \sqrt{4} \\[3ex] p = 2 \\[5ex] \text{Substitute for } p \text{ in } eqn.(1) \\[3ex] a\log_2{4} = 4 \\[3ex] a * 2 = 4 \\[3ex] a = \dfrac{4}{2} \\[5ex] a = 2 \\[5ex] \implies \\[3ex] y = 2\log_2{1[x - (-2)]} + 4 \\[3ex] y = 2\log_2{(x + 2)} + 4 \\[3ex] $ Student: Hmmmm...but that is not among the options.
But I do understand that any base would work.
Teacher: Would you like us to get the exact answer as the option?
Student: Yes, we probably need to do.
Teacher: Change of Base of Logarithms.

$ y = 2\log_2{(x + 2)} + 4 \\[3ex] @@@@@@@@@@@@@@@@@@@@@@@@@ \\[3ex] \text{For } \log_2{(x + 2)} \\[3ex] \text{Change from base 2 to base 4} \\[3ex] \log_2{(x + 2)} \\[5ex] = \dfrac{\log_4{(x + 2)}}{\log_4{2}}\quad\dots\text{Law 6 Log} \\[5ex] ........................................ \\[3ex] \log_4{2} = \dfrac{\log_2{2}}{\log_2{4}} \quad\dots\text{Law 6 Log} \\[5ex] = \dfrac{1}{2} \\[5ex] ........................................ \\[3ex] = \log_4{(x + 2)} \div \dfrac{1}{2} \\[5ex] = \log_4{(x + 2)} * 2 \\[3ex] = 2\log_4{(x + 2)} \\[3ex] @@@@@@@@@@@@@@@@@@@@@@@@@ \\[3ex] y = 2 * 2\log_4{(x + 2)} + 4 \\[3ex] y = 4\log_4{(x + 2)} + 4 $

$ n(W) = 14 \\[3ex] n(X) = 11 \\[3ex] n(W \cap X) = 0 \\[3ex] Y = W \cap X \\[3ex] \implies n(Y) = 0 \\[3ex] Z = W \cup X \\[3ex] n(Z) = n(W) + n(X) - n(W \cap X)\quad\dots\text{Addition Rule} \\[3ex] n(Z) = 14 + 11 - 0 \\[3ex] n(Z) = 25 \\[3ex] $ Given that event E consists of n simple events, $|E| = n$

$ |W| = 14 \\[3ex] |X| = 11 \\[3ex] |Y| = 0 \\[3ex] |Z| = 14 + 11 - 0 = 25 \\[3ex] \implies \\[3ex] 0 \lt 11 \lt 14 \lt 25 \\[3ex] |Y| \lt |X| \lt |W| \lt |Z| $

For Triangles ABC, ADH, DBE, DEH, HEC:
Triangle Midpoint Segment Theorem
States that: If a line segment joins the midpoints of any two sides of a triangle, then the line segment: (a.) is parallel to the third side.
(b.) is half the length of the third side.
This implies that:

$ |DE| = \dfrac{|AC|}{2} = |AH| = |HC| \\[5ex] |DH| = \dfrac{|BC|}{2} = |BE| = |EC| \\[5ex] |EH| = \dfrac{|AB|}{2} = |AD| = |DB| \\[5ex] $ Side – Side – Side Congruency Theorem
States that: Two triangles are congruent if the three sides of one triangle are equal to the three sides of another triangle.
This implies that:

$ \triangle DEH \cong \triangle ADH \cong \triangle EHC \cong \triangle DBE \\[3ex] \text{Consequently} \\[3ex] \text{Area of }\triangle DEH \\[3ex] = \text{Area of }\triangle ADH \\[3ex] = \text{Area of }\triangle EHC \\[3ex] = \text{Area of }\triangle DBE \\[5ex] \underline{\text{Diagram}} \\[3ex] \text{Area of }\triangle ABC \\[3ex] = \text{Area of }\triangle DEH \\[3ex] + \text{Area of }\triangle ADH \\[3ex] + \text{Area of }\triangle EHC \\[3ex] + \text{Area of }\triangle DBE \\[5ex] \text{Area of }\triangle ABC \\[3ex] = \text{Area of }\triangle DEH \\[3ex] + \text{Area of }\triangle DEH \\[3ex] + \text{Area of }\triangle DEH \\[3ex] + \text{Area of }\triangle DEH \\[5ex] 4 * \text{Area of }\triangle DEH = \text{Area of }\triangle ABC \\[3ex] \text{Area of }\triangle DEH = \dfrac{\text{Area of }\triangle ABC}{4}...eqn.(1) \\[5ex] $ For Triangles DEH, FGH:
Points F and G are the midpoints of $|DH|$ and $|EH|$, respectively.

Side – Angle – Side Similarity
States that: Two triangles are similar if two sides of one triangle are proportional to the two sides of the triangle and their included angles are equal.

$ \text{Because} \\[3ex] |HF| = \dfrac{|HD|}{2} \\[5ex] |HG| = \dfrac{|HE|}{2} \\[5ex] |HE| = 2 * |HG| ...eqn.(2) \\[5ex] \text{and} \\[3ex] m\angle FHG = m\angle DHE .\quad\dots\text{same vertex} \\[3ex] \triangle FGH \sim \triangle DEH \\[3ex] $ Similarity and Area Theorem
States that: If two polygons are similar, the ratio of their areas is equal to the square of the ratio of their corresponding linear side lengths.
This implies that:

$ \dfrac{\text{Area of }\triangle DEH}{\text{Area of }\triangle FGH} = \left(\dfrac{|HE|}{|HG|}\right)^2 \\[5ex] = \left(\dfrac{2 * |HG|}{|HG|}\right)^2 \quad\dots\text{From } eqn.(2) \\[5ex] = 2^2 \\[3ex] = 4 \\[3ex] \implies \\[3ex] \text{Area of }\triangle DEH = 4 * \text{Area of }\triangle FGH \\[5ex] \dfrac{\text{Area of }\triangle ABC}{4} = 4 * \text{Area of }\triangle FGH \quad\dots\text{From } eqn.(1) \\[5ex] \dfrac{\text{Area of }\triangle ABC}{\text{Area of }\triangle FGH} = 4 * 4 \\[5ex] = 16 \\[3ex] = \dfrac{16}{1} \\[5ex] = 16:1 $

Two non-vertical straight lines are perpendicular to each other if the product of their slopes is $-1$

Let us consider the minimum and maximum values of m and n to determine the greatest possible value of the absolute value of their difference.

$ -3 \le m \le 4 \\[3ex] \text{minimum } m = -3 \\[3ex] \text{maximum } m = 4 \\[5ex] -10 \le n \le 9 \\[3ex] \text{minimum } n = -10 \\[3ex] \text{maximum } n = 9 \\[5ex] \text{Check all possibilities} \\[3ex] |m - n| \\[3ex] = |-3 - (-10)| = |-3 + 10| = |7| = 7 \\[3ex] = |-3 - 9| = |-12| = 12 \\[3ex] = |4 - (-10)| = |4 + 10| = |14| = 14 \\[3ex] = |4 - 9| = |-5| = 5 \\[5ex] 14 \gt 12 \gt 7 \ht 5 \\[3ex] $ Therefore, the greatest possible value of $|m - n| = 14$

$ f(x) = 2.5x^2 - 30x + c \\[3ex] \text{Factor: } x + 4 \\[3ex] \text{Zero: } x + 4 = 0 \\[3ex] x = -4 \\[5ex] f(-4) = 2.5(4)^2 - 30(-4) + c = 0 \\[3ex] 2.5(16) + 120 + c = 0 \\[3ex] 40 + 120 + c = 0 \\[3ex] 160 + c = 0 \\[3ex] c = -160 $

The Generalized Multiplication Principle (Fundamental Counting Principle) states that the total number of ways of doing multiple tasks in succession is the product of the number of ways of doing each task individually.
5 questions with the answer choices of only “yes” or “no.”
1st question can be answered in two ways.
2nd question can be answered in two ways.
The same goes for the rest of the questions.
An employee can answer all 5 questions on the questionnaire in:
$2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^5 = 32$ ways.

$ 40\% = \dfrac{40}{100} = 0.4 \\[5ex] 112\% = \dfrac{112}{100} = 1.12 \\[5ex] 40\% \text{ of } m = a \\[3ex] a = 0.4m...eqn.(1) \\[5ex] b = \dfrac{7a}{10} \\[5ex] b = 0.7a \\[3ex] \text{Substitute for } b \\[3ex] b = 0.7(0.4m) \\[3ex] b = 0.28m \\[5ex] 112\% \text{ of } m \\[3ex] = 1.12m \\[3ex] ....................................... \\[3ex] \text{In terms of } b \\[3ex] 0.28m * what = 1.12m \\[3ex] what = \dfrac{1.12m}{0.28m} \\[5ex] what = 4 \\[3ex] ....................................... \\[3ex] = 4(0.28m) \\[3ex] = 4b $

This is an Arithmetic (Linear) Sequence because the first difference between two consecutive terms is the same (common difference).
Let the:
first term = a
common difference = d
number of terms = n
sum of the first n terms = $SAS_n$

$ \text{Using the Formula for Arithmetic Sequence} \\[3ex] \underline{\text{First 75 Positive Integers:}} 1, 2, 3, ..., 75 \\[3ex] \text{First Term, } a = 1 \\[3ex] \text{Common Difference, } d = 2 - 1 = 1 \\[3ex] \text{Number of Terms, } n = 75 \\[3ex] \text{Sum of the First n Terms, } SAS_n \\[5ex] SAS_{75} = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] = \dfrac{75}{2}[2(1) + 1(75 - 1)] \\[5ex] = \dfrac{75}{2}[2 + 74] \\[5ex] = \dfrac{75}{2} * 76 \\[5ex] = 2850 \\[3ex] $ Student: Is it not interesting that they will give us a value in a question and we don't get to use it?
Is it done to confuse us?
Teacher: Hmmm...not really.
But, some exams do.
For ACT, I don't think so.
Student: SamDom For Peace, what if I do not know the formula for the sum of an A.P off hand?
How can I solve this question?
Teacher: You asked a very good question.
ACT gave us that information so we can use it, rather than using the formula.
So, let's use it.

$ \underline{\text{First 15 terms}} \\[3ex] 1 + 2 + 3 + ... + 15 = 120 \\[5ex] \underline{\text{Next 15 terms}} \\[3ex] 16 + 17 + 18 + ... + 30 = 15(23) = 345 \\[5ex] \underline{\text{Next 15 terms}} \\[3ex] 31 + 32 + 33 + ... + 45 = 15(23 + 15) = 570 \\[5ex] \underline{\text{Next 15 terms}} \\[3ex] 15(23 + 15 + 15) = 15[23 + 2(15)] = 795 \\[5ex] \underline{\text{Next 15 terms}} \\[3ex] 15[23 + 3(15)] = 1020 \\[5ex] \underline{\text{First 75 terms}} \\[3ex] 120 + 345 + 570 + 795 + 1020 = 2850 \\[3ex] $ Student: Nice work.
But, I'm not sure if I can do this in a minute without having to use the formula.
And I really want to do it without the formula.
I understand these are the kind of questions I'll do at the end. Teacher: Well, guess what?
Use the calculator. The question can be done in less than a minute using the calculator.
Also, there is another formula that we can use that will take less than a minute.
The formula is easier to memorize but I would you suggest you use the calculator in this case to solve the question in less than a minute.
Student: Is the formula easier than the one we already used?
Teacher: Yes, it is easier.
It is known as:

The formula for the sum of the first $n$ positive integers.
Sum of the first $n$ positive integers, $Sum_n = \dfrac{n(n + 1)}{2}$

$ Sum_{75} = \dfrac{75(75 + 1)}{2} \\[5ex] = \dfrac{75(76)}{2} \\[5ex] = 2850 $

To Find the Vertical Asymptote (VA):
(1.) Simplify the function
(2.) Set the denominator to zero (if applicable after simplifying the function)
(3.) Solve for the value of $x$
(4.) $VA: x = value$

To Find the Horizontal Asymptote (HA):
(1.) Arrange the numerator in standard form
(2.) Arrange the denominator in standard form
(3.) If the degree of the numerator is less than the degree of the denominator, $HA: y = 0$
(4.) If the degree of the numerator is the same as the degree of the denominator,

$HA: y = \dfrac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}}$

$ \underline{\text{Vertical Asymptote}} \\[3ex] x^2 - 9 = 0 \\[3ex] x^2 = 9 \\[3ex] x = \pm\sqrt{9} \\[3ex] x = \pm 3 \\[5ex] \underline{\text{Horizontal Asymptote}} \\[3ex] y = \dfrac{3}{1} \\[5ex] y = 3 $