For in GOD we live, and move, and have our being.

- Acts 17:28

The Joy of a Teacher is the Success of his Students.

- Samuel Chukwuemeka

**Pre-requisite:** Linear Equations

For ACT Students

The ACT is a timed exam...$60$ questions for $60$ minutes

This implies that you have to solve each question in one minute.

Some questions will typically take less than a minute a solve.

Some questions will typically take more than a minute to solve.

The goal is to maximize your time. You use the time saved on those questions you
solved in less than a minute, to solve the questions that will take more than a minute.

So, you should try to solve each question __correctly__ and __timely__.

So, it is not just solving a question correctly, but solving it __correctly on time__.

Please ensure you attempt __all ACT questions__.

There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students

Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

**
Solve all questions.
Use at least two methods whenever applicable.
Show all work.
**

(1.) In January $2004$, the price of a gallon of milk was $$3.30$
per gallon.

In January $2015$, the price was $$3.76$ per gallon.

Calculate the average rate of change in the price of a gallon of milk from $2004$ to $2015$.

Interpret your solution. (Source: United States Department of Labor).

The average rate of change is also known as the slope.

From the information, we see that the input is the year and the output is the price of a gallon of milk.

*
Student: How did you know?
Teacher: What variable depends on what variable? Does the price of milk depend on
the year, or does the year depend on the price of milk? *

Student: The price of milk depends on the year.

Teacher: That is correct. This means that the price of milk is the ...

Student: Dependent variable also known as the output or the $y-value$

Teacher: That is correct. The year is the ...

Student: Independent variable also known as the input or the $x-value$

Teacher: Correct. Another reason is the based on the wording of the question: it asked for the average rate of change in the "price" based on the "year"

This means the changes in the output based on the changes in the input.

That describes the slope.

$ Point\:1 = (2004, 3.30) \\[3ex] x_1 = 2004 \\[3ex] y_1 = 3.30 \\[3ex] Point\:2 = (2015, 3.76) \\[3ex] x_2 = 2015 \\[3ex] y_2 = 3.76 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{3.76 - 3.30}{2015 - 2004} \\[5ex] = \dfrac{0.46}{11} \\[5ex] = 0.041818182 \\[3ex] $ The average rate of change in the price of a gallon of milk from $2004$ to $2015$ is about $\$0.0418$ per year.

This means that there was an increase of about $$0.04$ ($4$ cents) in the price of a gallon of milk per year from $2004$ to $2015$

$ \underline{Check\:\:Interpretation} \\[3ex] Use\:\:exact(not\:\:approximate)\:\:value\:\:of\:\:slope \\[3ex] 2004:\:\:Price = 3.30 \\[3ex] 2005:\:\:Price \approx 3.30 + 0.041818182 = 3.341818182 \\[3ex] 2006:\:\:Price \approx 3.341818182 + 0.041818182 = 3.38363636 \\[3ex] 2007:\:\:Price \approx 3.38363636 + 0.041818182 = 3.42545454 \\[3ex] 2008:\:\:Price \approx 3.42545454 + 0.041818182 = 3.46727272 \\[3ex] 2009:\:\:Price \approx 3.46727272 + 0.041818182 = 3.5090909 \\[3ex] 2010:\:\:Price \approx 3.5090909 + 0.041818182 = 3.55090908 \\[3ex] 2011:\:\:Price \approx 3.55090908 + 0.041818182 = 3.59272726 \\[3ex] 2012:\:\:Price \approx 3.59272726 + 0.041818182 = 3.63454544 \\[3ex] 2013:\:\:Price \approx 3.63454544 + 0.041818182 = 3.67636362 \\[3ex] 2014:\:\:Price \approx 3.67636362 + 0.041818182 = 3.7181818 \\[3ex] 2015:\:\:Price \approx 3.7181818 + 0.041818182 = \$3.75999998 $

In January $2015$, the price was $$3.76$ per gallon.

Calculate the average rate of change in the price of a gallon of milk from $2004$ to $2015$.

Interpret your solution. (Source: United States Department of Labor).

The average rate of change is also known as the slope.

From the information, we see that the input is the year and the output is the price of a gallon of milk.

Student: The price of milk depends on the year.

Teacher: That is correct. This means that the price of milk is the ...

Student: Dependent variable also known as the output or the $y-value$

Teacher: That is correct. The year is the ...

Student: Independent variable also known as the input or the $x-value$

Teacher: Correct. Another reason is the based on the wording of the question: it asked for the average rate of change in the "price" based on the "year"

This means the changes in the output based on the changes in the input.

That describes the slope.

$ Point\:1 = (2004, 3.30) \\[3ex] x_1 = 2004 \\[3ex] y_1 = 3.30 \\[3ex] Point\:2 = (2015, 3.76) \\[3ex] x_2 = 2015 \\[3ex] y_2 = 3.76 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{3.76 - 3.30}{2015 - 2004} \\[5ex] = \dfrac{0.46}{11} \\[5ex] = 0.041818182 \\[3ex] $ The average rate of change in the price of a gallon of milk from $2004$ to $2015$ is about $\$0.0418$ per year.

This means that there was an increase of about $$0.04$ ($4$ cents) in the price of a gallon of milk per year from $2004$ to $2015$

$ \underline{Check\:\:Interpretation} \\[3ex] Use\:\:exact(not\:\:approximate)\:\:value\:\:of\:\:slope \\[3ex] 2004:\:\:Price = 3.30 \\[3ex] 2005:\:\:Price \approx 3.30 + 0.041818182 = 3.341818182 \\[3ex] 2006:\:\:Price \approx 3.341818182 + 0.041818182 = 3.38363636 \\[3ex] 2007:\:\:Price \approx 3.38363636 + 0.041818182 = 3.42545454 \\[3ex] 2008:\:\:Price \approx 3.42545454 + 0.041818182 = 3.46727272 \\[3ex] 2009:\:\:Price \approx 3.46727272 + 0.041818182 = 3.5090909 \\[3ex] 2010:\:\:Price \approx 3.5090909 + 0.041818182 = 3.55090908 \\[3ex] 2011:\:\:Price \approx 3.55090908 + 0.041818182 = 3.59272726 \\[3ex] 2012:\:\:Price \approx 3.59272726 + 0.041818182 = 3.63454544 \\[3ex] 2013:\:\:Price \approx 3.63454544 + 0.041818182 = 3.67636362 \\[3ex] 2014:\:\:Price \approx 3.67636362 + 0.041818182 = 3.7181818 \\[3ex] 2015:\:\:Price \approx 3.7181818 + 0.041818182 = \$3.75999998 $

(2.) Onesimus bought a phone for $$84$ and signed up for a single-line
phone plan with $2000$ monthly anytime minutes.

The cost of the plan was $$116.99$ per month.

(a.) Write an equation that is used to determine the total cost, $C(t)$ of this phone plan for $t$ months.

(b.) Calculate the cost for $21$ months assuming that the number of minutes he uses does not exceed $2000$ per month.

Based on the question:

Total cost of the phone plan = $C(t)$

Number of months = $t$

Fixed cost = $84$

Variable cost = $116.99t$ ($$116.99$ per month)

Total cost = Fixed cost + Variable cost

$ C(t) = 84 + 116.99t \\[3ex] For\:\: t = 21 \\[3ex] C(21) = 84 + 116.99(21) \\[3ex] C(21) = 84 + 2456.79 \\[3ex] C(21) = 2540.79 \\[3ex] $ (a.) The equation is: $C(t) = 84 + 116.99t$

(b.) The cost for $21$ months is: $$2540.79$

The cost of the plan was $$116.99$ per month.

(a.) Write an equation that is used to determine the total cost, $C(t)$ of this phone plan for $t$ months.

(b.) Calculate the cost for $21$ months assuming that the number of minutes he uses does not exceed $2000$ per month.

Based on the question:

Total cost of the phone plan = $C(t)$

Number of months = $t$

Fixed cost = $84$

Variable cost = $116.99t$ ($$116.99$ per month)

Total cost = Fixed cost + Variable cost

$ C(t) = 84 + 116.99t \\[3ex] For\:\: t = 21 \\[3ex] C(21) = 84 + 116.99(21) \\[3ex] C(21) = 84 + 2456.79 \\[3ex] C(21) = 2540.79 \\[3ex] $ (a.) The equation is: $C(t) = 84 + 116.99t$

(b.) The cost for $21$ months is: $$2540.79$

(3.) **ACT** Students studying motion observed a cart rolling at a constant rate along
a straight line.

The table below gives the distance $d$ feet, the cart was from a reference point at $1-second$ intervals from $t=0$ seconds to $t=5$ seconds.

Which of the following equations represented this relationship between $d$ and $t$?

$ F.\:\: d = t + 15 \\[3ex] G.\:\: d = 3t + 12 \\[3ex] H.\:\: d = 3t + 15 \\[3ex] J.\:\: d = 15t + 3 \\[3ex] K.\:\: d = 33t \\[3ex] $

We can do this in two ways.

You may decide to use any method that is faster for you.

You can do either method in about a minute

__First Method:__ Slope-Intercept Form

Calculate the slope.

Use the first and last points.

Slope-Intercept form: $y = mx + b$

Slope-Intercept form: $d = mt + b$

Point $1$ is $(0, 15)$

$t_1 = 0$

$d_1 = 15$

Point $2$ is $(5, 30)$

$t_2 = 5$

$d_2 = 30$

$ m = \dfrac{d_2 - d_1}{t_2 - t_1} \\[5ex] m = \dfrac{30 - 15}{5 - 0} \\[5ex] m = \dfrac{15}{5} \\[5ex] m = 3 \\[3ex] b = d-intercept(value\:\: of\:\: d \:\:when\:\: t = 0) \\[3ex] b = 15 \\[3ex] d = 3t + 15 \\[3ex] $__Second Method:__ Testing and Eliminating

test each of the answer options and eliminate anyone that does not satisfy any of the points.

$ F.\:\: d = t + 15 \\[3ex] (0, 15) \implies 15 = 0 + 15 \:\:yes \\[3ex] (1, 18) \implies 18 = 1 + 15 \:\:no\:\: next\:\: option \\[3ex] G.\:\: d = 3t + 12 \\[3ex] (0, 15) \implies 15 = 3(0) + 12 = 0 + 12 \:\:no\:\: next\:\: option \\[3ex] H.\:\: d = 3t + 15 \\[3ex] (0, 15) \implies 15 = 3(0) + 15 = 0 + 15 \:\:yes \\[3ex] (1, 18) \implies 18 = 3(1) + 15 = 3 + 15 \:\:yes \\[3ex] (2, 21) \implies 21 = 3(2) + 15 = 6 + 15 \:\:yes \\[3ex] (3, 24) \implies 24 = 3(3) + 15 = 9 + 15 \:\:yes \\[3ex] (4, 27) \implies 27 = 3(4) + 15 = 12 + 15 \:\:yes \\[3ex] (5, 30) \implies 30 = 3(5) + 15 = 15 + 15 \:\:yes \\[3ex] d = 3t + 15 $

The table below gives the distance $d$ feet, the cart was from a reference point at $1-second$ intervals from $t=0$ seconds to $t=5$ seconds.

$t$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ |

$d$ | $15$ | $18$ | $21$ | $24$ | $27$ | $30$ |

Which of the following equations represented this relationship between $d$ and $t$?

$ F.\:\: d = t + 15 \\[3ex] G.\:\: d = 3t + 12 \\[3ex] H.\:\: d = 3t + 15 \\[3ex] J.\:\: d = 15t + 3 \\[3ex] K.\:\: d = 33t \\[3ex] $

We can do this in two ways.

You may decide to use any method that is faster for you.

You can do either method in about a minute

Calculate the slope.

Use the first and last points.

Slope-Intercept form: $y = mx + b$

Slope-Intercept form: $d = mt + b$

Point $1$ is $(0, 15)$

$t_1 = 0$

$d_1 = 15$

Point $2$ is $(5, 30)$

$t_2 = 5$

$d_2 = 30$

$ m = \dfrac{d_2 - d_1}{t_2 - t_1} \\[5ex] m = \dfrac{30 - 15}{5 - 0} \\[5ex] m = \dfrac{15}{5} \\[5ex] m = 3 \\[3ex] b = d-intercept(value\:\: of\:\: d \:\:when\:\: t = 0) \\[3ex] b = 15 \\[3ex] d = 3t + 15 \\[3ex] $

test each of the answer options and eliminate anyone that does not satisfy any of the points.

$ F.\:\: d = t + 15 \\[3ex] (0, 15) \implies 15 = 0 + 15 \:\:yes \\[3ex] (1, 18) \implies 18 = 1 + 15 \:\:no\:\: next\:\: option \\[3ex] G.\:\: d = 3t + 12 \\[3ex] (0, 15) \implies 15 = 3(0) + 12 = 0 + 12 \:\:no\:\: next\:\: option \\[3ex] H.\:\: d = 3t + 15 \\[3ex] (0, 15) \implies 15 = 3(0) + 15 = 0 + 15 \:\:yes \\[3ex] (1, 18) \implies 18 = 3(1) + 15 = 3 + 15 \:\:yes \\[3ex] (2, 21) \implies 21 = 3(2) + 15 = 6 + 15 \:\:yes \\[3ex] (3, 24) \implies 24 = 3(3) + 15 = 9 + 15 \:\:yes \\[3ex] (4, 27) \implies 27 = 3(4) + 15 = 12 + 15 \:\:yes \\[3ex] (5, 30) \implies 30 = 3(5) + 15 = 15 + 15 \:\:yes \\[3ex] d = 3t + 15 $

(4.) **ACT** A water tank that initially contained $200$ gallons of water is leaking water at a constant
rate of $4$ gallons per minute.

For the amount of time the tank has water, which of the following function models gives the number of gallons, $G$, in the tank $t$ minutes after the leak started?

$ A.\:\: G(t) = 196 -t \\[3ex] B.\:\: G(t) = 200 - 4t \\[3ex] C.\:\: G(t) = 200t - 4 \\[3ex] D.\:\: G(t) = 200t - 4t^2 \\[3ex] E.\:\: G(t) = 200\left(\dfrac{3}{4}\right)^t \\[5ex] $

Water is leaking at a constant rate of $4$ gallons per minute.

After $1$ minute, $4$ gallons leaks

After $2$ minutes, $4 * 2 = 8$ gallons leaks

Therefore, after $t$ minutes, $4 * t = 4t$ gallons leaks

$ Initial\:\:gallons = 200 \\[3ex] Leaked\:\:gallons = 4t \\[3ex] Remaining\:\:gallons = 200 - 4t $

For the amount of time the tank has water, which of the following function models gives the number of gallons, $G$, in the tank $t$ minutes after the leak started?

$ A.\:\: G(t) = 196 -t \\[3ex] B.\:\: G(t) = 200 - 4t \\[3ex] C.\:\: G(t) = 200t - 4 \\[3ex] D.\:\: G(t) = 200t - 4t^2 \\[3ex] E.\:\: G(t) = 200\left(\dfrac{3}{4}\right)^t \\[5ex] $

Water is leaking at a constant rate of $4$ gallons per minute.

After $1$ minute, $4$ gallons leaks

After $2$ minutes, $4 * 2 = 8$ gallons leaks

Therefore, after $t$ minutes, $4 * t = 4t$ gallons leaks

$ Initial\:\:gallons = 200 \\[3ex] Leaked\:\:gallons = 4t \\[3ex] Remaining\:\:gallons = 200 - 4t $

(5.) **ACT** Shasta is participating in a bike ride for charity.

The graph of speed $(s)$ versus time $(t)$ for the first $20$ seconds of her bike ride is shown in the coordinate plane below.

The graph is composed of $2$ line segments for which the endpoints are at $(0, 0), (10, 5),$ and $(20, 3)$.

Shasta traveled $25$ meters in the first $10$ seconds.

Beginning at $t = 20$ seconds, Shasta slows down as she approaches a familiar group of riders ahead of her and then travels at a constant speed with the group after joining them.

What is Shasta's speed, in meters per second, at $t = 3$ seconds?

$ F.\:\: 1.5 \\[3ex] G.\:\: 2.0 \\[3ex] H.\:\: 2.5 \\[3ex] J.\:\: 3.0 \\[3ex] K.\:\: 6.0 \\[3ex] $

The first $3$ seconds lies in the first line segment whose endpoints are $(0, 0)$ and $(10, 5)$

First: We need to find the slope

Second: We have to write the equation of the line joining those points in Slope-Intercept form

$ Point\:\:1 = (0, 0) \\[3ex] x_1 = 0 \\[3ex] y_1 = 0 \\[3ex] y-intercept, b = 0 \\[3ex] Point\:2 = (10, 5) \\[3ex] x_2 = 10 \\[3ex] y_2 = 5 \\[3ex] Slope, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{5 - 0}{10 - 0} \\[5ex] m = \dfrac{5}{10} \\[5ex] m = \dfrac{1}{2} \\[5ex] y = mx + b \\[3ex] y = \dfrac{1}{2}x + 0 \\[5ex] y = \dfrac{1}{2}x \\[3ex] This\:\:implies\:\:that \\[3ex] s = \dfrac{1}{2}t \\[5ex] when\:\:t = 3 \\[3ex] s = \dfrac{1}{2} * 3 \\[5ex] s = \dfrac{3}{2} \\[5ex] s = 1.5\:ms^{-1} $

The graph of speed $(s)$ versus time $(t)$ for the first $20$ seconds of her bike ride is shown in the coordinate plane below.

The graph is composed of $2$ line segments for which the endpoints are at $(0, 0), (10, 5),$ and $(20, 3)$.

Shasta traveled $25$ meters in the first $10$ seconds.

Beginning at $t = 20$ seconds, Shasta slows down as she approaches a familiar group of riders ahead of her and then travels at a constant speed with the group after joining them.

What is Shasta's speed, in meters per second, at $t = 3$ seconds?

$ F.\:\: 1.5 \\[3ex] G.\:\: 2.0 \\[3ex] H.\:\: 2.5 \\[3ex] J.\:\: 3.0 \\[3ex] K.\:\: 6.0 \\[3ex] $

The first $3$ seconds lies in the first line segment whose endpoints are $(0, 0)$ and $(10, 5)$

First: We need to find the slope

Second: We have to write the equation of the line joining those points in Slope-Intercept form

$ Point\:\:1 = (0, 0) \\[3ex] x_1 = 0 \\[3ex] y_1 = 0 \\[3ex] y-intercept, b = 0 \\[3ex] Point\:2 = (10, 5) \\[3ex] x_2 = 10 \\[3ex] y_2 = 5 \\[3ex] Slope, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{5 - 0}{10 - 0} \\[5ex] m = \dfrac{5}{10} \\[5ex] m = \dfrac{1}{2} \\[5ex] y = mx + b \\[3ex] y = \dfrac{1}{2}x + 0 \\[5ex] y = \dfrac{1}{2}x \\[3ex] This\:\:implies\:\:that \\[3ex] s = \dfrac{1}{2}t \\[5ex] when\:\:t = 3 \\[3ex] s = \dfrac{1}{2} * 3 \\[5ex] s = \dfrac{3}{2} \\[5ex] s = 1.5\:ms^{-1} $

(6.) **ACT** Based on Question $5$

Shasta's acceleration, $a$, over the interval from $t = 10$ seconds to $t = 20$ seconds, is equal to the slope of the graph over that interval, measured in meters per second per second.

What is the value of $a$?

$ A.\:\: -5 \\[3ex] B.\:\: -\dfrac{1}{5} \\[5ex] C.\:\: \dfrac{1}{5} \\[5ex] D.\:\: 2 \\[3ex] E.\:\: 5 \\[3ex] $

$ Point\:\:1 = (10, 5) \\[3ex] x_1 = 10 \\[3ex] y_1 = 5 \\[3ex] Point\:2 = (20, 3) \\[3ex] x_2 = 20 \\[3ex] y_2 = 3 \\[3ex] Slope, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{3 - 5}{20 - 10} \\[5ex] m = \dfrac{-2}{10} \\[5ex] m = \dfrac{-1}{5} \\[5ex] a = m \\[3ex] \therefore a = -\dfrac{1}{5}\:ms^{-2} $

Shasta's acceleration, $a$, over the interval from $t = 10$ seconds to $t = 20$ seconds, is equal to the slope of the graph over that interval, measured in meters per second per second.

What is the value of $a$?

$ A.\:\: -5 \\[3ex] B.\:\: -\dfrac{1}{5} \\[5ex] C.\:\: \dfrac{1}{5} \\[5ex] D.\:\: 2 \\[3ex] E.\:\: 5 \\[3ex] $

$ Point\:\:1 = (10, 5) \\[3ex] x_1 = 10 \\[3ex] y_1 = 5 \\[3ex] Point\:2 = (20, 3) \\[3ex] x_2 = 20 \\[3ex] y_2 = 3 \\[3ex] Slope, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{3 - 5}{20 - 10} \\[5ex] m = \dfrac{-2}{10} \\[5ex] m = \dfrac{-1}{5} \\[5ex] a = m \\[3ex] \therefore a = -\dfrac{1}{5}\:ms^{-2} $

(7.) **ACT** Based on Question $5$

Calen started his bike ride earlier than Shasta.

During the first $15$ seconds of Shasta's ride, Calen was traveling at a constant speed equal to $\dfrac{1}{2}$ of Shasta's maximum speed during that same time period.

How far, in meters, did Calen travel during the first $15$ seconds of Shasta's ride?

$ F.\:\: 22\dfrac{1}{2} \\[5ex] G.\:\: 25 \\[3ex] H.\:\: 37\dfrac{1}{2} \\[5ex] J.\:\: 45 \\[3ex] K.\:\: 75 \\[3ex] $

Shasta's maximum speed during the first $15$ seconds = $5$ meters per second

Calen's speed is half of that

Calen's speed = $\dfrac{1}{2} * 5 = \dfrac{5}{2}$ meters per second

This is the constant speed of Calen for the first $15$ seconds

This implies that time = $15$ seconds

__How far__ means __Distance__

distance = speed * time

Distance traveled by Calen during the first $15$ seconds:

$ = \dfrac{5}{2} * 15 \\[5ex] = \dfrac{75}{2} \\[5ex] = 37\dfrac{1}{2}\:meters $

Calen started his bike ride earlier than Shasta.

During the first $15$ seconds of Shasta's ride, Calen was traveling at a constant speed equal to $\dfrac{1}{2}$ of Shasta's maximum speed during that same time period.

How far, in meters, did Calen travel during the first $15$ seconds of Shasta's ride?

$ F.\:\: 22\dfrac{1}{2} \\[5ex] G.\:\: 25 \\[3ex] H.\:\: 37\dfrac{1}{2} \\[5ex] J.\:\: 45 \\[3ex] K.\:\: 75 \\[3ex] $

Shasta's maximum speed during the first $15$ seconds = $5$ meters per second

Calen's speed is half of that

Calen's speed = $\dfrac{1}{2} * 5 = \dfrac{5}{2}$ meters per second

This is the constant speed of Calen for the first $15$ seconds

This implies that time = $15$ seconds

distance = speed * time

Distance traveled by Calen during the first $15$ seconds:

$ = \dfrac{5}{2} * 15 \\[5ex] = \dfrac{75}{2} \\[5ex] = 37\dfrac{1}{2}\:meters $

(8.) **ACT** Based on Question $5$

Which of the following graphs best represents the portion of Shasta's ride beginning at $t = 20$ seconds?

Beginning at $t = 20$ seconds, Shasta slows down as she approaches a familiar group of riders ahead of her...

This eliminates Options $A$, $B$, and $E$

...and then travels at a constant speed with the group after joining them. This eliminates Option $C$

Option $D$ is the graph that best represents the portion of Shasta's ride beginning at $t = 20$ seconds

Which of the following graphs best represents the portion of Shasta's ride beginning at $t = 20$ seconds?

Beginning at $t = 20$ seconds, Shasta slows down as she approaches a familiar group of riders ahead of her...

This eliminates Options $A$, $B$, and $E$

...and then travels at a constant speed with the group after joining them. This eliminates Option $C$

Option $D$ is the graph that best represents the portion of Shasta's ride beginning at $t = 20$ seconds

(9.) **ACT** The length, $L$ in meters, of a spring is given by the equation
$L = \dfrac{2}{3}F + 0.03$

where $F$ is the applied force in newtons.

What force, in newtons, must be applied for the spring's length to be $0.18$ meters?

$ F.\:\: 0.13 \\[3ex] G.\:\: 0.15 \\[3ex] H.\:\: 0.225 \\[3ex] J.\:\: 0.255 \\[3ex] K.\:\: 0.27 \\[3ex] $

$ L = \dfrac{2}{3}F + 0.03 \\[5ex] \dfrac{2}{3}F + 0.03 = L \\[5ex] \dfrac{2}{3}F = L - 0.03 \\[5ex] LCD = 3 \\[3ex] 3 * \dfrac{2}{3}F = 3(L - 0.03) \\[5ex] 2F = 3(L - 0.03) \\[3ex] F = \dfrac{3(L - 0.03)}{2} \\[5ex] L = 0.18\:\:meters \\[3ex] F = \dfrac{3(0.18 - 0.03)}{2} \\[5ex] F = \dfrac{3(0.15)}{2} \\[5ex] F = \dfrac{0.45}{2} \\[5ex] F = 0.225\:\:Newtons $

where $F$ is the applied force in newtons.

What force, in newtons, must be applied for the spring's length to be $0.18$ meters?

$ F.\:\: 0.13 \\[3ex] G.\:\: 0.15 \\[3ex] H.\:\: 0.225 \\[3ex] J.\:\: 0.255 \\[3ex] K.\:\: 0.27 \\[3ex] $

$ L = \dfrac{2}{3}F + 0.03 \\[5ex] \dfrac{2}{3}F + 0.03 = L \\[5ex] \dfrac{2}{3}F = L - 0.03 \\[5ex] LCD = 3 \\[3ex] 3 * \dfrac{2}{3}F = 3(L - 0.03) \\[5ex] 2F = 3(L - 0.03) \\[3ex] F = \dfrac{3(L - 0.03)}{2} \\[5ex] L = 0.18\:\:meters \\[3ex] F = \dfrac{3(0.18 - 0.03)}{2} \\[5ex] F = \dfrac{3(0.15)}{2} \\[5ex] F = \dfrac{0.45}{2} \\[5ex] F = 0.225\:\:Newtons $

(10.) **ACT** The relationship between temperatures in degrees Fahrenheit, $F$, and temperatures
in degrees Celsius, $C$, is expressed by the formula $F = \dfrac{9}{5}C + 32$

Calvin reads a temperature of $38^\circ$ on a Celsius thermometer.

To the nearest degree, what is the equivalent temperature on a Fahrenheit thermometer?

$ A.\:\: 36^\circ \\[3ex] B.\:\: 53^\circ \\[3ex] C.\:\: 68^\circ \\[3ex] D.\:\: 70^\circ \\[3ex] E.\:\: 100^\circ \\[3ex] $

$ F = \dfrac{9}{5}C + 32 \\[5ex] C = 38^\circ \\[3ex] F = \dfrac{9}{5} * 38 + 32 \\[5ex] F = \dfrac{342}{5} + 32 \\[5ex] F = \dfrac{342}{5} + \dfrac{160}{5} \\[5ex] F = \dfrac{342 + 160}{5} \\[5ex] F = \dfrac{502}{5} \\[5ex] F = 100.4^\circ $

Calvin reads a temperature of $38^\circ$ on a Celsius thermometer.

To the nearest degree, what is the equivalent temperature on a Fahrenheit thermometer?

$ A.\:\: 36^\circ \\[3ex] B.\:\: 53^\circ \\[3ex] C.\:\: 68^\circ \\[3ex] D.\:\: 70^\circ \\[3ex] E.\:\: 100^\circ \\[3ex] $

$ F = \dfrac{9}{5}C + 32 \\[5ex] C = 38^\circ \\[3ex] F = \dfrac{9}{5} * 38 + 32 \\[5ex] F = \dfrac{342}{5} + 32 \\[5ex] F = \dfrac{342}{5} + \dfrac{160}{5} \\[5ex] F = \dfrac{342 + 160}{5} \\[5ex] F = \dfrac{502}{5} \\[5ex] F = 100.4^\circ $

(11.) **ACT** When Jeff starts a math assignment, he spends $5$ minutes getting out his book and a
sheet of paper, sharpening his pencil, looking up the assignment in his assignment notebook, and turning
to the correct page in his book.

The equation $t = 10p + 5$ models the time, $t$ minutes, Jeff budgets for a math assignment with $p$ problems.

Which of the following statements is necessarily true according to Jeff's model?

F. He budgets $15$ minutes per problem

G. He budgets $10$ minutes per problem

H. He budgets $5$ minutes per problem

J. He budgets $10$ minutes per problem for the hard problems and $5$ minutes per problem for the easy problems.

K. He budgets a $5-minute$ break after each problem.

Let us analyze this question.

Look at the function: $t = 10p + 5$

Compare to $y = mx + b$

The $slope, m = 10$ minutes per problem

The $y-intercept, b = 5$ minutes

(1.) Jeff takes $5$ minutes to get prepared prior to solving his math assignment problems.

That $5$ minutes is the $y-intercept$

(2.) But, he actually spends $10$ minutes on EACH problem.

That $10$ minutes is the slope.

So, he budgets $10$ minutes per problem....Correct answer

However, he budgets $t = 10(1) + 5 = 10 + 5 = 15$ minutes to get started and solve a problem

He budgets $t = 10(2) + 5 = 20 + 5 = 25$ minutes to get started and solve two problems....and so on and so forth

The equation $t = 10p + 5$ models the time, $t$ minutes, Jeff budgets for a math assignment with $p$ problems.

Which of the following statements is necessarily true according to Jeff's model?

F. He budgets $15$ minutes per problem

G. He budgets $10$ minutes per problem

H. He budgets $5$ minutes per problem

J. He budgets $10$ minutes per problem for the hard problems and $5$ minutes per problem for the easy problems.

K. He budgets a $5-minute$ break after each problem.

Let us analyze this question.

Look at the function: $t = 10p + 5$

Compare to $y = mx + b$

The $slope, m = 10$ minutes per problem

The $y-intercept, b = 5$ minutes

(1.) Jeff takes $5$ minutes to get prepared prior to solving his math assignment problems.

That $5$ minutes is the $y-intercept$

(2.) But, he actually spends $10$ minutes on EACH problem.

That $10$ minutes is the slope.

So, he budgets $10$ minutes per problem....Correct answer

However, he budgets $t = 10(1) + 5 = 10 + 5 = 15$ minutes to get started and solve a problem

He budgets $t = 10(2) + 5 = 20 + 5 = 25$ minutes to get started and solve two problems....and so on and so forth

(12.) **ACT** Mr. Cleary's algebra class is discussing slopes of lines.

The class is to graph the total cost, $C$, of buying $h$ hamburgers that cost $99¢$ each.

Mr. Cleary asks the class to describe the slope between any $2$ points $(h, C)$ on the graph.

Devon gives a correct response that the slope between any $2$ points on this graph is always:

F. zero

G. the same positive value.

H. the same negative value.

J. a positive value, but the value varies.

K. a negative value, but the value varies.

A graph of the total cost, $C$, of buying $h$ hamburgers that cost $99¢$ each indicates that:

the total cost, $C$ is the dependent variable on the $y-axis$

the hamburgers, $h$ is the independent variable on the $x-axis$

hamburgers that cost $99¢$ each (means that the cost per hamburger), $99¢$ is the slope

This implies that the slope between any $2$ points on this graph is always that same positive value, $99¢$

The class is to graph the total cost, $C$, of buying $h$ hamburgers that cost $99¢$ each.

Mr. Cleary asks the class to describe the slope between any $2$ points $(h, C)$ on the graph.

Devon gives a correct response that the slope between any $2$ points on this graph is always:

F. zero

G. the same positive value.

H. the same negative value.

J. a positive value, but the value varies.

K. a negative value, but the value varies.

A graph of the total cost, $C$, of buying $h$ hamburgers that cost $99¢$ each indicates that:

the total cost, $C$ is the dependent variable on the $y-axis$

the hamburgers, $h$ is the independent variable on the $x-axis$

hamburgers that cost $99¢$ each (means that the cost per hamburger), $99¢$ is the slope

This implies that the slope between any $2$ points on this graph is always that same positive value, $99¢$

(13.) **ACT** At a certain airline company, the cost to transfer mileage points from one
person's account to another person's account is $\$0.75$ for every $100$ mileage points transferred
plus a onetime $\$20$ processing fee.

What is the cost to transfer $7,000$ mileage points from one account to another at that airline company?

$ F.\:\: \$25.25 \\[3ex] G.\:\: \$67.50 \\[3ex] H.\:\: \$72.50 \\[3ex] J.\:\: \$75.00 \\[3ex] K.\:\: \$95.00 \\[3ex] $

We shall use the Equation of a Straight Line in Slope-Intercept Form

$y = mx + b$

$m$ is the slope.

The slope is the cost to transfer for every mileage point (every $1$ mileage point).

Let us find the slope.

__Proportional Reasoning Method__

$ \dfrac{m}{1} = \dfrac{0.75}{100} \\[5ex] m = \$0.0075\:\:per\:\:mileage\:\:point \\[3ex] $ The $y-intercept$ is the one-time processing fee.

$b$ is the $y-intercept$

$x$ is the mileage point

$y$ is the cost to transfer mileage points from one account to another

$ b = \$20 \\[3ex] m = 0.0075 \\[3ex] x = 7000 \\[3ex] y = ? \\[3ex] y = mx + b \\[3ex] y = 0.0075(7000) + 20 \\[3ex] y = 52.5 + 20 \\[3ex] y = \$72.50 $

What is the cost to transfer $7,000$ mileage points from one account to another at that airline company?

$ F.\:\: \$25.25 \\[3ex] G.\:\: \$67.50 \\[3ex] H.\:\: \$72.50 \\[3ex] J.\:\: \$75.00 \\[3ex] K.\:\: \$95.00 \\[3ex] $

We shall use the Equation of a Straight Line in Slope-Intercept Form

$y = mx + b$

$m$ is the slope.

The slope is the cost to transfer for every mileage point (every $1$ mileage point).

Let us find the slope.

cost($\$$) | mileage |
---|---|

$0.75$ | $100$ |

$m$ | $1$ |

$ \dfrac{m}{1} = \dfrac{0.75}{100} \\[5ex] m = \$0.0075\:\:per\:\:mileage\:\:point \\[3ex] $ The $y-intercept$ is the one-time processing fee.

$b$ is the $y-intercept$

$x$ is the mileage point

$y$ is the cost to transfer mileage points from one account to another

$ b = \$20 \\[3ex] m = 0.0075 \\[3ex] x = 7000 \\[3ex] y = ? \\[3ex] y = mx + b \\[3ex] y = 0.0075(7000) + 20 \\[3ex] y = 52.5 + 20 \\[3ex] y = \$72.50 $

(14.) **ACT** The graph below gives the speed, in *knots* (nautical miles per hour), of a
cruise ship during a $5-hour$ period.

Which of the following values is closest to the rate of change, in knots per hour, of the speed of the ship between hours $2$ and $4$?

$
A.\:\: 2 \\[3ex]
B.\:\: 3 \\[3ex]
C.\:\: 5 \\[3ex]
D.\:\: 10 \\[3ex]
E.\:\: 25 \\[3ex]
$

The rate of change is the slope

$ Point = (time, speed) \\[3ex] Unit = (hour, knot) \\[3ex] Point\:1 = (2, 15) \\[3ex] x_1 = 2 \\[3ex] y_1 = 15 \\[3ex] Point\:2 = (4, 25) \\[3ex] x_2 = 4 \\[3ex] y_2 = 25 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{25 - 15}{4 - 2} \\[5ex] = \dfrac{10}{2} \\[5ex] = 5\:knots\:per\:hour $

Which of the following values is closest to the rate of change, in knots per hour, of the speed of the ship between hours $2$ and $4$?

The rate of change is the slope

$ Point = (time, speed) \\[3ex] Unit = (hour, knot) \\[3ex] Point\:1 = (2, 15) \\[3ex] x_1 = 2 \\[3ex] y_1 = 15 \\[3ex] Point\:2 = (4, 25) \\[3ex] x_2 = 4 \\[3ex] y_2 = 25 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{25 - 15}{4 - 2} \\[5ex] = \dfrac{10}{2} \\[5ex] = 5\:knots\:per\:hour $

(15.) **CSEC** The amount a plumber charges for services depends on the time taken to complete the
repairs plus a fixed charge.

The graph below shows the charges in dollars $(d)$ for repairs in terms of the number of minutes $(t)$ taken to complete the repairs.

(a) What was the charge for a plumbing job which took $20$ minutes?

(b) How many minutes were spent completing repairs that cost:

(i) $\$38.00$

(ii) $\$20.00$?

(c) What is the amount of the fixed charge?

(d) Calculate the gradient of the line.

(e) Write down the equation of the line in terms of $d$ and $t$

(f) Determine the length of time taken to complete a job for which the charge was $\$78.00$

$ Based\:\:\:on\:\:the\:\:Graph \\[3ex] (a) \\[3ex] t = 20\:\:minutes \\[3ex] d = \$30.00 \\[3ex] $ The charge for a plumbing job which took $20$ minutes is $\$30.00$

$ (b) \\[3ex] (i) \\[3ex] d = \$38.00 \\[3ex] t = 35\:\:minutes \\[3ex] (ii) \\[3ex] d = \$20.00 \\[3ex] t = 0\:\:minutes \\[3ex] $ $35$ minutes were spent completing repairs that cost $\$38.00$

$0$ minutes were spent completing repairs that cost $\$20.00$

(c) The fixed charge is $\$20.00$

$ (d) \\[3ex] Point\:1:(20, 30) \\[3ex] x_1 = 20 \\[3ex] y_1 = 30 \\[3ex] Point\:2: (60, 50) \\[3ex] x_2 = 60 \\[3ex] y_2 = 50 \\[3ex] Gradient, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{50 - 30}{60 - 20} \\[5ex] m = \dfrac{20}{40} \\[5ex] m = \dfrac{1}{2}\$/minutes \\[5ex] $ The gradient of the line is $\dfrac{1}{2}$ dollars per minute

$ (e) \\[3ex] y = mx + b...Slope-Intercept\:\:Form \\[3ex] d = mt + b \\[3ex] b = 20 \\[3ex] m = \dfrac{1}{2} \\[5ex] d = \dfrac{1}{2}t + 20 \\[5ex] (f) \\[3ex] d = \dfrac{1}{2}t + 20 \\[5ex] d = \$78.00 \\[3ex] 78 = \dfrac{1}{2}t + 20 \\[5ex] LCD = 2 \\[3ex] 2(78) = 2\left(\dfrac{1}{2}t\right) + 2(20) \\[5ex] 156 = t + 40 \\[3ex] t + 40 = 156 \\[3ex] t = 156 - 40 \\[3ex] t = 116 \\[3ex] $ The time taken to complete a job for which the charge was $\$78.00$ is $116$ minutes

The graph below shows the charges in dollars $(d)$ for repairs in terms of the number of minutes $(t)$ taken to complete the repairs.

(a) What was the charge for a plumbing job which took $20$ minutes?

(b) How many minutes were spent completing repairs that cost:

(i) $\$38.00$

(ii) $\$20.00$?

(c) What is the amount of the fixed charge?

(d) Calculate the gradient of the line.

(e) Write down the equation of the line in terms of $d$ and $t$

(f) Determine the length of time taken to complete a job for which the charge was $\$78.00$

$ Based\:\:\:on\:\:the\:\:Graph \\[3ex] (a) \\[3ex] t = 20\:\:minutes \\[3ex] d = \$30.00 \\[3ex] $ The charge for a plumbing job which took $20$ minutes is $\$30.00$

$ (b) \\[3ex] (i) \\[3ex] d = \$38.00 \\[3ex] t = 35\:\:minutes \\[3ex] (ii) \\[3ex] d = \$20.00 \\[3ex] t = 0\:\:minutes \\[3ex] $ $35$ minutes were spent completing repairs that cost $\$38.00$

$0$ minutes were spent completing repairs that cost $\$20.00$

(c) The fixed charge is $\$20.00$

$ (d) \\[3ex] Point\:1:(20, 30) \\[3ex] x_1 = 20 \\[3ex] y_1 = 30 \\[3ex] Point\:2: (60, 50) \\[3ex] x_2 = 60 \\[3ex] y_2 = 50 \\[3ex] Gradient, m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] m = \dfrac{50 - 30}{60 - 20} \\[5ex] m = \dfrac{20}{40} \\[5ex] m = \dfrac{1}{2}\$/minutes \\[5ex] $ The gradient of the line is $\dfrac{1}{2}$ dollars per minute

$ (e) \\[3ex] y = mx + b...Slope-Intercept\:\:Form \\[3ex] d = mt + b \\[3ex] b = 20 \\[3ex] m = \dfrac{1}{2} \\[5ex] d = \dfrac{1}{2}t + 20 \\[5ex] (f) \\[3ex] d = \dfrac{1}{2}t + 20 \\[5ex] d = \$78.00 \\[3ex] 78 = \dfrac{1}{2}t + 20 \\[5ex] LCD = 2 \\[3ex] 2(78) = 2\left(\dfrac{1}{2}t\right) + 2(20) \\[5ex] 156 = t + 40 \\[3ex] t + 40 = 156 \\[3ex] t = 156 - 40 \\[3ex] t = 116 \\[3ex] $ The time taken to complete a job for which the charge was $\$78.00$ is $116$ minutes

(16.) **ACT** Jayla plotted the data from her science project as a scatterplot in the standard
$(x, y)$ coordinate plane.

She found the line containing $2$ of the points to be $y = 0.28x + 6$.

The scatterplot and the line are shown below.

Jayla decided that this line was not a good fit for her data.

To transform her line into the regression line for her data, Jayla*must:*

A. increase both the slope and the y-intercept.

B. increase the slope and decrease the y-intercept.

C. decrease both the slope and the y-intercept.

D. decrease the slope and the increase the y-intercept.

E. use either a horizontal or vertical line.

__Line in Slope-Intercept Form__

The slope of the line is $0.28$

The y-intercept of the line is $6$

__Regression Line__

*The y-intercept of the regression line seems to be $2$.*

In that regard, it is important to note that the y-intercept of the line in slope-intercept form must be decreased.

This eliminates Options $A$ and $D$.

Option $E$ is also incorrect because the regression line is not a horizontal line or a vertical line.

Using a horizontal or vertical line is not a good fit for the regression line.

Let us analyze the remaining options:

Option $B:$ increase the slope and decrease the y-intercept.

The regression is steeper than the slope-intercept line

This implies that it has a greater slope than the slope-intercept line

*
*__Student:__ How? May you explain?

__Teacher:__ Remember we discussed steepness here

We can also look at an example to prove what I said.

Let us find the slope of the regression line ...not by formula...but by picking any two points in that regression line

__Student:__ Point $1$ = $(1, 3)$ and Point $2$ = $(7, 10)$

__Teacher__ Okay. Find the slope

__Student:__

$ x_1 = 1 \\[3ex] y_1 = 3 \\[3ex] x_2 = 7 \\[3ex] y_2 = 10 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{10 - 3}{7 - 1} \\[5ex] = \dfrac{7}{6} \\[5ex] = 1.16666667 \\[3ex] $__Teacher:__ $1.16666667 \gt 0.28$

Because the regression line is steeper than the slope-intercept line, it has a higher slope.

Therefore, Jayla needs to increase the slope of her slope-intercept line in order for the line to fit her regression line.

Option $B$ is correct.

Option $C$ is incorrect because decreasing the slope of the slope-intercept line would cause it to be less steep. It would not fit the regression line.

She found the line containing $2$ of the points to be $y = 0.28x + 6$.

The scatterplot and the line are shown below.

Jayla decided that this line was not a good fit for her data.

To transform her line into the regression line for her data, Jayla

A. increase both the slope and the y-intercept.

B. increase the slope and decrease the y-intercept.

C. decrease both the slope and the y-intercept.

D. decrease the slope and the increase the y-intercept.

E. use either a horizontal or vertical line.

The slope of the line is $0.28$

The y-intercept of the line is $6$

*The y-intercept of the regression line seems to be $2$.*

In that regard, it is important to note that the y-intercept of the line in slope-intercept form must be decreased.

This eliminates Options $A$ and $D$.

Option $E$ is also incorrect because the regression line is not a horizontal line or a vertical line.

Using a horizontal or vertical line is not a good fit for the regression line.

Let us analyze the remaining options:

Option $B:$ increase the slope and decrease the y-intercept.

The regression is steeper than the slope-intercept line

This implies that it has a greater slope than the slope-intercept line

We can also look at an example to prove what I said.

Let us find the slope of the regression line ...not by formula...but by picking any two points in that regression line

$ x_1 = 1 \\[3ex] y_1 = 3 \\[3ex] x_2 = 7 \\[3ex] y_2 = 10 \\[3ex] m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{10 - 3}{7 - 1} \\[5ex] = \dfrac{7}{6} \\[5ex] = 1.16666667 \\[3ex] $

Because the regression line is steeper than the slope-intercept line, it has a higher slope.

Therefore, Jayla needs to increase the slope of her slope-intercept line in order for the line to fit her regression line.

Option $B$ is correct.

Option $C$ is incorrect because decreasing the slope of the slope-intercept line would cause it to be less steep. It would not fit the regression line.