Question 1

Solve.
-A vendor has learned that, by pricing carmel apples at $\$ 1.25$, sales will reach 91 carmel apples per day. Raising the price to $\$ 2.25$ will cause the sales to fall to 43 carmel apples per day. Let $y$ be the number of carmel apples the vendor sells at $x$ dollars each. Write a linear equation that relates the number sold $y$ to the price $x$.
A) $y = - 48 x - 151$
B) $y = 48 x + 31$
C) $y = - 48 x + 151$
D) $y = - \frac { 1 } { 48 } x + \frac { 17467 } { 192 }$

Accepted Answer

The relationship between the price $x$ and the number of carmel apples sold $y$ is linear, and we can use the two given points to find the slope: $\frac{43 - 91}{2.25 - 1.25} = -48$. Then, using one of the points, say $x = 1.25, y = 91$, we can find the y-intercept: $91 = -48(1.25) + b$, solving for $b$ gives $b = 151$. Thus, the equation is $y = -48x + 151$.

Question 2

Solve.
-An investment is worth $\$ 3358$ in 2006. By 2011 it has grown to $\$ 6193$. Let y be the value of the investment in the year $x$, where $x = 0$ represents 2006 . Write a linear equation that relates value $y$ to year x.
A) $y = \frac { 1 } { 567 } x + 3358$
B) $y = - 567 x + 9028$
C) $y = - 567 x + 3358$
D) $y = 567 x + 3358$

Accepted Answer

The increase in investment value from 2006 to 2011 (5 years) is $6193 - $3358 = $2835. The annual increase is $2835 / 5 = $567. Thus, the equation is $y = 567x + 3358$, where $x = 0$ in 2006.

Question 3

Solve.
-A faucet is used to add water to a large bottle that already contained some water. After it has been filling for 5 seconds, the gauge on the bottle indicates that it contains 14 ounces of water. After it has been filling for 12 seconds, the gauge indicates the bottle contains 28 ounces of water. Let $y$ be the amount of water in the bottle $x$ seconds after the faucet was turned on. Write a linear equation that relates amount $y$ to time $x$.
A) $y = 2 x + 16$
B) $y = \frac { 1 } { 2 } x + \frac { 23 } { 2 }$
C) $y = - 2 x + 24$
D) $y = 2 x + 4$

Accepted Answer

The rate of water flow is $(28 - 14) / (12 - 5) = 14 / 7 = 2$ ounces per second. Using the point (5, 14), the equation is $y = 2x + b$. Substituting $x = 5$ and $y = 14$ gives $14 = 2(5) + b$, solving for $b$ gives $b = 4$. Thus, the equation is $y = 2x + 4$.

Question 4

Find an equation of the line graphed. Write the equation using function notation.
-
A) $f ( x ) = 2 x - 9$
B) $f ( x ) = 2 x + 9$
C) $f ( x ) = - 2 x - 9$
D) $f ( x ) = - 2 x + 9$

Accepted Answer

The answer of Find an equation of the line graphed....

Question 5

Solve.
-When making a telephone call using a calling card, a call lasting 4 minutes costs $\$ 2.10$. A call lasting 11 minutes costs $\$ 4.90$. Let $y$ be the cost of making a call lasting $x$ minutes using a calling card. Write a linear equation that relates cost $y$ to time $x$.
A) $y = 0.4 x - 6.1$
B) $y = 0.4 x + 0.5$
C) $y = - 0.4 x + 3.7$
D) $y = \frac { 5 } { 2 } x - \frac { 79 } { 10 }$

Accepted Answer

To find the linear equation, we first calculate the slope ($m$) using the given points (4, 2.10) and (11, 4.90). The slope $m = \frac{4.90 - 2.10}{11 - 4} = \frac{2.80}{7} = 0.4$. Then, using one of the points and the slope, we can find the y-intercept ($b$) by plugging into the equation $y = mx + b$. Using the point (4, 2.10), $2.10 = 0.4(4) + b$, solving for $b$ gives $b = 0.5$. Thus, the equation is $y = 0.4x + 0.5$.

Question 6

Find an equation of the line graphed. Write the equation using function notation.
-
A) $f ( x ) = - \frac { 7 } { 4 } x + 7$
B) $f ( x ) = \frac { 7 } { 4 } x + 7$
C) $f ( x ) = - \frac { 7 } { 4 } x - 7$
D) $f ( x ) = \frac { 7 } { 4 } x - 7$

Accepted Answer

The answer of Find an equation of the line graphed....

Question 7

Solve.
-The total sales made by a salesperson was $25,000 after 3 months and $68,000 after 23 months. Find an equation relating time and sales. Use the equation to predict the total sales after 42 months.
A) $108,850
B) $108,820
C) $108,892
D) $108,950

Accepted Answer

The sales can be modeled by a linear equation. The slope is calculated as (68000 - 25000) / (23 - 3) = 2150. Using the point (3, 25000), the equation is Sales = 2150 * Time + 18,550. For 42 months, Sales = 2150 * 42 + 18,550 = $108,850.

Question 8

Find an equation of the line graphed. Write the equation using function notation.
-
A) $f ( x ) = - 5 x - 2$
B) $f ( x ) = 5 x - 2$
C) $f ( x ) = 5 x + 2$
D) $f ( x ) = - 5 x + 2$

Accepted Answer

The answer of Find an equation of the line graphed....

Question 9

Find the equation of the line. Write the equation using standard notation.
-Through $( 2,2 )$; perpendicular to the line $2 x - 7 y = 2$
A) $2 x + 7 y = 18$
B) $7 x - 2 y = 18$
C) $2 x - 7 y = 18$
D) $7 x + 2 y = 18$

Accepted Answer

The slope of the line $2x - 7y = 2$ is $\frac{2}{7}$. A line perpendicular to it will have a slope of $-\frac{7}{2}$. Using the point-slope form with point $(2,2)$, the equation is $y - 2 = -\frac{7}{2}(x - 2)$. Converting to standard form gives $7x + 2y = 18$.

Question 10

Solve.
-In a certain city, the cost of a taxi ride is computed as follows: There is a fixed charge of $\$ 2.35$ as soon as you get in the taxi, to which a charge of $\$ 2.05$ per mile is added. Find an equation that can be used to determine the cost of an $x$-mile taxi ride. Use this equation to find the cost of a 8-mile taxi ride.
A) $\$ 18.75$
B) $\$ 18.63$
C) $\$ 19.65$
D) $\$ 18.93$

Accepted Answer

The cost of a taxi ride is computed as follows: There is a fixed charge of $2.35 as soon as you get in the taxi, to which a charge of $2.05 per mile is added. So, the cost of an x-mile taxi ride is given by the equation:Cost = 2.35 + 2.05xFor an 8-mile taxi ride, the cost would be:Cost = 2.35 + 2.05(8) = 2.35 + 16.40 = $18.75

Question 11

Solve.
-In 2010 a certain county recovered 27% of its municipal wastes through recycling, up from 17% in 2005. Find a linear equation that fits this data. Use this equation to predict the percentage recycled
In 2021.
A) 45.4%
B) 47.1%
C) 50.7%
D) 49%

Accepted Answer

The answer of Solve.
-In 2010 a certain county recovered 27%...

Question 12

Solve.
-In 2005, John invested $19,000 in the stock market. By 2010 his investment had grown to $23,700. Find an equation relating time and the value of the investment. If the market continues to grow at
The same rate, how much will be in his account in 2017? Round to the nearest dollar.
A) $28,400
B) $30,280
C) $25,581
D) $23,701

Accepted Answer

The investment grows linearly from $19,000 in 2005 to $23,700 in 2010, which is a growth of $4,700 over 5 years. The annual growth rate is $4,700 / 5 = $940 per year. The equation for the investment value V at time t years after 2005 is V = 19,000 + 940t. In 2017, t = 12, so V = 19,000 + 940*12 = $30,280.

Question 13

Find the equation of the line. Write the equation using standard notation.
-Through $( - 12,10 )$; parallel to $\mathrm { y } = 3$
A) $x = 10$
B) $\mathrm { y } = 10$
C) $x = - 12$
D) $\mathrm { y } = - 12$

Accepted Answer

A line parallel to $y = 3$ is a horizontal line. Since it passes through $(-12, 10)$, the equation is $y = 10$.

Question 14

Find the equation of the line. Write the equation using standard notation.
-Through $( 12,8 )$; perpendicular to $x = 3$
A) $x = 8$
B) $y = 8$
C) $x = 12$
D) $y = 12$

Accepted Answer

The answer of Find the equation of the line. Write...

Question 15

Find the equation of the line. Write the equation using standard notation.
-Through $( 10 , - 10 )$; perpendicular to $\mathrm { y } = 8$
A) $y = 10$
B) $x = 10$
C) $x = - 10$
D) $y = - 10$

Accepted Answer

The given line $y = 8$ is horizontal. A line perpendicular to a horizontal line is vertical. A vertical line passing through $(10, -10)$ has the equation $x = 10$, which is option B.

Question 16

Solve.
-The average value of a certain type of automobile was $\$ 20,640$ in 2006 and depreciated to $\$ 11,340$ in 2010 . Let $y$ be the average value of the automobile in the year $x$, where $x = 0$ represents 2006 . Write a linear equation that relates value $\mathrm { y }$ to year $\mathrm { x }$.
A) $y = - 2325 x + 2040$
B) $y = - 2325 x + 11,340$
C) $y = - \frac { 1 } { 2325 } x - 11340$
D) $y = - 2325 x + 20,640$

Accepted Answer

The depreciation from 2006 to 2010 is $20,640 - 11,340 = 9,300$. This occurred over 4 years, so the annual depreciation is $9,300 / 4 = 2,325$. The equation is $y = -2325x + 20,640$, where $x = 0$ corresponds to the year 2006, and $y$ is the value of the automobile.

Question 17

Find the equation of the line. Write the equation using standard notation.
-Through $( - 9 , - 1 )$; parallel to $x = 8$
A) $\mathrm { y } = - 1$
B) $x = - 1$
C) $x = - 9$
D) $y = - 9$

Accepted Answer

The line parallel to $x = 8$ is also a vertical line, which means its equation will have the form $x = k$, where $k$ is a constant. Since it passes through $(-9, -1)$, the equation is $x = -9$.

Question 18

Solve.
-Marty's Tee Shirt \& Jacket Company is to produce a new line of jackets with a embroidery of a Great Pyrenees dog on the front. There are fixed costs of $\$ 530$ to set up for production, and variable costs of $\$ 35$ per jacket. Write an equation that can be used to determine the total cost encountered by Marty's Company in producing $x$ jackets. Use the equation to find the total cost of producing 94 jackets.
A) $\$ 3800$
B) $\$ 3812$
C) $\$ 3832$
D) $\$ 3820$

Accepted Answer

The answer of Solve.
-Marty's Tee Shirt \& Jacket Company is...

Question 19

Solve.
-A gas station sells 4820 gallons of regular unleaded gasoline in a day when they charge $\$ 1.35$ per gallon, whereas they sell 3922 gallons on a day that they charge $\$ 1.40$ per gallon. Find a linear equation that relates gallons sold to price. Use this equation to predict the number of gallons sold at a price of $\$ 1.23$ per gallon. Round to the nearest tenth when appropriate.
A) $6979.3 \mathrm { gal }$
B) $6975.2 \mathrm { gal }$
C) $6984.2 \mathrm { gal }$
D) $6971.9 \mathrm { gal }$

Accepted Answer

The answer of Solve.
-A gas station sells 4820 gallons of...

Question 20

Solve.
-Persons taking a 30-hour review course to prepare for a standardized exam average a score of 620 on that exam. Persons taking a 70-hour review course average a score of 800 . Find a linear equation which fits this data, and which relates score to review time. Use this equation to predict an average score for persons taking a 46-hour review course. Round to the nearest tenth.
A) $684.7$
B) $692.0$
C) $706.0$
D) $696.2$

Accepted Answer

The answer of Solve.
-Persons taking a 30-hour review course to...

Quiz 8: More on Functions and Graphs

Solve. -An investment is worth $\$ 3358$ in 2006. By 2011 it has grown to $\$ 6193$ . Let y be the value of the investment in the year $x$ , where $x = 0$ represents 2006 . Write a linear equation that relates value $y$ to year x.

Find an equation of the line graphed. Write the equation using function notation. -

Solve. -When making a telephone call using a calling card, a call lasting 4 minutes costs $\$ 2.10$ . A call lasting 11 minutes costs $\$ 4.90$ . Let $y$ be the cost of making a call lasting $x$ minutes using a calling card. Write a linear equation that relates cost $y$ to time $x$ .

Solve. -The total sales made by a salesperson was $25,000 after 3 months and $68,000 after 23 months. Find an equation relating time and sales. Use the equation to predict the total sales after 42 months.

Find an equation of the line graphed. Write the equation using function notation. -

Find the equation of the line. Write the equation using standard notation. -Through $( 2,2 )$ ; perpendicular to the line $2 x - 7 y = 2$

Solve. -In 2010 a certain county recovered 27% of its municipal wastes through recycling, up from 17% in 2005. Find a linear equation that fits this data. Use this equation to predict the percentage recycled In 2021.

Solve. -In 2005, John invested $19,000 in the stock market. By 2010 his investment had grown to $23,700. Find an equation relating time and the value of the investment. If the market continues to grow at The same rate, how much will be in his account in 2017? Round to the nearest dollar.

Find the equation of the line. Write the equation using standard notation. -Through $( - 12,10 )$ ; parallel to $\mathrm { y } = 3$

Find the equation of the line. Write the equation using standard notation. -Through $( 12,8 )$ ; perpendicular to $x = 3$

Find the equation of the line. Write the equation using standard notation. -Through $( 10 , - 10 )$ ; perpendicular to $\mathrm { y } = 8$

Find the equation of the line. Write the equation using standard notation. -Through $( - 9 , - 1 )$ ; parallel to $x = 8$

Quiz 8: More on Functions and Graphs

Solve. -An investment is worth $3358\$ 3358$3358 in 2006. By 2011 it has grown to $6193\$ 6193$6193 . Let y be the value of the investment in the year xxx , where x=0x = 0x=0 represents 2006 . Write a linear equation that relates value yyy to year x.

Find an equation of the line graphed. Write the equation using function notation. -

Find an equation of the line graphed. Write the equation using function notation. -

Solve. -The total sales made by a salesperson was $25,000 after 3 months and $68,000 after 23 months. Find an equation relating time and sales. Use the equation to predict the total sales after 42 months.

Find an equation of the line graphed. Write the equation using function notation. -

Find the equation of the line. Write the equation using standard notation. -Through (2,2) ( 2,2 ) (2,2) ; perpendicular to the line 2x−7y=22 x - 7 y = 22x−7y=2

Solve. -In 2010 a certain county recovered 27% of its municipal wastes through recycling, up from 17% in 2005. Find a linear equation that fits this data. Use this equation to predict the percentage recycled In 2021.

Solve. -In 2005, John invested $19,000 in the stock market. By 2010 his investment had grown to $23,700. Find an equation relating time and the value of the investment. If the market continues to grow at The same rate, how much will be in his account in 2017? Round to the nearest dollar.

Find the equation of the line. Write the equation using standard notation. -Through (−12,10) ( - 12,10 ) (−12,10) ; parallel to y=3\mathrm { y } = 3y=3

Find the equation of the line. Write the equation using standard notation. -Through (12,8) ( 12,8 ) (12,8) ; perpendicular to x=3x = 3x=3

Find the equation of the line. Write the equation using standard notation. -Through (10,−10) ( 10 , - 10 ) (10,−10) ; perpendicular to y=8\mathrm { y } = 8y=8

Find the equation of the line. Write the equation using standard notation. -Through (−9,−1) ( - 9 , - 1 ) (−9,−1) ; parallel to x=8x = 8x=8

Solve. -An investment is worth $\$ 3358$ in 2006. By 2011 it has grown to $\$ 6193$ . Let y be the value of the investment in the year $x$ , where $x = 0$ represents 2006 . Write a linear equation that relates value $y$ to year x.

Find the equation of the line. Write the equation using standard notation. -Through $( 2,2 )$ ; perpendicular to the line $2 x - 7 y = 2$

Find the equation of the line. Write the equation using standard notation. -Through $( - 12,10 )$ ; parallel to $\mathrm { y } = 3$

Find the equation of the line. Write the equation using standard notation. -Through $( 12,8 )$ ; perpendicular to $x = 3$

Find the equation of the line. Write the equation using standard notation. -Through $( 10 , - 10 )$ ; perpendicular to $\mathrm { y } = 8$

Find the equation of the line. Write the equation using standard notation. -Through $( - 9 , - 1 )$ ; parallel to $x = 8$