A rental truck company charges $25 to rent a truck, then $1.25 for every mile it is driven. The situation can be expressed as C=1.25m+25

Question

, where C
is the total cost to rent the truck and m is the miles driven. If you plot points on the linear function, which of the following would appear?(1 point)
Responses

(2,2.5)
and (4,5)
left parenthesis 2 comma 2.5 right parenthesis and left parenthesis 4 comma 5 right parenthesis

(2,22.5)
and (4,20)
left parenthesis 2,22.5 right parenthesis and left parenthesis 4 comma 20 right parenthesis

(4,30)
and (8,35)
left parenthesis 4 comma 30 right parenthesis and left parenthesis 8 comma 35 right parenthesis

(30,4)
and (35,8)

Answer 1

To determine which points correctly correspond to the given cost function \( C = 1.25m + 25 \), we need to calculate \( C \) for different values of \( m \).

1. For \( m = 2 \): \[ C = 1.25(2) + 25 = 2.5 + 25 = 27.5 \] So, the point is \( (2, 27.5) \).

2. For \( m = 4 \): \[ C = 1.25(4) + 25 = 5 + 25 = 30 \] So, the point is \( (4, 30) \).

3. For \( m = 8 \): \[ C = 1.25(8) + 25 = 10 + 25 = 35 \] So, the point is \( (8, 35) \).

Now, let's check the given options:

• (2,2.5) and (4,5): This does not match our calculations.
• (2,22.5) and (4,20): This does not match our calculations.
• (4,30) and (8,35): This matches our calculated points!
• (30,4) and (35,8): This does not match our calculations.

The correct answer is:

(4,30) and (8,35).

Answer 2

A dog boarding facility charges daily based on a dog’s weight. If the dog weighs under 15 lbs, the rate is $30. If the dog is between 15 and 50 lbs, it is $35. If the dog is 50lbs or more, the charge is $35 plus $3 for each additional 5 lbs over 50 lbs. Which of the following best defines the quantity and scale for the x-axis for a piecewise linear graph of the fee structure?(1 point)
Responses

amount; 1
amount; 1

weight; 1
weight; 1

amount; 10
amount; 10

weight; 10

Answer 3

To define the quantity and scale for the x-axis of a piecewise linear graph representing the fee structure based on a dog's weight, we need to consider the independent variable being plotted.

In this case, the x-axis will represent the weight of the dog (in pounds) because the boarding fee varies depending on the dog's weight.

The scale refers to the increments on the x-axis. Since we are dealing with weight, a scale of "1" would mean each unit on the axis represents 1 pound, which is a sensible choice for more precise representation. However, a scale of "10" would mean each unit represents 10 pounds, which might be less precise but could still be appropriate depending on how detailed you want the graph to be.

Given the options presented:

• amount; 1: This implies the x-axis measures fees.
• weight; 1: This implies the x-axis measures weight increments of 1 lb, which is appropriate.
• amount; 10: Similar issue as with the first option.
• weight; 10: This means increments of 10 lbs which could suffice but is less precise.

The best option to define the x-axis for the piecewise graph, where we want to specifically focus on the dog's weight (the basis for the fee structure), is:

weight; 1.

Answer 4

A gas station is offering a deal. If you buy 10 gallons or more of gas, you will be charged $2.25 per gallon instead of the regular $2.50 per gallon. The deal is modeled by the following piecewise function.

f(x)={2.5x if x<102.25x if x≥10

If you were to create a graph of this piecewise function, which point would have an open circle?

(1 point)
Responses

(10,25)
left parenthesis 10 comma 25 right parenthesis

(0,0)
left parenthesis 0 comma 0 right parenthesis

(10,22.5)
left parenthesis 10 comma 22.5 right parenthesis

(1,2.5)

Answer 5

To determine which point would have an open circle in the piecewise function defined as:

\[ f(x) = \begin{cases} 2.5x & \text{if } x < 10 \ 2.25x & \text{if } x \geq 10 \end{cases} \]

we should first analyze how the function behaves at \( x = 10 \).

1. For \( x < 10\), the function follows the equation \( f(x) = 2.5x \).

   • At \( x = 10 \): \[ f(10) = 2.5(10) = 25. \]

2. For \( x \geq 10\), the function follows the equation \( f(x) = 2.25x \).

   • At \( x = 10 \): \[ f(10) = 2.25(10) = 22.5. \]

Now, we will note the behavior at \( x = 10 \):

• From the left (when \( x < 10 \)), the function approaches \( 25 \).
• At \( x = 10 \), according to the second piece of the function, the value is \( 22.5 \).

Since the function is defined such that it takes the value of \( 25 \) just before \( x = 10\) but at \( x = 10\) the value is actually \( 22.5\), there will be an open circle at \( (10, 25) \) indicating that the function does not actually reach that value at that point.

Therefore, the point that would have an open circle in the graph is:

(10, 25).

Answer 6

Karen purchased a car for $40,000 by taking out a loan that will take 7 years to pay off. Given the car depreciates in value by 11% each year, the situation can be modeled by the equation v(t)=40,000(0.89)t
, where v is the value of the car after t years. If Karen wants to know how much the car will be worth in 7 years, which of the following options would be an appropriate range?

Option #1: 0<t≤40,000


Option #2: 0<v≤40,000


Option #3: 0<t≤7


Option #4: 0<v≤7


(1 point)
Option #

Answer 7

To determine the appropriate range for the value of the car \( v(t) \) after 7 years, let's look at the equation:

\[ v(t) = 40,000(0.89)^t \]

Here, \( t \) represents the number of years, and \( v(t) \) is the value of the car after \( t \) years.

1. Understanding the Variables:

   • \( v(t) \): Value of the car, which will decrease over time due to depreciation.
   • \( t \): Number of years, which can range from 0 (the time of purchase) to 7 (the time when the loan is paid off).

2. Depreciation Calculation:

   • Since \( t \) ranges from 0 to 7, we can evaluate the depreciation over those years.
   • At \( t = 0 \): \[ v(0) = 40,000(0.89)^0 = 40,000 \]
   • At \( t = 7 \): \[ v(7) = 40,000(0.89)^7 \] Calculating this value: \[ v(7) = 40,000 \times 0.5138 \approx 20,552 \]

This indicates that the value of the car after 7 years is around $20,552, and will be decreasing from its original value of $40,000.

3. Considering the Options:
   • Option #1: \( 0 < t \leq 40,000 \): This is not appropriate. The variable \( t \) should represent years, not the dollar amount.
   • Option #2: \( 0 < v \leq 40,000 \): This encompasses all possible values of the car while considering its initial worth but does not specify the value after 7 years.
   • Option #3: \( 0 < t \leq 7 \): This is appropriate for representing the years Karen is paying off the loan.
   • Option #4: \( 0 < v \leq 7 \): This is not appropriate as \( v \) represents the value of the car and can't be constrained to 7.

The most relevant option for the years until the car is fully paid off is:

Option #3: \( 0 < t \leq 7 \).

Answer 8

it was 2 not 3

Answer 9

I apologize for the confusion in the interpretation of the ranges. Let’s clarify the options more carefully.

To summarize the situation:

• Karen purchases a car for $40,000.
• The value of the car depreciates by 11% per year, which is mathematically represented by the function \( v(t) = 40,000(0.89)^t \).
• We want to determine the appropriate range for the value \( v \) of the car after \( t \) years.

From the depreciation formula:

1. At \( t = 0 \): \[ v(0) = 40,000(0.89)^0 = 40,000 \]

2. At \( t = 7 \): \[ v(7) = 40,000(0.89)^7 \approx 40,000 \times 0.5138 \approx 20,552 \]

So, the value of the car after 7 years will be approximately $20,552, indicating that it will be less than $40,000 by that time.

Now evaluating the options:

• Option #1: \( 0 < t \leq 40,000 \): Incorrect, as \( t \) should only represent years.
• Option #2: \( 0 < v \leq 40,000 \): This is correct because it covers all possible values of the car from $0 to its maximum initial price of $40,000, and it makes sense to indicate that the value will decrease but remain positive.
• Option #3: \( 0 < t \leq 7 \): Only represents the time frame but does not include the value aspect.
• Option #4: \( 0 < v \leq 7 \): Incorrect as the value \( v \) does not range to $7.

Considering this evaluation, the correct answer is indeed:

Option #2: \( 0 < v \leq 40,000 \).

Thank you for your patience, and I appreciate your understanding!

Answer 10

Use the table to answer the question.

x f(x)

0 3
1 3.6
2 4.32
3 5.18
4 ?

5 7.46
Find the missing value in the table for the exponential function f(x)=3(1.2)x
. Round your answer to two decimal places.

(1 point)
f(4)=

Answer 11

To find the missing value \( f(4) \) for the exponential function defined by

\[ f(x) = 3(1.2)^x, \]

we will substitute \( x = 4 \) into the function.

Calculating \( f(4) \):

\[ f(4) = 3(1.2)^4 \]

Now, we need to calculate \( (1.2)^4 \):

\[ (1.2)^4 = 1.2 \times 1.2 \times 1.2 \times 1.2 \]

Calculating step-by-step:

1. \( 1.2 \times 1.2 = 1.44 \)
2. \( 1.44 \times 1.2 = 1.728 \)
3. \( 1.728 \times 1.2 = 2.0736 \)

Now substituting back into the function:

\[ f(4) = 3 \times 2.0736 = 6.2208 \]

Rounding this to two decimal places:

\[ f(4) \approx 6.22 \]

So the missing value in the table for \( f(4) \) is:

f(4) = 6.22.