QuestionThe cost $C(x)=0.20x+45$ gives truck rental costs. Find costs for $x=80$ miles, solve for $C=60$ , max miles for $C \leq 150$ , domain, slope, and intercept.

Studdy Solution

STEP 1

Assumptions1. The cost function for renting a moving truck for a day is $C(x)=0.20 x+45$ , where $x$ is the number of miles driven. . The cost function is linear, and the cost increases with the number of miles driven.
3. The cost includes a fixed cost and a variable cost that depends on the number of miles driven.

STEP 2

(a) To find the cost if a person drives $x=80$ miles, we substitute $x=80$ into the cost function $C(x)$ .
$C(80)=0.20 \cdot80+45$

STEP 3

Calculate the cost when $x=80$ .
$C(80)=0.20 \cdot80+45 = \$61$ So, the cost is $61$ dollars if a person drives $80$ miles.

STEP 4

(b) To find out how many miles the person drove if the cost of renting the moving truck is $60$ , we set the cost function $C(x)$ equal to $60$ and solve for $x$ .
$60=0.20x+45$

STEP 5

Subtract $45$ from both sides of the equation to isolate the term with $x$ .
$60-45=0.20x$

STEP 6

olve for $x$ by dividing both sides of the equation by $0.20$ .
$x=\frac{60-45}{0.20}$

STEP 7

Calculate the value of $x$ .
$x=\frac{60-45}{0.20} =75$ So, the person drove $75$ miles.

STEP 8

(c) To find the maximum number of miles the person can drive if the cost is to be no more than $150$ , we set the cost function $C(x)$ equal to $150$ and solve for $x$ .
$150=0.20x+45$

STEP 9

Subtract $45$ from both sides of the equation to isolate the term with $x$ .
$150-45=.20x$

STEP 10

olve for $x$ by dividing both sides of the equation by $0.20$ .
$x=\frac{150-45}{0.20}$

STEP 11

Calculate the value of $x$ .
$x=\frac{150-45}{0.20} =525$ So, the person can drive a maximum of $525$ miles.

STEP 12

(d) The implied domain of $C$ is all non-negative real numbers, because the number of miles driven cannot be negative. In interval notation, this is $[0, \infty)$ .

STEP 13

(e) The slope of the cost function $C(x)$ is $0.20$ . This means that for each additional mile driven, the cost increases by $0.20$ dollars. This is the variable cost per mile.

STEP 14

(f) The $y$ -intercept of the cost function $C(x)$ is $45$ . This means that even if no miles are driven (i.e., $x=0$ ), the cost is $45$ dollars. This is the fixed cost of renting the moving truck for a day.

Was this helpful?

QuestionThe cost $C(x)=0.20x+45$ gives truck rental costs. Find costs for $x=80$ miles, solve for $C=60$ , max miles for $C \leq 150$ , domain, slope, and intercept.

Studdy Solution

STEP 1

STEP 2

(a) To find the cost if a person drives $x=80$ miles, we substitute $x=80$ into the cost function $C(x)$ .
$C(80)=0.20 \cdot80+45$

STEP 3

Calculate the cost when $x=80$ .
$C(80)=0.20 \cdot80+45 = \$61$ So, the cost is $61$ dollars if a person drives $80$ miles.

STEP 4

(b) To find out how many miles the person drove if the cost of renting the moving truck is $60$ , we set the cost function $C(x)$ equal to $60$ and solve for $x$ .
$60=0.20x+45$

STEP 5

Subtract $45$ from both sides of the equation to isolate the term with $x$ .
$60-45=0.20x$

STEP 6

olve for $x$ by dividing both sides of the equation by $0.20$ .
$x=\frac{60-45}{0.20}$

STEP 7

Calculate the value of $x$ .
$x=\frac{60-45}{0.20} =75$ So, the person drove $75$ miles.

STEP 8

(c) To find the maximum number of miles the person can drive if the cost is to be no more than $150$ , we set the cost function $C(x)$ equal to $150$ and solve for $x$ .
$150=0.20x+45$

STEP 9

Subtract $45$ from both sides of the equation to isolate the term with $x$ .
$150-45=.20x$

STEP 10

olve for $x$ by dividing both sides of the equation by $0.20$ .
$x=\frac{150-45}{0.20}$

STEP 11

Calculate the value of $x$ .
$x=\frac{150-45}{0.20} =525$ So, the person can drive a maximum of $525$ miles.

STEP 12

(d) The implied domain of $C$ is all non-negative real numbers, because the number of miles driven cannot be negative. In interval notation, this is $[0, \infty)$ .

STEP 13

(e) The slope of the cost function $C(x)$ is $0.20$ . This means that for each additional mile driven, the cost increases by $0.20$ dollars. This is the variable cost per mile.

STEP 14

(f) The $y$ -intercept of the cost function $C(x)$ is $45$ . This means that even if no miles are driven (i.e., $x=0$ ), the cost is $45$ dollars. This is the fixed cost of renting the moving truck for a day.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionThe cost C(x)=0.20x+45C(x)=0.20x+45C(x)=0.20x+45 gives truck rental costs. Find costs for x=80x=80x=80 miles, solve for C=60C=60C=60, max miles for C≤150C \leq 150C≤150, domain, slope, and intercept.

Studdy Solution

STEP 1

STEP 2

(a) To find the cost if a person drives x=80x=80x=80 miles, we substitute x=80x=80x=80 into the cost function C(x)C(x)C(x).C(80)=0.20⋅80+45C(80)=0.20 \cdot80+45C(80)=0.20⋅80+45

STEP 3

Calculate the cost when x=80x=80x=80.C(80)=0.20⋅80+45=$61C(80)=0.20 \cdot80+45 = \$61C(80)=0.20⋅80+45=$61So, the cost is 616161 dollars if a person drives 808080 miles.

STEP 4

(b) To find out how many miles the person drove if the cost of renting the moving truck is 606060, we set the cost function C(x)C(x)C(x) equal to 606060 and solve for xxx.60=0.20x+4560=0.20x+4560=0.20x+45

STEP 5

Subtract 454545 from both sides of the equation to isolate the term with xxx.60−45=0.20x60-45=0.20x60−45=0.20x

STEP 6

olve for xxx by dividing both sides of the equation by 0.200.200.20.x=60−450.20x=\frac{60-45}{0.20}x=0.2060−45​

STEP 7

Calculate the value of xxx.x=60−450.20=75x=\frac{60-45}{0.20} =75x=0.2060−45​=75So, the person drove 757575 miles.

STEP 8

(c) To find the maximum number of miles the person can drive if the cost is to be no more than 150150150, we set the cost function C(x)C(x)C(x) equal to 150150150 and solve for xxx.150=0.20x+45150=0.20x+45150=0.20x+45

STEP 9

Subtract 454545 from both sides of the equation to isolate the term with xxx.150−45=.20x150-45=.20x150−45=.20x

STEP 10

olve for xxx by dividing both sides of the equation by 0.200.200.20.x=150−450.20x=\frac{150-45}{0.20}x=0.20150−45​

STEP 11

Calculate the value of xxx.x=150−450.20=525x=\frac{150-45}{0.20} =525x=0.20150−45​=525So, the person can drive a maximum of 525525525 miles.

STEP 12

(d) The implied domain of CCC is all non-negative real numbers, because the number of miles driven cannot be negative. In interval notation, this is [0,∞)[0, \infty)[0,∞).

STEP 13

(e) The slope of the cost function C(x)C(x)C(x) is 0.200.200.20. This means that for each additional mile driven, the cost increases by 0.200.200.20 dollars. This is the variable cost per mile.

STEP 14

(f) The yyy-intercept of the cost function C(x)C(x)C(x) is 454545. This means that even if no miles are driven (i.e., x=0x=0x=0), the cost is 454545 dollars. This is the fixed cost of renting the moving truck for a day.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionThe cost $C(x)=0.20x+45$ gives truck rental costs. Find costs for $x=80$ miles, solve for $C=60$ , max miles for $C \leq 150$ , domain, slope, and intercept.

(a) To find the cost if a person drives $x=80$ miles, we substitute $x=80$ into the cost function $C(x)$ .
$C(80)=0.20 \cdot80+45$

Calculate the cost when $x=80$ .
$C(80)=0.20 \cdot80+45 = \$61$ So, the cost is $61$ dollars if a person drives $80$ miles.

(b) To find out how many miles the person drove if the cost of renting the moving truck is $60$ , we set the cost function $C(x)$ equal to $60$ and solve for $x$ .
$60=0.20x+45$

Subtract $45$ from both sides of the equation to isolate the term with $x$ .
$60-45=0.20x$

olve for $x$ by dividing both sides of the equation by $0.20$ .
$x=\frac{60-45}{0.20}$

Calculate the value of $x$ .
$x=\frac{60-45}{0.20} =75$ So, the person drove $75$ miles.

(c) To find the maximum number of miles the person can drive if the cost is to be no more than $150$ , we set the cost function $C(x)$ equal to $150$ and solve for $x$ .
$150=0.20x+45$

Subtract $45$ from both sides of the equation to isolate the term with $x$ .
$150-45=.20x$

olve for $x$ by dividing both sides of the equation by $0.20$ .
$x=\frac{150-45}{0.20}$

Calculate the value of $x$ .
$x=\frac{150-45}{0.20} =525$ So, the person can drive a maximum of $525$ miles.

(d) The implied domain of $C$ is all non-negative real numbers, because the number of miles driven cannot be negative. In interval notation, this is $[0, \infty)$ .

(e) The slope of the cost function $C(x)$ is $0.20$ . This means that for each additional mile driven, the cost increases by $0.20$ dollars. This is the variable cost per mile.

(f) The $y$ -intercept of the cost function $C(x)$ is $45$ . This means that even if no miles are driven (i.e., $x=0$ ), the cost is $45$ dollars. This is the fixed cost of renting the moving truck for a day.