Ask question

Ask question

All categories

3 years ago

13

Paul Kim estimates that he will drive about 10,000 miles during the year and will have $3,460 in annual fixed costs. He projects

a cost per mile of $0.54 or less for his truck. What is the maximum annual variable cost he can have to reach his projection?

1 answer:

Mama L [17]3 years ago

5 0

Answer:

The maximum annual variable cost he can have to reach his projection is $1,940

Step-by-step explanation:

Given;

Number of miles drive per year N = 10,000 miles

Total annual Fixed cost F = $3,460

cost per mile(rate) r = $0.54 or less

Total cost = fixed cost + variable cost

Total cost = cost per mile × number of miles

Total cost = r × N = $0.54 × 10,000 = $5,400

Let V represent the total variable cost per year;

F + V ≤ r × N

Substituting the values;

3,460 + V ≤ 5,400

V ≤ 5,400 - 3,460

V ≤ 1,940

The maximum annual variable cost he can have to reach his projection is $1,940

You might be interested in

What is the nth term of the linear sequence shown below? -9, -5, -1, 3, 7.

frutty [35]

Answer:

4n - 13

Step-by-step explanation:

There's a +4 between each number in the sequence.

4n

The previous number before the first given (-9) is -13.

4n - 13

6 0

2 years ago

The indicated function y1(x) is a solution of the given differential equation. use reduction of order or formula (5) in section

Ber [7]

Given that $y_1=e^{2x/3}$ , we can use reduction of order to find a solution $y_2=v(x)y_1=ve^{2x/3}$ .

$\implies {y_2}'=\dfrac23ve^{2x/3}+v'e^{2x/3}=\left(\dfrac23v+v'\right)e^{2x/3}$
$\implies{y_2}''=\dfrac23\left(\dfrac23v+v'\right)e^{2x/3}+v''e^{2x/3}=\left(\dfrac49v+v'+v''\right)e^{2x/3}$

$\implies9y''-12y'+4y=0$
$\implies 9\left(\dfrac49v+v'+v''\right)e^{2x/3}-12\left(\dfrac23v+v'\right)e^{2x/3}+4ve^{2x/3}=0$
$\implies9v''-3v'=0$

Let $u=v'$ , so that

$9u'-3u=0\implies 3u'-u=0\implies u'-\dfrac13u=0$
$e^{-x/3}u'-\dfrac13e^{-x/3}u=0$
$\left(e^{-x/3}u\right)'=0$
$e^{-x/3}u=C_1$
$u=C_1e^{x/3}$

$\implies v'=C_1e^{x/3}$
$\implies v=3C_1e^{x/3}+C_2$

$\implies y_2=\left(3C_1e^{x/3}+C_2\right)e^{2x/3}$
$\implies y_2=3C_1e^x+C_2e^{2x/3}$

Since $y_1$ already accounts for the $e^{2x/3}$ term, we end up with

$y_2=e^x$

as the remaining fundamental solution to the ODE.

7 0

3 years ago

Need help on this please giving brainlest! :)

masha68 [24]

A is 9,4 B is -9,4 all you do is flip the sign according to the axis

7 0

3 years ago

Factor -1/2out of -1/2x+6

Volgvan

$-\dfrac{1}{2}x+6=-\dfrac{1}{2}\left(\dfrac{-\frac{1}{2}x}{-\frac{1}{2}}+\dfrac{6}{-\frac{1}{2}}\right)=-\dfrac{1}{2}\left(x-\dfrac{6\cdot2}{1}\right)=\boxed{-\dfrac{1}{2}\left(x-12\right)}$

3 0

4 years ago

Is 6+2 +4x equivalent to 6- ( 2-4x) . if not , why not ?

Xelga [282]

Well lets solve for both.
6 + 2 + 4x
Combine like terms.
8 + 4x
Now solve the other one.
6 - (2 - 4x)
Distribute.
6 - 2 + 4x
Combine like terms.
4 + 4x.
These equations are not equivalent because instead of adding two, like in the first one, the second problem is subtracting. So you get 8 + 4x and 4 + 4x. I hope this helps love! :)

6 0

4 years ago