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3 years ago

13

If you cannot see the photo the question is A rental car charges 160.00 per week plus 0.25 per mile to rent a car. How many mile

s can you travel in one week for $317.50

1 answer:

Rzqust [24]3 years ago

8 0

Answer:

You can travel 630 miles.

Step-by-step explanation:

317.5 - 160 = 157.5

157.5 / .25 = 630

You might be interested in

1. The midpoint of KL is M(5, -7). One endpoint is K(9,-7). Find the coordinates of the other endpoint L.

kirill115 [55]

Answer:

L (1, - 7 )

Step-by-step explanation:

give endpoints (x₁, y₁ ( and (x₂, y₂ ) then the midpoint is

( $\frac{x_{1}+x_{2} }{2}$ , $\frac{y_{1}+y_{2} }{2}$ )

here (x₁, y₁ ) = K (9, - 7 ) and (x₂, y₂ ) = L (x, y )

use the midpoint formula and equate to corresponding coordinates of M

$\frac{9+x}{2}$ = 5 ( multiply both sides by 2 to clear the fraction )

9 + x = 10 ( subtract 9 from both sides )

x = 1

and

$\frac{-7+y}{2}$ = - 7 ( multiply both sides by 2 )

- 7 + y = - 14 ( add 7 to both sides )

y = - 7

Then L = (1, - 7 )

8 0

1 year ago

Ive been doing this for 20 minutes now please help!!! I will give you brainliest if you give an actual answer!! SHOW YOUR WORK S

yanalaym [24]

Answer:

v= 3

Step-by-step explanation:

-19 + v = -8 ( v - 1 )

multiply -8(v - 1)= -8v + 8

-8v + 8v and then add the v with the +8v

Subtract -19 +19 and add the +19 with the +8= 27

27 divided by 9= 3

6 0

3 years ago

Read 2 more answers

Explain how to get that answer!!

ra1l [238]

We need to simplify $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$

First lets factor $\sqrt{14x^3}$

$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$
$\sqrt{14} = \sqrt{2} \sqrt{7}$ by applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$
$\sqrt{x^3} = x^{3/2}$ By applying the radical rule $\sqrt[n]{x^m} = x^{m/n}$

So
$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$ = $\sqrt{2} \sqrt{7}x^{3/2}$

Now let's factor $\sqrt{18x}$
By applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$ ,
$\sqrt{18x} = \sqrt{18} \sqrt{x}$
$\sqrt{18} = \sqrt{2} * 3$

So $\sqrt{18x}$ = $\sqrt{2}*3 \sqrt{x}$

So $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x} }$

We know that $\sqrt[n]{x} = x^{1/n}$ so $\sqrt{x} = x^{1/2}$

We now have $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}} = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}}$

We know that $\frac{x^a}{x^b} = x^{a-b}$
So $\frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x$

We now got $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}} = \frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3}
$

We can notice that the numerator and the denominator both got √2 in a multiplication, so we can simplify them, and we get:
$\frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3} = \frac{ \sqrt{7}x }{3}$

All in All, we get $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{7}x }{3}$

Hope this helps! :D

6 0

3 years ago

Which ordered pairs in the form (x, y)(x, y) are solutions to the equation 3x−4y=213x−4y=21 ?

Natalija [7]

None of these couples are solutions, (11, 3)(11, 3); (−1, −6)(−1, −6); (−3, 3)(−3, 3); <span>(7, 0). Perhaps the choice of answer are insufficient. we can add (1, 48) the couple (7, 0) and (7, 0)(1, 48) is a true answer, why? because it verifies the equation.</span>

5 0

3 years ago

Read 2 more answers

If 3/4 and 12/x are equivalent what is the value of x <br><br>A. 48<br>B.36<br>C.34<br>D.16

ahrayia [7]

Your answer for that is

D.16

Hope this helped

5 0

3 years ago

Read 2 more answers