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g100num [7]
1 year ago
8

A truck can be rented from Company A for ​$40 a day plus ​$0.40 per mile. Company B charges ​$20 a day plus ​$0.50 per mile to r

ent the same truck. Find the number of miles in a day at which the rental costs for Company A and Company B are the same.
Mathematics
1 answer:
torisob [31]1 year ago
6 0

The number of miles in day at which the rental cost for the company A and Company B are the same is 200 miles

The Company A rented the truck for $40 a day plus $0.40 per miles

The expression will be

40+0.40x

Where x  is the number of miles

The company B rented the truck for $20 a day plus $0.50 per miles

20+0.50x

To find the number of miles in a day at which the rental costs for Company A and Company B are the same, the linear equation will be

40+0.40x = 20+0.50x

40-20 = 0.50x-0.40x

0.1x = 20

x = 200 miles

Hence, the number of miles in day at which the rental cost for the company A and Company B are the same is 200 miles

Learn more about linear equation here

brainly.com/question/13738061

#SPJ1

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Answer:

y=-\dfrac{5}{3}x+6.5

Step-by-step explanation:

Given: trapezoid RSTU with vertices R(-1, 5), S(1, 8), T(7, -2), and U(2, 0).

Plot these points on the coordinate plane. As you can see, lines RU and ST are parallel lines and segments ST and RU are bases of the trapezoid.

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M\left(\dfrac{x_S+x_R}{2},\dfrac{y_S+y_R}{2}\right)\Rightarrow M(0,6.5)\\ \\N\left(\dfrac{x_T+x_U}{2},\dfrac{y_T+y_U}{2}\right)\Rightarrow N(4.5,-1)

The equation of the line MN is

\dfrac{x-0}{4.5-0}=\dfrac{y-6.5}{-1-6.5}\\ \\\dfrac{x}{4.5}=\dfrac{y-6.5}{-7.5}\\ \\y=-\dfrac{5}{3}x+6.5

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