Question

Question about equations, please help!

Answer 1

(a) y = 0.80x + 50

(b) Plugging x = 5 into the equation from part (a), we get y = 0.80(5) + 50 = 54, so the ordered pair associated with x = 5 is (5, 54). This means that if the car is driven for 5 miles, the total charge to the renter is $54.

(c) Let y be the total charge and solve for x:

y = 0.80x + 50

187.60 = 0.80x + 50

137.60 = 0.80x

x ≈ 172

Therefore, the car must have been driven approximately 172 miles

Answer 2

(a) y = 0.8x + 50

(b) D. The ordered pair associated with the equation x = 5 is (5, 54) and it means that the charge for driving the car for 5 miles is $54.

(c) If the renter paid $187.60, the car must have been driven for 172 miles.

Explanation:

Part (a)

In the given problem, we are told that the rental car costs a flat fee of $50 plus an additional charge of $0.80 per mile driven.

Let x be the number of miles driven.

Let y be the total charge to the renter (in dollars).

We know that the total charge y will depend on the number of miles driven x, and we can express this relationship using a linear equation in the form y = mx + b, where m is the slope of the line and b is the y-intercept (the value of y when x = 0).

In this case, we know that the cost per mile is $0.80 so the slope of the line is m = 0.8, and the flat fee is $50 so the y-intercept is b = 50.

Substitute these values into the equation to get:

`y = 0.8x + 50`

`\hrulefill`

Part (b)

To find the ordered pair associated with x = 5, substitute x = 5 into the equation:

`\begin{aligned}x=5\implies y &= 0.8(5) + 50\\&=4+54\\&=54\end{aligned}`

The ordered pair associated with the equation x = 5 is (5, 54) and it means that the charge for driving the car for 5 miles is $54.

`\hrulefill`

Part (c)

To find how many miles the car was driven if the renter paid $187.60, set y = 187.60 and solve for x:

`\begin{aligned}\implies 0.8x + 50&=187.60\\0.8x + 50-50&=187.60-50\\0.8x&=137.60\\(0.8x)/(0.8)&=(137.60)/(0.8)\\x &= 172\end{aligned}`

Therefore, if the renter paid $187.60, the car must have been driven for 172 miles.