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egoroff_w [7]
3 years ago
13

The monthly cost of driving a car depends on the number of miles driven. Lynn found that in May her driving cost was $380 for 48

0 mi and in June her cost was $460 for 800 mi. Assume that there is a linear relationship between the monthly cost C of driving a car and the distance x driven. (a) Find a linear function C that models the cost of driving x miles per month. (b) Draw a graph of C. What is the slope of this line? (c) At what rate does Lynn's cost increase for every additional mile she drives?

Mathematics
1 answer:
dlinn [17]3 years ago
5 0

Answer:

a) see attachment

b) Cost f(distance) = 0.25d + 260

c) $0.25

Step-by-step explanation:

b) The linear relationship between two variables is as follows:

C f(d) = m*d + k .... Eq 1

Where, m and k are constants to be evaluated

Find m: we will find slope of the line between two points (480,$380) & (800,$460)

Slope of line (m) = \frac{460-380}{800-480} = $0.25 / mile

Find k: consider one of the two points and input in Eq 1 using (480,$380)

k = 380 - 0.25(480) = $260

The equation of line is C f(d) = 0.25*d +260.

We can use this equation for plotting the graph for part a

c) Rate of cost / mile is given by the slope of the graph if scruutinized on the units; hence, $0.25

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Answer:

The all polar coordinates of P are:

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Step-by-step explanation:

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# Ex: the following four points are all coordinates for the same point.

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∵ P = (3 , -π/3)

∵ (r , θ + 2πn)

∴ r = 3 an d Ф = -π/3

- let n = 1

∴ P = (3 , -π/3 + 2π)

∴ P = (3 , 5π/3)

∵ P =(3 , -π/3)

∵ P = (-r , θ + (2n + 1) π)

Let n = 0

∴ P = (-3 , -π/3 + (2×0 + 1) π)

∴ P = (-3 , -π/3 + (0 + 1) π)

∴ P = (-3 , -π/3 + π)

∴ P = (-3 , 2π/3)

∵ P =(3 , -π/3)

∵ P = (-r , θ + (2n + 1) π)

Let n = -1

∴ P = (-3 , -π/3 + (2(-1) + 1) π)

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∴ P = (-3 , -π/3 + -π) = (-3 , -4π/3)

∴ P = (-3 , -4π/3)

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