Starting from 300 feet away, a car drives toward you. it then passes by you at a speed of 48 feet per second. the distance dd (in feet) of the car from you after tt seconds is given by the equation d=|300−48t|d=|300−48t|. at what times is the car 60 feet from you?

d = |300 -48t|

t when d=60

Put in the value for d and solve.

60 = |300 -48t|

This resolves to two equations:

First

-60 = 300 -48t . . . . the absolute value changes the sign of its content

0 = 360 -48t . . . . . add 60

0 = 7.5 -t . . . . . . . . divide by 48

t = 7.5

Second

60 = 300 -48t . . . . . the absolute value leaves the sign of its content unchanged

0 = 240 - 48t . . . . . . subtract 60

0 = 5 - t . . . . . . . . . . divide by 48

t = 5

Summary

The car is 60 feet from you at times t=5 seconds and t=7.5 seconds.

(2,9)

i think thats the answer i checked it on google

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step-by-step explanation:

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the expression is   or