Question

Starting from 1.5 miles away, a car drive toward a speed checkpoint and then passes it. The car travels at a constant rate of 53 miles per hour. The distance of the car from the checkpoint is given by d=l.5-53tl. At what times is the car 0.1 miles from the check point? calculate your answer in seconds.

A. 95.1 s and 108.7 s
B. 10.2 s and 101.9 s
C. 108.7 s and 10.2 s
D. 95.1 s and 10.2 s

Answer 1

@kewlgeek555
@aligallegos
@Euler271

Answer 2

I can't help right now, but I'll call @agent0smith for you. He is like the second best. ;]

Answer 3

thanks!

Answer 4

I'm guessing those l's are absolute value signs\[\large d=|.5-53t|\]
"At what times is the car 0.1 miles from the check point? "

plug in 0.1 for d, then try to solve for t

Answer 5

yes, they are absolute value signs.

Answer 6

What? Sorry, i don't understand

Answer 7

plug in 0.1 for d, then try to solve for t

\[\Large 0.1=|.5−53t|\]

Answer 8

when you drop absolute value signs you have to solve two equations, like this \[\Large |x| = a\]then \[\Large x = a\]and \[\Large x = -a\]

Answer 9

so what'll you get when you drop them\[\Large |.5−53t|=0.1\]

Answer 10

i don't understand.

Answer 11

what is the answer?

Answer 12

Do you understand how to solve absolute value equations? Like any idea at all how to solve this \[\large |x| =2\]

Answer 13

x= -2 or 2

Answer 14

Then do this the same way \[\Large |.5−53t|=0.1\]

Answer 15

"what is the answer?" The point is for you to learn how to find the answer. I won't give it to you.