SOLUTION: I attempted to try this problem for myself, yet I still got the wrong answer. So can you figure out what I did wrong? I multiplied 0.04 by 2500 and 0.7 by 20. And I still got the w

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Question 1124993: I attempted to try this problem for myself, yet I still got the wrong answer. So can you figure out what I did wrong? I multiplied 0.04 by 2500 and 0.7 by 20. And I still got the wrong answer.

For problem 121 the polynomial 0.04v^2 + 0.7v models the number of feet it takes for a car traveling v miles per hour to stop on dry, level concrete.

a) How many feet will it take a car traveling 50 miles per hour to stop on dry, level concrete?
b) How many feet will it take a car traveling 20 miles per hour through a school zone to stop on dry, level concrete?
Found 2 solutions by rothauserc, MathTherapy:
Answer by rothauserc(4718)   (Show Source): You can put this solution on YOUR website!
f(v) = 0.04v^2 + 0.7v
:
a) 50 mph stopping distance = 0.04(50)^2 + 0.7(50) = 135 feet
:
b) 20 mph stopping distance = 0.04(20)^2 + 0.7(20) = 30 feet
:
Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!
I attempted to try this problem for myself, yet I still got the wrong answer. So can you figure out what I did wrong? I multiplied 0.04 by 2500 and 0.7 by 20. And I still got the wrong answer.

For problem 121 the polynomial 0.04v^2 + 0.7v models the number of feet it takes for a car traveling v miles per hour to stop on dry, level concrete.

a) How many feet will it take a car traveling 50 miles per hour to stop on dry, level concrete?
b) How many feet will it take a car traveling 20 miles per hour through a school zone to stop on dry, level concrete?
For a), you're supposed to ADD .04 times 2,500 TO .7 times 50.

For b), you're supposed to ADD .04 times 400 TO .7 times 20.

This should gives you, 135 feet, and 30 feet, respectively.

What's so DIFFICULT about that? 

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