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Sally drove a distance of 520 mile. Had she averaged 15mph faster, the trip would of taken 2.4 hours less. find the average speed that she drove.
Sally drove a distance of 520 mile. Had she averaged 15mph faster, the trip would of taken 2.4 hours less. find the average speed that she drove.
Answers
Answered by
Reiny
actual trip:
speed x mph
distance = 520 miles
time = 520/x
imaginary trip:
speed = x+15
distance = 520
time = 520/(x+15)
520/x - 520/(x+15) = 2.4
times x(x+15)
520(x+15) - 520x = 2.4x^2 + 36x
520x + 7800 - 520x = 2.4x^2 + 36x
2.4x^2 + 36x - 7800 = 0
divide by 2.4
x^2 + 15x - 3250 = 0
(x - 50)(x + 65) = 0
x = 50 or a negative
She went 50 mph
check:
at 50 mph, time to go 520 mi = 10.4 hrs
at 65 mph, time to go 520 mi = 8 hrs
which is a difference of 2.4 hours, YEAHH
speed x mph
distance = 520 miles
time = 520/x
imaginary trip:
speed = x+15
distance = 520
time = 520/(x+15)
520/x - 520/(x+15) = 2.4
times x(x+15)
520(x+15) - 520x = 2.4x^2 + 36x
520x + 7800 - 520x = 2.4x^2 + 36x
2.4x^2 + 36x - 7800 = 0
divide by 2.4
x^2 + 15x - 3250 = 0
(x - 50)(x + 65) = 0
x = 50 or a negative
She went 50 mph
check:
at 50 mph, time to go 520 mi = 10.4 hrs
at 65 mph, time to go 520 mi = 8 hrs
which is a difference of 2.4 hours, YEAHH
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