1. An amateur cyclist training for a road race rode the first 20 mile
portion of his workout at a constant rate. For the nest 16 mile
cool down portion of his workout he reduced his speed by 2 miles
per hour. Each portion of the workout took the same amount of
time. Find the cyclist's rate during the first portion and his rate
during his cool down.

the cyclist's speed during the first portion is 10 mph and his speed during the cool down portion is 8 mph.

Step-by-step explanation:

Let x represent the cyclist's rate during the first portion.

The amateur cyclist training for a road race rode the first 20 mile

portion of his workout at a constant rate.

Time = distance/speed

Time taken to ride through the first portion is

20/x

For the next 16 mile cool down portion of his workout, he reduced his speed by 2 miles per hour. It meas that his speed during the cool down portion is (x - 2) mph. Time taken to ride through the cool down portion is

16/(x - 2)

Since each portion of the workout took the same time, it means that

20/x = 16/(x - 2)

Cross multiplying, it becomes

20(x - 2) = 16x

20x - 40 = 16x

20x - 16x = 40

4x = 40

x = 40/4

x = 10 mph

His speed during the cool down portion is

10 - 2 = 8 mph

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

9x