If a 10.0-ft waterwheel (see Figure 1.42) is rotating.
at a speed of 5.0 rev/min, what is the speed of the
current (in mi/h, rounded to the nearest tenth)?

Sounds like 10.0 ft refers to the height/diameter of the wheel, in which case its radius is 5.0 ft.

In one complete revolution, a point on the edge of the wheel travels 2π*(5.0 ft) = 10π ft, so the point's linear speed (and thus the current's speed) is

(5.0 rev/min) * (10π ft/rev) = 50π ft/min

Convert this to mi/h using

1 mi = 5280 ft

1 h = 60 min

==>  (50π ft/min) * (1/5280 mi/ft) * (60 min/hr) = 1.8 mi/h

y = i/2 xtnop + (3i/2 tnop+ 1)

step-by-step explanation:

the picture is the step by step explanation.