A river is flowing at the speed of 2 miles per hour. A kayaker moves 528 feet upstream in 20 minutes. At this rate, how fast would the kayaker be paddling in calm water?

D=rt

r = (boatrate+riverrate) = (x+2) mph

d=528ft=528/5280mi=1/10mi

t=1/3hr

remember to keep the same units

so

1/10 = (x+2) (1/3)

times both sides by 3

3/10=x+2

minus 2 or 20/10

-17/10=x

-1.7=x

it would be going - 1.7mph (means going backwards)