Rowing upstream for 5 mi against a strong current, your speed is 20% less than your normal rowing speed. On the return trip downstream, your speed is 20% greater than your normal rowing speed. Will you finish the whole trip in the same time as if you had rowed at normal speed throughout? How much longer than usual will your total time be?

it would be the same as rowing normal speed because you loose 20% for 5 mi and gain 20% back for 5 mi

Step-by-step explanation:

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation:

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.