) At a certain instant of time, a cube has side length 2 inches and its volume is increasing at a rate of 0.5 inches cubed per hour. How fast is its surface area increasing at that instant

1 inches per cube

As required.

Step-by-step explanation:

Solution:

Given:

Volume = l3

dV / dt = 3 l2dl / dt

dv / dt = 0.5 inches cubed / hour

Surface area = 6 l2

Differentiate with respect to t: s = 6 l2

We get:

ds / dt = 12 l dl / dt

and dv / dt = = 0.5 inches cubed / hour

we know that:

dv / dt = 20 = 3 l2dl / dt

0.5 = 3 l2 (dl / dv)(dv / dt)

0.5 = 3l2 dl / dt

0.5 / 3l2 = dl / dt

Put 0.5 / 3l2 = dl / dt  in eq ds / dt = 12 l dl / dt

ds / dt = 12 l (0.5 / 3l2)

        =  4(0.5) / l

       = 2 / 2 inches pr cube    [ l = 2 inches per cube]

    = 1 inches per cube

As required.

susie has 7 apples

b. y = 4x - 2

step-by-step explanation:

to find a parallel equation to an equation, make sure both equations have the same slope but different y-intercepts.

y = 4x + 1

4 = slope

1 = y-intercept

y = 4x - 2

4 = slope

-2 = y-intercept

they both have the same slope but different y-intercepts, so, b. is the correct answer.

and to check your work, you can graph both equations. : )