CPhill

Nope, saseflower....you were right the first time...it's just to the first power.....

Total Area = area of the square + area of the triangle = y^2 + (1/2)(4)(y)  =

(y^2 + 2y)cm^2

You lost me at the words, "null hypothesis".....

BJ just wants to have some......

a) and c) have the same answer, 2.25 ft

b)= 12 ft/s

d) = NO

Here's a graph subbing y for h and x  for t

HAHAHA, BJ!!!...I just hope it's correct !!! LOL!!!

I think the function should be h = -16t^2 + 32t + 5

The position of the grappling hook at time 0 = 5 ft.

And the max height it reaches is 21 ft.....but it needs to reach a height of 25ft....so, no, it won't be thrown high enough to reach the ledge....

Here's a graph that substitutes y for h and x for t....

False.....twice a year....

Notice that the top will factor as  [4 - √a] [4 + √a]

So we have

lim as  a → 16     [4 - √a] [4 + √a] / [4 - √a]  =

lim as  a → 16     [4 + √a]  =

4 + √16  =  4 + 4   = 8