If a ball is thrown straight up into the air with an initial velocity of 45 ft/s, it height in feet after t second is given by y=45t−16t2. Find the average velocity for the time period begining when t=2 and lasting. 0.1 seconds?

step-by-step explanation:

entering that problem on the tube might get you an answer

b. would be the answer : )

step-by-step explanation:

step 1: add -16x to both sides.

16x−4y+−16x=64+−16x  

−4y=−16x+64

step 2: divide both sides by -4.

−4y /-4 = -16x + 64/-4

answer:   y=4x−16

hope i'm right

exactly you go it

step-by-step explanation: