Rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, a in seconds, by the given equation. Using this equation, find out
the time at which the rocket will reach its max, to the nearest 100th of a second.
y=16x^2+ +225x+79

7.03 seconds

Step-by-step explanation:

First, you need to differentiate the equation (The derivative of y represents the rocket's velocity):

y' = 32x + 225

Then, set y' to 0 and solve for x

0 = 32x + 225

x = 7.03s

answer: a) (0,0), (1,1), (2,8), and (3,27).

step-by-step explanation:

0^3=0

1^3=1

2^3=8

3^3=27

in effect, the exponential equation for this would me y=3^x. each x-coordinate cubed is giving the y coordinate, so a is your answer : )

n=9

step-by-step explanation:

7 times a number = 7n

decreased by 13 = 7n-13

same as is =

5 times the number = 5n

increased by 5 = 5n+5

7n-13 = 5n+5

2n=18

n=9