The projectile motion of an object can be modeled using
s(t) = gt2 + v0t + s0, where g is the acceleration due to gravity, t is the time in seconds since launch, s(t) is the height after t seconds, v0 is the initial velocity, and s0 is the initial height. The acceleration due to gravity is –4.9 m/s2.
A rocket is launched from the ground at an initial velocity of 39.2 meters per second. Which equation can be used to model the height of the rocket after t seconds?

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.

step-by-step explanation:

well if you round, things become much more simple. round 5.95 to 6, 4.25 to 4, and 1.05 to 1. now, add 6+4+1 and that would equal 11. so the items together would equal to about $11.