The height ofthe saddle ofa horse above the base ofa carousel can be modeled 4t by the equation f-rr) : 12 sin ^ r 42, where I represents seconds after the ride started. I How much time does to take for the horse to complete one cycle of motion and return to its starting height. What is the maximum height and the minimum height ofthe horse's saddle above the base ofthe carousel

(a) The time to complete 1 cycle and return is 16/3

(b) The minimum height is 30 inches and the maximum is 54 inches

Step-by-step explanation:

Given

Solving (a): Time to complete 1 cycle and return

This implies that we calculate the period.

This is calculated using:

Where:

So, we have:

Rewrite as:

Solving (b): The maximum and the minimum height

To do this, we have:

Which means:

So, the maximum and the minimum of are:

--- minimum

--- maximum

Given that:

So, the minimum height is:

The maximum is:

#15. square

step-by-step explanation:

because quadrilatural (idk how to spell it) means somthing that is not 3d or like, equal sides (or somthing like that   you know what i meanz)

hope this halps!

bai!

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.