The path of an object launched above ground can be modeled by h(t)=-3t^2+30t+4 where h(t) represents the height in feet at any given time in seconds Prior to launch, the initial height of the object is feet After launch, the object reached its maximum height of feet in seconds The height of the object above the ground 10 seconds after it was launched is feet The object will strike the ground in seconds (round answer to nearest tenth)

answer: 2x + 2 − 2 = 10 − 2; subtraction property of equality

step-by-step explanation:

a: inscribed angles

step-by-step explanation:

i reviewed this question with a peer and teacher, seems to be correct, me later

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.