Which of the following steps were applied to AABC obtain AA'B'C' ? 12
10
8
B'
6
C'
A
2+
+
-6
-4
-2
N-
8
10
-2+
4
O A. Shifted 4 units left and 2 units down
O B. Shifted 3 units left and 2 units down
O C. Shifted 4 units left and 3 units down
O D. Shifted 3 units left and 3 units down

1/3

step-by-step explanation:

to find this, you can divide 12 by 4 which is 3 or you can multiply 4 until you reach 12 (4x3=12). both should give you 3, meaning theres three times less bananas than there are oranges.

r = 78539458390583095803458309458039485093485 or just 3

step-by-step explanation:

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.