Answer:
a_n = 2^(n - 1) 3^(3 - n)
Step-by-step explanation:
9,6,4,8/3,…
a1 = 3^2
a2 = 3 * 2
a3 = 2^2
As we can see, the 3 ^x is decreasing and the 2^ y is increasing
We need to play with the exponent in terms of n
Lets look at the exponent for the base of 2
a1 = 3^2 2^0
a2 = 3^1 2^1
a3 = 3^ 0 2^2
an = 3^ 2^(n-1)
I picked n-1 because that is where it starts 0
n = 1 (1-1) =0
n=2 (2-1) =1
n=3 (3-1) =2
Now we need to figure out the exponent for the 3 base
I will pick (3-n)
n =1 (3-1) =2
n =2 (3-2) =1
n=3 (3-3) =0
Step-by-step explanation:
y = 2x
y = x + 1
y-1 = x
y = 2x
y = 2 (y - 1)
y = 2y - 1
y = -1
I believe it would be the second option....
I hope this hepls you
f⁻¹(x)=3x+7
f⁻¹(4)=3.4+7=19
$45+$18+$26-$21-$93
45+18=63
63+18=81
81+26=107
107-21=86
86-93=-7
