Let S be the subset of the set of ordered pairs of integers defined recursively by: Base case: (0, 0) ∈ S Recursive step: If (a, b) ∈ S, then (a + 1, b + 3) ∈ S and (a + 3, b + 1) ∈ S. (1) List the elements of S produced by the first four applications of the recursive definition (this should produce 14 new elements).
(2) Use structural induction to show for all (a, b) ∈ S that (a+b).

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step-by-step explanation:

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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.

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