Which of the following transformations are linear? a. t(f(t))=f′(t)+4f(t) t(f(t))=f′(t)+4f(t) from c[infinity] c[infinity] to c[infinity] c[infinity] b. t(x+iy)=9x−iy t(x+iy)=9x−iy from c c to c c c. t(x0,x1,x2,…)=(1,x0,x1,x2,…) t(x0,x1,x2,…)=(1,x0,x1,x2,…) from the space of infinite sequences into itself d. t(f(t))=f′(t)+6f(t)+8 t(f(t))=f′(t)+6f(t)+8 from c[infinity] c[infinity] to c[infinity] c[infinity] e. t(f(t))=f(7) t(f(t))=f(7) from p2 p2 to r r f. t(f(t))=f(−t) t(f(t))=f(−t) from p3 p3 to p3 p3

b is the answer that makes most sense

step-by-step explanation:

typically if you want , you would actually put questions here for us to you. but hey, that's just me. maybe someone will magically answer the questions for you without knowing what they are.

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.