find the solution f (n) when n = 2^k, where f(n) satisfies the recurrence relation f (n) = f (n/2) 3 with f (1) = 2.

i think is 3 and 6. this took me 10 min to think.hope to and hope that my answer is brainliest.(follow me )

answer: 7. if the functions had a negative leading coefficient and is of even degree, which statement about the graph is true?

step-by-step explanation:

if e is the midpoint of bd, then bd is 14.

step-by-step explanation:

you would 18 *0.25 = 4.5