Find the first five terms of the sequence in which a(1) = –10 and a(n) = 4a(n) – 1 + 7, if n ≥ 2.
options:
a –33, –125, –493, –1965, –7853
b 11, 15, 19, 23, 27
c 47, –10, –33, –125, –493
d –10, –33, –125, –493, –1965

hello : A

Step-by-step explanation:

a(1) = –10 and a(n) = 4a(n– 1) + 7, if n ≥ 2.

n=2 : a(2)=4a(1)+7        a(2) =4(-10)+7 = -33

n=3 : a(2)=4a(2)+7        a(3) =4(-33)+7 = -125

n=4 : a(2)=4a(3)+7        a(4) =4(-125)+7 = -493

n=5 : a(2)=4a(4)+7        a(5) =4(-493)+7 = -1965

n=6 : a(2)=4a(5)+7        a(6) =4(-1965)+7 = -7853

area = length x width

length = 2 x width

32 = 2 x width x width

32 = 2 x width^2

36 = width ^2  

width = 6

length = 12

area is 32

perimeter = 2 x 12 + 2 x 6 = 24 + 12 = 36

step-by-step explanation:

$450 is the answer.

step-by-step explanation:

30% of 450 = 135

450-135=315