What are the missing values in the table representing the recursively defined geometric sequence f(x+1)=4(f(x)) x_f(x) 1_4 2_a 3_b 4_c 5_1,024

Hence, the values of unknowns A,B and C are:

A=16

B=64

C=256

We are given a recursive defined geometric sequence f(x+1)=4(f(x))

X       f(x)

1         4

2        A  

3        B

4        C

5      1,024

When x=1,

we have f(x)=4.

when so,

f(2)=4f(1)=4×4=16.

Hence, when x=2 f(x)=16=A.

Similarly,

f(3)=4×f(2)

f(3)=4×16=64

Hence, when x=3    f(x)=64=B.

Similarly,

f(4)=4×f(3)=4×64=256

i..e when x=4   f(x)=256=C.

similarly,

f(5)=4×f(4)=4×256=1024

when x=5   f(x)=1024

Hence, the values of unknowns A,B and C are:

A=16

B=64

C=256

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

example 2 – simplify: step 1: completely factor both the numerators and denominators of all fractions. step 2: change the division sign to a multiplication sign and flip (or reciprocate) the fraction after the division sign; essential you need to multiply by the reciprocal. step 3 : cancel or reduce the fractions.