What is the 11th term of the sequence 2, 8, 32, 128 A. 262,144

B. 524,288

C. 1,048,576

D. 2,097,152

Step-by-step explanation:

Given the sequence

2, 8, 32, 128..

The sequence is a geometric sequence. The nth term of the geometric sequence is expressed as;

Tn = arⁿ⁻¹

a is the first term

r is the common ratio

n is the number of terms

From the sequence;

a = 2

r = 8/2 = 32/8 = 4

n = 11 (since we are to find the 11th sequence)

Substitute into the formula;

T₁₁ = 2(4)¹¹⁻¹

T₁₁ = 2(4)¹⁰

T₁₁ = 2(1,048,576)

T₁₁ = 2,097,152

Hence the 11th term of the sequence is 2,097,152

yes   because as the functions are graphed they eventually will intersect unless otherwise

step-by-step explanation: