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

5.25 ounces

step-by-step explanation:

divided 3/4 and 6

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3