Question

VERY CONFUSED!! PLEASE HELP!!

For j(x) = 4x − 2, find j of the quantity x plus h end quantity minus j of x all over h period

4 to the power of the quantity x minus 2 end quantity times the quantity 4 to the power of h end quantity all over h

4 to the power of the quantity x minus 2 end quantity times the quantity 4 to the power of h plus 1 end quantity all over h

4 to the power of the quantity x minus 2 end quantity times the quantity 4 to the power of h minus 1 end quantity all over h

the quantity x minus 2 end quantity times the quantity 4 to the power of h plus 1 end quantity all over h

Answer 1

Answer: it is C

Explanation:

I got it right

Answer 2

j(x) = 4x - 2

To find j of the quantity x plus h minus j of x all over h period 4 to the power of the quantity x minus 2 end quantity times the quantity 4 to the power of h end quantity all over h, we need to follow the order of operations (PEMDAS):

1. Evaluate x plus h minus j of x:

x + h - j(x) = x + h - 4x + 2 = 3x + h - 2

2. Raise 4 to the power of the quantity x minus 2 end quantity:

4^x - 2 = 4^x \* 4^(-2) = 4^x \* 1/16 = 4^x / 16

3. Multiply the results of steps 1 and 2:

3x + h - 2 \* 4^x / 16 = 3x + h - 2 \* 4^x / 16

4. Simplify the expression:

3x + h - 2 \* 4^x / 16 = 3x + h - 2 \* (4^x / 16)

5. Evaluate the expression:

3x + h - 2 \* (4^x / 16) = 3x + h - 2 \* (4^x / 16)

The final answer is:

3x + h - 2 \* (4^x / 16)

Explanation: