Question

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Answer 1

`(4f)^3 = 64*f^3`

`(-(3)/(2)t^2)^2 = (9)/(4)t^4`

Explanation:

Solving (a):

`(4f)^3`

Apply law of indices:

`x^3 = x * x * x`

So, we have:

`(4f)^3 = 4f * 4f * 4f`

Rewrite as:

`(4f)^3 = 4 * 4 * 4*f*f*f`

`(4f)^3 = 64*f^3`

Solving (b):

`(-(3)/(2)t^2)^2`

Apply law of indices:

`x * x = x^2`

So, we have:

`(-(3)/(2)t^2)^2 = (-(3)/(2)t^2) * (-(3)/(2)t^2)`

Remove brackets

`(-(3)/(2)t^2)^2 = -(3)/(2)t^2 * -(3)/(2)t^2`

Rewrite as:

`(-(3)/(2)t^2)^2 = -(3)/(2)*-(3)/(2)*t^2 * *t^2`

`(-(3)/(2)t^2)^2 = (3)/(2)*(3)/(2)*t^4`

`(-(3)/(2)t^2)^2 = (9)/(4)*t^4`

`(-(3)/(2)t^2)^2 = (9)/(4)t^4`