Question

Some math problems so yep help pls

Answer 1

Explanation:

Take this one thing at a time...baby steps to help us keep our sanity. Math does tend to drive a person around the bend...

The first term is the tricky one. If the exponent is negative, it can be rewritten as a positive by putting whatever the exponent is on under a 1. For example:

`x^(-2)=(1)/(x^2)`

So we'll use that to make this simpler.

`(-(1)/(2))^(-2)=(1)/((-(1)/(2))^2 ) }` Squaring that negative makes it a positive, so we can rewrite as

`(1)/(((1)/(2))^2 )` which simplifies to

`(1)/((1)/(4) )` What we have there is a fraction over a fraction; namely:

`((1)/(1) )/((1)/(4) )` and the rule for that is to bring the bottom fraction up, flip it and multiply:

`(1)/(1)` ×
`(4)/(1)` which is 4.

Now for the second part. All we are doing here is squaring that fraction. Remember that squaring a negative makes it a positive, so

`(-(1)/(2))^2=(1)/(4)` . Now we'll put these together to get the sum

`4+(1)/(4)` and with a common denominator of 4, this becomes

`(16)/(4)+(1)/(4)` and that equals

`(17)/(4)`, the seond choice down.