Question

Need help on these. Don't answer if you don't know. I'm looking at YOU, bmdaniel1

Answer 1

use m a t h w a y

really easy to use

works for all algebraic equations

Explanation:

Answer 2

I assume you're trying to simplify these.

For all of these, divide the coefficients separately from the x's and that separately from the y's.

#5:
`(16t^4)/(8t)`

Start with 16/8 = 2.

Then move to
`(t^4)/(t)`.

`(t^4)/(t) = (t \cdot t \cdot t \cdot t)/(t)`

So, you'll cancel out one t from the top with the one t in the bottom and be left with
`t^3`.

Putting that all together:

`(16t^4)/(8t) = 2t^3`

#6:
`(x^6y^(14))/(x^4y^9)`

You have 6 x's in the top and 4 x's in the bottom. When you cancel out 4 from the top to cancel the 4 in the bottom, you'll be left with 2 x's in the top:

`(x^6)/(x^4)=x^2`

Similarly, 14 y's up top and 9 in the bottom. When you cancel one-for-one, you'll be left with 5 y's up top:

`(y^(14))/(y^9)=y^5`

Putting that all together:

`(x^6y^(14))/(x^4y^9)=x^2y^5`

#7:
`(3^4x^4)/(3x^2)`

This is the same as #6, just with 3's and x's.

You have four 3's up top and one in the bottom. When you cancel, you'll be left with three 3's up top:

`(3^4)/(3)=3^3`

You have 4 x's up top and 2 x's down below. That will leave 2 x's up top once you cancel them out:

`(x^4)/(x^2)=x^2`

Putting that together:

`(3^4x^4)/(3x^2)=3^3x^2`