Studying for an exam please help me solve these problems

there is 4 problems that i need help with!

1) 9x^9 / 16y^4 divided by 6x^8 / 4y^8

2) 4 / x+5 divided by 8 / x^2 - 25

3) x-5 / x^2 divided by x-10 / x^3

4) -1 / x-1 divided by -1 / x-2

1) (9x^9/16y^4)(4y^8/6x^8) = (3xy^4)/8

2) (4/(x+5))((x+5)(x-5)/8) = (x-5)/2

3) ((x-5)/x^2)(x^3/(x-10)) = x(x-5)/(x-10)

4) (-1/(x-1))(-(x-2)/1 = (x-2)/(x-1)

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Realize that these are just fractions. If you were asked to compute 1/2 divided by 3/5 you would invert and multiply, as they say: 1/2 * 5/3. You just do the same thing here.

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( (9x^9)/(16y^4) ) / ( 6x^8 / (4y^8) ) =

( 9x^9 ) /( 16y^4 ) * ( (4y^8) / (6x^8) ) =

(36x^9y^8) / (96x^8y^4) =

(3xy^4)/8


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( 4/(x+5) ) / ( 8/(x^2-25) ) =

( 4/(x+5) ) * ( (x-5)(x+5) ) / 8 ) =

(x-5) /2

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Work the rest the same way.

1. When you divide fraction you take the denominator and invert it and then just multiply the numerator and cancel what you can. in the first problem the answer is 3x/8
2. Do just like 1 except factor the X^2-25 to (x+5)*(x-5). then you can cancel several parts to give (x-5)/2.

3 and 4 can be handled the same way.

1) flip the seconed fraction, then cancel anything you can you get (3xy^4)/(8)

2)factor that second one: (x+5)(x-5) so (x+5) cancels 4(x-5)/(2)

3)x(x-5)/(x-10)

4) (x-2)/(x-1)


Just flip the one, cancel and multiply across

9x^9 / 16y^4 divided by 6x^8 / 4y^8
9x^9 / 16y^4 times 4y^8 / 6x^8
(3x/2)(y^4/4)
3xy^4/8