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I NEED HELP! !!!!!!!!!!!

I NEED HELP! !!!!!!!!!!!-example-1
User REALSOFO
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1 Answer

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14)

(3x^2y)/(4x^3y^5) / (6y^2)/(2xy^3)

Okay first you need to realize that (a^b)/(a^c)=a^(b-c)

And you also need to know that (a^b)*(a^c)=a^(b+c), with that out of the way..

3/(4xy^4) / 3/(xy)

Now you need to know that (a/b) / (c/d) = (a/b)*(d/c) = (ad)/(bc) now we have:

3/(4xy^4) * xy/3 the 3's cancel out leaving

(xy)/(4xy^3)

1/(4y^2)

...

15)

(x^2-3x-4)/(x^2-3x-18) * (x-6)/(x+1) factor the numerator and denominator

[(x-4)(x+1)]/[(x-6)(x+3)] * (x-6)/(x+1) note that the (x+1)s and (x-6)s cancel

(x-4)/(x+3)

...

16)

(x^2-8x+15)/(x^2+12x+32) * (x+4)/(x^2-25) do the factoring...

[(x-3)(x-5)]/[(x+4)(x+8)] * (x+4)/[(x-5)(x+5)] the (x-5)s and (x+4)s cancel

(x-3)/(x+8) * 1/(x+5)

(x-3)/[(x+8)(x+5)]

...

19) maybe for some variety...

4/(x+5) + 2x/(x^2-25) factor

4/(x+5) + 2x/[(x+5)(x-5)] so to have a common denominator we multiply left term by (x-5)...

[4(x-5)+2x]/(x^2-25)

(4x-20+2x)/(x^2-25)

(8x+20)/(x^2-25)




User Bithin Alangot
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