Algebra Tutor: None Selected Time limit: 1 Day

Aug 7th, 2015

Hello!

18. [(m*x*y^2*p)^2*p^(-2)*x^2]/[(p^0*x*m*y^3)^(-2)*x*p^(-2)] =
= [m^2*x^2*y^4*p^2*p^(-2)*x^2]/[x^(-2)*m^(-2)*y^(-6)*x*p^(-2)] =
= [m^2*x^4*y^4]/[m^(-2)*p^(-2)*x^(-1)*y^(-6)] =
= m^4*p^2*x^5*y^10.

20. [p^2*x^(-4)*k^5*(p^2*k)^(-2)]/[(p^2*x^(-3))^(-2)] =
= [p^2*x^(-4)*k^5*p^(-4)*k^(-2)]/[p^(-4)*x^6] =
= [k^3*p^(-2)*x^(-4)]/[p^(-4)*x^6] =
= k^3*p^2*x^(-10).

22. [(x^2*y^(-2)*p^0)^(-3)*p^2]/[x^2*(x^(-4))^0*(p^(-2)*y^5)^(-2)] =
= [x^(-6)*y^6*p^2]/[x^2*p^4*y^(-10)] =
= p^(-2)*x^(-8)*y^16.

Aug 7th, 2015

I still don't understand how you got to the answer.. I thought you'd still need to keep it as a simplified fraction

Aug 7th, 2015

but thank you i will keep studying it

Aug 7th, 2015

You posted no text for these tasks so I don't know what form of answer you need.

Consider the first example in more details:

18. Rewrite by symbols from the picture:
[(m*x*y^2*p)^2*p^(-2)*x^2] / [(p^0*x*m*y^3)^(-2)*x*p^(-2)] =

treat numerator and denominator separately,
use (a*b)^n = a^n*b^n and also a^0 = 1:

= [m^2*x^2*y^4*p^2*p^(-2)*x^2] / [x^(-2)*m^(-2)*y^(-6)*x*p^(-2)] =

now gather variables in degrees using a^n*a^m = a^(m+n):

= [m^2*x^4*y^4] / [m^(-2)*p^(-2)*x^(-1)*y^(-6)] =

now, move denominator to the numerator using 1/[a^n] = a^(-n)
(yes, I now see that I omitted this step):

= [m^2*x^4*y^4] * [m^2*p^2*x^1*y^6] =

and finally gather variables again:

m^4*p^2*x^5*y^10.

Aug 7th, 2015

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Aug 7th, 2015
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Aug 7th, 2015
Dec 11th, 2016
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