Simplify and evaluate: (4^2x^3y^-2z^-3/x^-5y^8z^-9)^2

[flora33]

Hey all, I'm pretty sure I understand this Unit better then the last... Could someone please take a look at this problem and let me know if you see any errors. I'm REALLY hoping you dont. Thanks.

(4^2x^3y^-2z^-3/x^-5y^8z^-9)^2

4^2(2)x^3(2)y^-2(2)z^-3(2)/x^-5(2)y^8(2)z^-9(2)
4^4x^6y^-4z^-6/x^-10y^16z^-18
4^4x^6x^10z^18/y^4y^16z^6
256x^16z^12/y^20

 

[Loren]

Re: Simplify and evaluate

As I read the problem as written, you have lost a (-5) which was the coefficient of y^8 in the denominator. Of course, it is squared as per the exponent attached to the entire expression. Therefore, my answer has a "25" as a coefficient in the denominator. Did I miss something?

 

[flora33]

Re: Simplify and evaluate

As I read the problem as written, you have lost a (-5) which was the coefficient of y^8 in the denominator. Of course, it is squared as per the exponent attached to the entire expression. Therefore, my answer has a "25" as a coefficient in the denominator. Did I miss something?

4^2x^3y^-2z^-3/x^-5y^8z^-9)^2

4^2(2)x^3(2)y^-2(2)z^-3(2)/x^-5(2)y^8(2)z^-9(2)
4^4x^6y^-4z^-6/x^-10y^16z^-18
4^4x^6x^10z^18/y^4y^16z^6
256x^16z^12/y^20

-5 is the exponent that belongs to x. There is only one coefficient, which is the 4 in the numerator. Do you know how I would attach a word document? It is much easier to show the work with actual exponents and in the fraction form! I tried to attach a doc. but for some reason they are not allowed?

 

[Loren]

Sorry. I misread.

(4^2x^3y^-2z^-3/x^-5y^8z^-9)^2 should be typed as ((4^2 x^3 y^-2 z^-3)/(x^-5 y^8 z^-9))^2 (spaces optional) for more clarity to yield

\(\displaystyle (\frac{4^2x^3y^{-2}z^{-3}}{x^{-5}y^8z^{-9}})^2 = (\frac{4^2 x^8 z^{6}}{y^{10}})^2 = \frac{256 x^{16} z^{12}}{y^{20}}\).

Note that the entire denominator in the original expression is enclosed in parenthesis. Without the parenthesis we would have...

\(\displaystyle (\frac{4^2x^3y^{-2}z^{-3}}{x^{-5}}y^8z^{-9})^2\)

Final answer.. 256x^16 z^12/y^20 as you have indicated or possibly 256x^16 z^12 y^-20. Good job.

 

[Deleted member 4993]

Sorry. I misread.

(4^2x^3y^-2z^-3/x^-5y^8z^-9)^2 should be typed as

better yet

[{(4^2)(x^3)(y^-2)(z^-3)}/{(x^-5)(y^8)(z^-9)}]^2

Now there is no chance of confusion


((4^2 x^3 y^-2 z^-3)/(x^-5 y^8 z^-9))^2 (spaces optional) for more clarity to yield

\(\displaystyle (\frac{4^2x^3y^{-2}z^{-3}}{x^{-5}y^8z^{-9}})^2 = (\frac{4^2 x^8 z^{6}}{y^{10}})^2 = \frac{256 x^{16} z^{12}}{y^{20}}\).

Note that the entire denominator in the original expression is enclosed in parenthesis. Without the parenthesis we would have...

\(\displaystyle (\frac{4^2x^3y^{-2}z^{-3}}{x^{-5}}y^8z^{-9})^2\)

Final answer.. 256x^16 z^12/y^20 as you have indicated or possibly 256x^16 z^12 y^-20. Good job.

 

[flora33]

Sorry. I misread.

(4^2x^3y^-2z^-3/x^-5y^8z^-9)^2 should be typed as ((4^2 x^3 y^-2 z^-3)/(x^-5 y^8 z^-9))^2 (spaces optional) for more clarity to yield

\(\displaystyle (\frac{4^2x^3y^{-2}z^{-3}}{x^{-5}y^8z^{-9}})^2 = (\frac{4^2 x^8 z^{6}}{y^{10}})^2 = \frac{256 x^{16} z^{12}}{y^{20}}\).

Note that the entire denominator in the original expression is enclosed in parenthesis. Without the parenthesis we would have...

\(\displaystyle (\frac{4^2x^3y^{-2}z^{-3}}{x^{-5}}y^8z^{-9})^2\)

Final answer.. 256x^16 z^12/y^20 as you have indicated or possibly 256x^16 z^12 y^-20. Good job.

Great, thank you. I need to figure out how to use the code to write out this type of problem using the regular fraction form. If I need to enter another problem, I will try to use the other method mentioned by Subhotash or use the html... Thanks again.