#### janekommer

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(22x5y-3z4/40x-2y-8z5)3

- Thread starter janekommer
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(22x5y-3z4/40x-2y-8z5)3

\(\displaystyle \L\\\left(\frac{22x^{5}y-3z^{4}}{40x-2y-8z^{5}}\right)^{3}\)

If this is the case, please use approriate sybols to represent 'powers', such as ^.

In general, few are going to know what you mean and, therefore, you won't get an accurate response.

Most would think you mean:

\(\displaystyle \L\\3(\frac{22x\cdot{5y}-4z\cdot{3}}{40x-2y-8z\cdot{5}})\)

Unless of course, that is what you mean. Which seems unlikely.

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Sorry it should read. (2^2x^5y^-3z^4/4^0x^-2y^-8z^5)^3

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So you mean the following...?janekommer said:Sorry it should read. (2^2x^5y^-3z^4/4^0x^-2y^-8z^5)^3

. . . . .\(\displaystyle \L \left(\frac{2^2\, x^5\, y^{-3}\, z^4}{4^0\, x^{-2}\, y^{-8}\, z^5}\right)^3\)

Then this is a simple matter of taking the cube through, and simplifying, using the addition and subtraction rules you memorized for exponents, namely:

. . . . .(x<sup>m</sup>)(x<sup>n</sup>) = x<sup>m+n</sup>

. . . . .(x<sup>m</sup>) / (x<sup>n</sup>) = x<sup>m-n</sup>

. . . . .x<sup>-m</sup> = 1/x<sup>m</sup>

Where are you stuck? Please reply showing all of your steps so far. For instance, you squared the 2, converted 4<sup>0</sup> to 1, and... then what?

Thank you.

Eliz.

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If I divide here is where I am.

2^2/4^0=1

X^5/x^-2=x^-2.5

Y^-3/y^-8=y^.375

Z^4/z^5=z^20

adding would look like this

2^2+4^0=6^2

X^5/x^-2=x^3

Y^-3/y^-8=y^-11

Z^4/z^5=z^9

W

where x is any number

so 2^2/4^0 = 4/1=4

when dividing x^a/x^b you just subtract the powers so ypu get x^(a-b)

so X^5/x^-2 = x^(5--2) = x^7

and just do that for y and z and multiply it all together and you got the answer.

Jane, Stapel told you to ADD or SUBTRACT; why are you DIVIDING?janekommer said:Here is where I am stuck. Do I divide or add these up.

You obviously need classroom help: not provided here; try:

http://mathrefresher.blogspot.com/2005/ ... nents.html

Follow along:

See that 3 on the outside of the 'big' parentheses?. Multiply all your exponents inside by it. You get:

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

Becomes:

\(\displaystyle \L\\\frac{64x^{15}y^{-9}z^{12}}{x^{-6}y^{-24}z^{15}}\)

Rearrange so they're all positive exponents(if you want):

\(\displaystyle \L\\\frac{64x^{15}z^{12}x^{6}y^{24}}{y^{9}z^{15}}\)

Remember, when you multiply add exponents, and when you divide subtract exponents:

\(\displaystyle \H\\\frac{64x^{21}y^{15}}{z^{3}}\)