Here's the question you clicked on:
JessicaNicoleCraven
k^3(k7/5)^-5
|dw:1352048933|dw:1352048968202:dw|095:dw|
? Is \(\large k^3 (\frac{k^7}{5})^{-5}\) ?
:O where did my solution go :O
^ ? I dunno, looks like you were trying to post a drawing.
I am so confused. Its suppose to be k with an exponent of 3, times k(7/5) with an exponent of negative 5.
Then is it \(\large k^3 \cdot (k^{7/5})^{-5} \) ?
Ok. Order of operations, handle the exponent first. A power-to-a-power, multiply the exponents.
All u have to know is that x^a*x^b=x^(a+b) (x^a)^b=x^ab and x^-a=1/x^a so u'll have |dw:1352049321095:dw|
\(\huge \rightarrow k^3 \cdot k^{(\frac{7}{5} \cdot -5)} \) \(\huge \rightarrow k^3 \cdot k^{-7}\)
Oh wow, thats a lot more simple than I was trying to make it. Thank you!