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- anonymous

i have no clue how to solve this

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- anonymous

i have no clue how to solve this

- schrodinger

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- anonymous

- abb0t

\[\frac{ g+3 }{ 4f^3 }\times \frac{ 6f }{ 8g+24 }\]

- anonymous

then simplify...

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- anonymous

would i cross multiply?

- abb0t

no. just simplify n multiply

- anonymous

i dont know how i would simplify that my teacher never explained it

- abb0t

\[\frac{ g+3 }{ 4f^3 }\times \frac{ 6f }{ 8(g+3) }\]
you can cancel out the (g+3).
remember:
\[\frac{ T }{ T } = 1\]
the same thing with:
\[\frac{ 2a^{10} }{ 4a^{11} } = \frac{ 1 }{ 2a }\]
I basically simplified 2/4 and the a's were subtracted
\[\frac{ a^{10} }{ a^{11} } = (10-11 = -1) = a^{-1} = \frac{ 1 }{ a^1 }= \frac{ 1 }{ }\]

- abb0t

Appy that to your problem and you should get an answer.

- anonymous

\[4f ^{2}* \frac{ 6f }{ 8 }\]?

- abb0t

Yes. But now your answer is:
\[\frac{ 6f }{ 4f^2(8) }\]
can you simplify that answer more?

- anonymous

i dont think so

- abb0t

You actually can. Oh and I made a mistake, i didn't provide you the correct answer with that last one. It should actually be:
\[\frac{ 6 }{ 32f^2 }\]
and that CAN be simplified furhter to:
\[\frac{ 3 }{ 16f^2 }\]
this is because you can divide 6 and 32 by 2. 2 is the common number between them. It works the same as I explained before:
\[\frac{ 3 \times 2 }{ (16 \times 2) f^2 }= (\frac{ 3 }{ 16 })(\frac{ 2 }{ 2 })(\frac{ 1 }{ f^2 })\]
and remember:
\[\frac{ 2 }{ 2 } = 1\]
and 1 multiplied by anything is simply the number or letter itself.

- abb0t

The answer you provided:
\[(\frac{ 1 }{ 4f^2 })(\frac{ 6f }{ 8 }) \] is half right. I'm not sure if you just messed up bu accident, but just wanted to clarify incase you didn't notice and where I got my final correct answer from.

- anonymous

ok thank you so much !! = )

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