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i have no clue how to solve this

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

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Other answers:

would i cross multiply?
no. just simplify n multiply
i dont know how i would simplify that my teacher never explained it
\[\frac{ g+3 }{ 4f^3 }\times \frac{ 6f }{ 8(g+3) }\] 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 \[\frac{ a^{10} }{ a^{11} } = (10-11 = -1) = a^{-1} = \frac{ 1 }{ a^1 }= \frac{ 1 }{ }\]
Appy that to your problem and you should get an answer.
\[4f ^{2}* \frac{ 6f }{ 8 }\]?
Yes. But now your answer is: \[\frac{ 6f }{ 4f^2(8) }\] can you simplify that answer more?
i dont think so
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.
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.
ok thank you so much !! = )

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