## Spartan_Of_Ares Group Title i have no clue how to solve this one year ago one year ago

1. Spartan_Of_Ares Group Title

2. abb0t Group Title

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

3. dpaInc Group Title

then simplify...

4. Spartan_Of_Ares Group Title

would i cross multiply?

5. abb0t Group Title

no. just simplify n multiply

6. Spartan_Of_Ares Group Title

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

7. abb0t Group Title

$\frac{ g+3 }{ 4f^3 }\times \frac{ 6f }{ 8(g+3) }$ 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 $\frac{ a^{10} }{ a^{11} } = (10-11 = -1) = a^{-1} = \frac{ 1 }{ a^1 }= \frac{ 1 }{ }$

8. abb0t Group Title

9. Spartan_Of_Ares Group Title

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

10. abb0t Group Title

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

11. Spartan_Of_Ares Group Title

i dont think so

12. abb0t Group Title

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.

13. abb0t Group Title

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.

14. Spartan_Of_Ares Group Title

ok thank you so much !! = )