Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Spartan_Of_Ares

  • one year ago

i have no clue how to solve this

  • This Question is Closed
  1. Spartan_Of_Ares
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    1 Attachment
  2. abb0t
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  3. dpaInc
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    then simplify...

  4. Spartan_Of_Ares
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    would i cross multiply?

  5. abb0t
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    no. just simplify n multiply

  6. Spartan_Of_Ares
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  7. abb0t
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    \[\frac{ g+3 }{ 4f^3 }\times \frac{ 6f }{ 8(g+3) }\] 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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  9. Spartan_Of_Ares
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  10. abb0t
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  11. Spartan_Of_Ares
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i dont think so

  12. abb0t
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok thank you so much !! = )

  15. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.