Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

aceace Group Title

Arithmetic and geometric means with geometry!!!

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. aceace Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Question 11a and b

    • 2 years ago
    1 Attachment
  2. aceace Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Sorry b and c actually

    • 2 years ago
  3. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    know similarity of triangles ?

    • 2 years ago
  4. aceace Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yah just need help with b adn c not a

    • 2 years ago
  5. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    can u prove ABC and ACM are similar?

    • 2 years ago
  6. aceace Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yes ... its equilangular basically

    • 2 years ago
  7. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    so do you know that corresponding sides of similar triangles are proportional ?

    • 2 years ago
  8. aceace Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

    • 2 years ago
  9. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    can you apply that in triangles ACM and BCM ?

    • 2 years ago
  10. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    oh, sorry i mentioned ABC and ACM earlier it should be ACM and BCM

    • 2 years ago
  11. aceace Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i see what you're getting at

    • 2 years ago
  12. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    if you equate corresponding sides as equal , you directly get CM^2 as AM*BM

    • 2 years ago
  13. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    sorry , proportional*

    • 2 years ago
  14. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    for part c) you take triangles ABC and BCM

    • 2 years ago
  15. aceace Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i got AM=BM * x^2 where x was the reduction factor or ratio

    • 2 years ago
  16. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    what? how ?

    • 2 years ago
  17. aceace Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    oh i did substitution... but i also got your answer

    • 2 years ago
  18. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    good :)

    • 2 years ago
  19. aceace Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    thnx so much XD

    • 2 years ago
    • Attachments:

See more questions >>>

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.