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

Jonask

which one is larger

  • one year ago
  • one year ago

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

    \[\sqrt[3]{60}\] \[2+\sqrt[3]{7}\]no machines allowed

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

    \[2=\sqrt[3]{8}\]

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

    \[60=(\frac{ 7+8 }{ 2 })10\]

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

    \[\sqrt[3]{60}= \sqrt[3]{5}(\sqrt[3]{7+8})\]

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

    almost simlar now to\[\sqrt[3]{8}+\sqrt[3]{7}\] HALT

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

    !!!!!!

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

    is comparing 60 with \[(2+\sqrt[3]{7})^3\]the same idea(tautology)

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

    i think we have to use inequalities \[(2+\sqrt[3]{7})^3=8+3(4)\sqrt[3]{7}+3(2)(\sqrt[3]{7})^2+7\]

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

    \[\sqrt[3]{1}<\sqrt[3]{7}<\sqrt[3]{8}\]

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

    \[15+12a+6a^2\] and 60

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

    Interesting question!

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

    \[a=\sqrt[3]{7}\]

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

    we can allow machines

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

    \[a<2\]

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

    \[\frac{ x+y+z }{ 3 } \leq \sqrt[3]{xyz}\] useful ineq

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

    geometric mean

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

    exhausted

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

    Try to do like this. How to evaluate \(\sqrt[3]{7}\). \(1<{7}<8 \Rightarrow 1<\sqrt[3]{7}<2\). As 7 is closer to 8 than to 1, I hope \(\sqrt[3]{7}\) is closer to 2. Lets put \(\sqrt[3]{7}=2+\alpha\). Then \(7=8+12\alpha+6\alpha^2+\alpha^3\). Since \(\alpha <1\) we can neglect \(\alpha^2 ,\alpha^3\). So, \(\alpha=-\frac1{12}\). So the approximation for \(\sqrt[3]{7}\) is \(2-\frac1{12}=\frac{23}{12}\). Do this once again, because it is not a good accuracy for this problem.

    • one year 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.