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jackoo

Hey, please help:)

  • one year ago
  • one year ago

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  1. jackoo
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    \[3 \sqrt[3]{24}\]

    • one year ago
  2. klimenkov
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    Hint:\[\sqrt[3]{24}=\sqrt[3]{8\cdot3}=\sqrt[3]8\cdot\sqrt[3]{3}\]

    • one year ago
  3. jackoo
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    Thank you!

    • one year ago
  4. klimenkov
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    Can you write the answer?

    • one year ago
  5. jackoo
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    \[6\sqrt[3]{3}\]

    • one year ago
  6. klimenkov
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    Very nice.

    • one year ago
  7. jackoo
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    Could you help me with another question?

    • one year ago
  8. jackoo
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    \[3\sqrt{x+4}-2=13\]

    • one year ago
  9. klimenkov
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    What is the problem?

    • one year ago
  10. klimenkov
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    What cant you do?

    • one year ago
  11. klimenkov
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    Solve for \(\sqrt{x+4}\).

    • one year ago
  12. klimenkov
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    Hey! Are you here?

    • one year ago
  13. jackoo
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    Yes, I am sorry

    • one year ago
  14. jackoo
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    couldn't I just do 4-2 giving me 2 so then I do x+2=13?

    • one year ago
  15. jackoo
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    x+2=13 -2 -2 x=11

    • one year ago
  16. jackoo
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    idk

    • one year ago
  17. klimenkov
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    Can you solve \(3y-2=13\) ?

    • one year ago
  18. jackoo
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    y=5

    • one year ago
  19. klimenkov
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    Yes. Think if \(y=\sqrt{x+4}\).

    • one year ago
  20. klimenkov
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    If you make a substitution \(y=\sqrt{x+4}\), you will get \(\sqrt{x+4}=5\). Do you know what \(\sqrt{\phantom{a}}\) sign means?

    • one year ago
  21. jackoo
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    square root

    • one year ago
  22. klimenkov
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    What does square root means?

    • one year ago
  23. jackoo
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    a number multiplied by itself?

    • one year ago
  24. klimenkov
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    Can you tell me what does \(\sqrt{16}\) mean?

    • one year ago
  25. jackoo
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    4?

    • one year ago
  26. klimenkov
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    Yes. But Also -4 too. -4*(-4)=16. So what can you say about \(\sqrt{x+4}=5\) ?

    • one year ago
  27. jackoo
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    x=1?

    • one year ago
  28. jackoo
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    I don't think I know

    • one year ago
  29. klimenkov
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    See the analogy. Try to think. Compute \(\sqrt{36}, \sqrt{25}, \sqrt{\frac14}\).

    • one year ago
  30. jackoo
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    alright. 6,5,2

    • one year ago
  31. klimenkov
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    Check the last one.

    • one year ago
  32. klimenkov
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    \(\sqrt x\) is the inverse function for \(x^2\). \(\sqrt x\) is a number that squared must be \(x\). \((\sqrt x )^2=x\). Think about last one.

    • one year ago
  33. jackoo
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    I still think it's1

    • one year ago
  34. klimenkov
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    Compute \(\sqrt{\frac14}\).

    • one year ago
  35. jackoo
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    2

    • one year ago
  36. klimenkov
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    \(2^2\neq\frac14\)

    • one year ago
  37. klimenkov
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    Try again.

    • one year ago
  38. jackoo
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    1/4

    • one year ago
  39. klimenkov
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    \(\left(\frac14\right)^2\neq\frac14\). Try again.

    • one year ago
  40. jackoo
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    So I have to square it?

    • one year ago
  41. klimenkov
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    You have to find \(\sqrt{\frac14}\) using the definition I wrote above.

    • one year ago
  42. jackoo
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    .5

    • one year ago
  43. jackoo
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    .5*.5=1/4

    • one year ago
  44. klimenkov
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    Yes. Thats it. Now back to \(\sqrt{x+4}=5\). Can you solve it?

    • one year ago
  45. jackoo
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    Would I plug something in x?

    • one year ago
  46. klimenkov
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    Ok. Can you solve this: \(\sqrt x = 2\) ?

    • one year ago
  47. jackoo
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    4

    • one year ago
  48. klimenkov
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    Yes. Thats it. Now back to \(\sqrt{x+4}=5\). Can you solve it?

    • one year ago
  49. jackoo
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    25?

    • one year ago
  50. klimenkov
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    \(\sqrt{25+4}\neq5\)

    • one year ago
  51. jackoo
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    21

    • one year ago
  52. klimenkov
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    Bingo. You solved this by your own. Congratulations!

    • one year ago
  53. jackoo
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    It has to be that's why thought of 25 =5

    • one year ago
  54. jackoo
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    Thank you:)!!

    • one year ago
  55. klimenkov
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    You are welcome.

    • one year ago
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