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hiralpatel121 Group Title

How can I solve the following? (attached)

  • 2 years ago
  • 2 years ago

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  1. hiralpatel121 Group Title
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    • 2 years ago
  2. hiralpatel121 Group Title
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    Sample problem:

    • 2 years ago
  3. Romero Group Title
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    What step are you confused on the sample problem?

    • 2 years ago
  4. hiralpatel121 Group Title
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    I understand the sample problem. I just don't know how to use it for the actual problem.

    • 2 years ago
  5. Romero Group Title
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    do the same...instead of using 7 use 5

    • 2 years ago
  6. hiralpatel121 Group Title
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    I did, but it shows up as wrong when I put it in.

    • 2 years ago
  7. Romero Group Title
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    then you did something wrong. ...show me the steps

    • 2 years ago
  8. Romero Group Title
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    what is the final answer that you got when you did the problem?

    • 2 years ago
  9. hiralpatel121 Group Title
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    Nevermind, I found my error, instead of 2/9 I put 2/5. Sorry. Can you help me with another problem though?

    • 2 years ago
  10. Romero Group Title
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    yes sure....remember to follow all the steps and make sure you don't make a mistake. One small mistake can make the answer wrong

    • 2 years ago
  11. hiralpatel121 Group Title
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    • 2 years ago
  12. hiralpatel121 Group Title
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    Sample:

    • 2 years ago
  13. Romero Group Title
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    What's the first step? Let's make the square root into an exponent \[\int\limits_{3}^{7}\frac{1}{\sqrt{6x+7}}dx\]\[\int\limits_{3}^{7}\frac{1}{(6x+7)^\frac{1}{2}}dx\]

    • 2 years ago
  14. Romero Group Title
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    then we bring the denominator to the top. But in order to do this we have to make the exponent a negative. Then we get \[\int\limits_{3}^{7}(6x+7)^{-\frac{1}{2}}dx\]

    • 2 years ago
  15. Romero Group Title
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    You just integrate what's inside first....can you tell me what you get?

    • 2 years ago
  16. hiralpatel121 Group Title
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    3x^2+7x+C

    • 2 years ago
  17. Romero Group Title
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    No that's not it. Look back in the problem you did before and see how they integrated a square root after it became \[(-------)^\frac{1}{2}\]

    • 2 years ago
  18. Romero Group Title
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    you have to do the chain rule.

    • 2 years ago
  19. Romero Group Title
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    Try it one more time....what is the integral?

    • 2 years ago
  20. hiralpatel121 Group Title
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    I have no idea here..

    • 2 years ago
  21. Romero Group Title
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    I thought you know how to integrate something that was a square root. Look at the example problem you did before and see what happens.

    • 2 years ago
  22. hiralpatel121 Group Title
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    Would it be: 1/12 root of 6x+7?

    • 2 years ago
  23. hiralpatel121 Group Title
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    Is the answer 1/4? If yes, then I've done my work correct.

    • 2 years ago
  24. Romero Group Title
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    I think you made a mistake \[\frac{1}{6} *\frac{1}{2}=\frac{1}{12}\] that's how you got the 12 right?

    • 2 years ago
  25. hiralpatel121 Group Title
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    Yes

    • 2 years ago
  26. Romero Group Title
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    it's suppose to be\[\frac{1}{6}*2=\frac{2}{3}=\frac{1}{3}\]

    • 2 years ago
  27. Romero Group Title
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    1/3 root of 6x+7 not 1/12 root of 6x+7

    • 2 years ago
  28. hiralpatel121 Group Title
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    Oh, so then the answer would be: 1.

    • 2 years ago
  29. Romero Group Title
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    what no I got something else

    • 2 years ago
  30. Romero Group Title
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    \[\frac{1}{3}\sqrt{6x+7}\]\[\frac{1}{3}\sqrt{6(7)+7}-\frac{1}{3}\sqrt{6(3)+7}\]

    • 2 years ago
  31. Romero Group Title
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    \[\frac{1}{3}(\sqrt{6(7)+7}-\sqrt{6(3)+7})\]\[\frac{1}{3}(\sqrt{42+7}-\sqrt{18+7})\]

    • 2 years ago
  32. Romero Group Title
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    can you do the rest?

    • 2 years ago
  33. hiralpatel121 Group Title
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    Yes, I got it.

    • 2 years ago
  34. Romero Group Title
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    what did you get?

    • 2 years ago
  35. hiralpatel121 Group Title
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    Im doing the math right now.

    • 2 years ago
  36. Romero Group Title
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    ok tell me what you get just to check if you got it right.

    • 2 years ago
  37. hiralpatel121 Group Title
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    2/3

    • 2 years ago
  38. Romero Group Title
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    yeah that's what I got good job.

    • 2 years ago
  39. Romero Group Title
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    If you have more problems post them in a different question.

    • 2 years ago
  40. Romero Group Title
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    I want to give you more medals :) for the good job

    • 2 years ago
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