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hiralpatel121

  • 2 years ago

How can I solve the following? (attached)

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

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

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

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

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

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

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

  9. hiralpatel121
    • 2 years ago
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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?

  10. Romero
    • 2 years ago
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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

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

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

  14. Romero
    • 2 years ago
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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 \[\int\limits_{3}^{7}(6x+7)^{-\frac{1}{2}}dx\]

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

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

  17. Romero
    • 2 years ago
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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 \[(-------)^\frac{1}{2}\]

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

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

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

  21. Romero
    • 2 years ago
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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.

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

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

  24. Romero
    • 2 years ago
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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?

  25. hiralpatel121
    • 2 years ago
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    Yes

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

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

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

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

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

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

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

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

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

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

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

  37. hiralpatel121
    • 2 years ago
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    2/3

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

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

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

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