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anonymous
 4 years ago
How can I solve the following? (attached)
anonymous
 4 years ago
How can I solve the following? (attached)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What step are you confused on the sample problem?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I understand the sample problem. I just don't know how to use it for the actual problem.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do the same...instead of using 7 use 5

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I did, but it shows up as wrong when I put it in.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then you did something wrong. ...show me the steps

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what is the final answer that you got when you did the problem?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Nevermind, I found my error, instead of 2/9 I put 2/5. Sorry. Can you help me with another problem though?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes sure....remember to follow all the steps and make sure you don't make a mistake. One small mistake can make the answer wrong

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What's the first step? Let's make the square root into an exponent \[\int\limits_{3}^{7}\frac{1}{\sqrt{6x+7}}dx\]\[\int\limits_{3}^{7}\frac{1}{(6x+7)^\frac{1}{2}}dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then we bring the denominator to the top. But in order to do this we have to make the exponent a negative. Then we get \[\int\limits_{3}^{7}(6x+7)^{\frac{1}{2}}dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You just integrate what's inside first....can you tell me what you get?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No that's not it. Look back in the problem you did before and see how they integrated a square root after it became \[()^\frac{1}{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you have to do the chain rule.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Try it one more time....what is the integral?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have no idea here..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I thought you know how to integrate something that was a square root. Look at the example problem you did before and see what happens.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Would it be: 1/12 root of 6x+7?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is the answer 1/4? If yes, then I've done my work correct.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think you made a mistake \[\frac{1}{6} *\frac{1}{2}=\frac{1}{12}\] that's how you got the 12 right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's suppose to be\[\frac{1}{6}*2=\frac{2}{3}=\frac{1}{3}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01/3 root of 6x+7 not 1/12 root of 6x+7

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, so then the answer would be: 1.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what no I got something else

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{3}\sqrt{6x+7}\]\[\frac{1}{3}\sqrt{6(7)+7}\frac{1}{3}\sqrt{6(3)+7}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{3}(\sqrt{6(7)+7}\sqrt{6(3)+7})\]\[\frac{1}{3}(\sqrt{42+7}\sqrt{18+7})\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Im doing the math right now.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok tell me what you get just to check if you got it right.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah that's what I got good job.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If you have more problems post them in a different question.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I want to give you more medals :) for the good job
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