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RomeroBest ResponseYou've already chosen the best response.1
What step are you confused on the sample problem?
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
I understand the sample problem. I just don't know how to use it for the actual problem.
 one year ago

RomeroBest ResponseYou've already chosen the best response.1
do the same...instead of using 7 use 5
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
I did, but it shows up as wrong when I put it in.
 one year ago

RomeroBest ResponseYou've already chosen the best response.1
then you did something wrong. ...show me the steps
 one year ago

RomeroBest ResponseYou've already chosen the best response.1
what is the final answer that you got when you did the problem?
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
Nevermind, I found my error, instead of 2/9 I put 2/5. Sorry. Can you help me with another problem though?
 one year ago

RomeroBest ResponseYou've already chosen the best response.1
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
 one year ago

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

RomeroBest ResponseYou've already chosen the best response.1
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 \[\int\limits_{3}^{7}(6x+7)^{\frac{1}{2}}dx\]
 one year ago

RomeroBest ResponseYou've already chosen the best response.1
You just integrate what's inside first....can you tell me what you get?
 one year ago

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

RomeroBest ResponseYou've already chosen the best response.1
you have to do the chain rule.
 one year ago

RomeroBest ResponseYou've already chosen the best response.1
Try it one more time....what is the integral?
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
I have no idea here..
 one year ago

RomeroBest ResponseYou've already chosen the best response.1
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.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
Would it be: 1/12 root of 6x+7?
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
Is the answer 1/4? If yes, then I've done my work correct.
 one year ago

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

RomeroBest ResponseYou've already chosen the best response.1
it's suppose to be\[\frac{1}{6}*2=\frac{2}{3}=\frac{1}{3}\]
 one year ago

RomeroBest ResponseYou've already chosen the best response.1
1/3 root of 6x+7 not 1/12 root of 6x+7
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
Oh, so then the answer would be: 1.
 one year ago

RomeroBest ResponseYou've already chosen the best response.1
what no I got something else
 one year ago

RomeroBest ResponseYou've already chosen the best response.1
\[\frac{1}{3}\sqrt{6x+7}\]\[\frac{1}{3}\sqrt{6(7)+7}\frac{1}{3}\sqrt{6(3)+7}\]
 one year ago

RomeroBest ResponseYou've already chosen the best response.1
\[\frac{1}{3}(\sqrt{6(7)+7}\sqrt{6(3)+7})\]\[\frac{1}{3}(\sqrt{42+7}\sqrt{18+7})\]
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.1
Im doing the math right now.
 one year ago

RomeroBest ResponseYou've already chosen the best response.1
ok tell me what you get just to check if you got it right.
 one year ago

RomeroBest ResponseYou've already chosen the best response.1
yeah that's what I got good job.
 one year ago

RomeroBest ResponseYou've already chosen the best response.1
If you have more problems post them in a different question.
 one year ago

RomeroBest ResponseYou've already chosen the best response.1
I want to give you more medals :) for the good job
 one year ago
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