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anonymous
 5 years ago
And you WANT another problem?
anonymous
 5 years ago
And you WANT another problem?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let me attach the one I am doing now that doesn't have an answer attached to it.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0now that I know what im looking at yes lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0thats the exact same one; only with a sqrt7 lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, there you go. Let me try to work it too and see what I come up with and then you can let me know where I'VE SCREWED UP

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know it's the same one basically, but it doesn't matter, cuz I keep getting them wrong consistently!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cheat like crazy: use the computer.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we know the integral goes to \[s+s^{1/2}\] tho so thats a start

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when we use the bottom limit we get: \[12 = 1 right?\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[s 2s^{1/2}\] i meant to type

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[s  \frac{2}{\sqrt{s}}+1 \rightarrow s=\sqrt{7}\] is?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{7} \frac{2}{\sqrt[4]{7}}+1\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0essentally its just: \[F(\sqrt{7})F(1)\] \[F(s) = s{2\over\sqrt{s}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, you went WAYYYYYY too fast for me! I just got F(x)! Let me work this and see if I can get what you got, ok?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i just successfully got sqrt7  2/ 4th root 7 for F(sqrt7). That's where I am in the figuring. Now for F(1)...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and for shorthand: 4rt(7) is easier to read thru

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, then here's my answer: Let me get the equation thing and do it that way.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{7}2/\sqrt[4]{7}+1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0YAY!!! Couldn't have done it without you! Did I tell you I have been working on this particular problem (#31) since noon today?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lets try one that is not 'exactly' the same lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, so does this mean, now that you have taught me how to send you my problems, that you are willing to look at any of them I cannot do on my own (as long as you are online and available with the time?)?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, ow that i aint gotta read your mind it should be easier ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That particular problem will have all the same types with the same integrand. That's the point, to let ua do the same ones over and over to get the precise idea of how top do them. I will see if the next problem is different and let you know, ok?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0interactmath.com is just like your math program except its free. you can practice on it all you like.
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