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anonymous
 5 years ago
Factoring Advice: Goal is to eliminate the x's.
[(x+y)^2  x^2] / y
anonymous
 5 years ago
Factoring Advice: Goal is to eliminate the x's. [(x+y)^2  x^2] / y

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's equal to 2x+y , eliminate the x's ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh that's different than the first one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think there is a way to do it, but I am not sure how I can get them to all cancel out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[((x+y)2−y2)/y \] goal is to eliminate the x's and be left with only y's to deal with...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thoughts? I have tried the usual things I can think of, but unless I made an error I am not catching, I am stumped.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is this part of a LIM problem ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, how on earth did you guess?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean are you looking for: \[\lim_{y \rightarrow 0} \frac{(x+y)^2x^2}{y}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you help me get unstuck? I really want to learn how to do this...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have it down to y + 2x and I don't know if I can use x in my lim answer, I've never seen a variable in a lim.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{y \rightarrow 0} \frac{\cancel{x^2}+2xy+y^2\cancel{x^2}}{y}=\lim_{y \rightarrow 0}2x+y=2x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is it acceptable to have a variable in the lim solution?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, coz it's basically the derivative: \[(x^2)'=2x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the final answer is 2x which is the derivative of x^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't know enough to understand that in the context of this problem. I understand what a first derivative is, and how you showed that, but I do not understand it in relation to my first "evaluate lim's" homework assignment.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how was the problem worded ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So the lim is really 2x?!?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The lim stmt you wrote, preceded by the word "Evaluate:"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, so the answer would be 2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We haven't covered derivatives yet  not until tomorrow. So this is kinda tricky!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is how they introduce you to derivatives

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you don't have to know what a derivative is to solve the LIM problem. But in effect, they will reveal that what you were doing was calculating the derivative.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have you seen Gilbert Strang's high school calculus course on MIT OCW, called "The Big Picture of Calculus?" THAT's how to teach calculus.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you a student or faculty?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that formula is basically calculating rise over run (where they make the run smaller and smaller)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0neither. I studied math about 20 years ago.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What do you do now that allows you to stay fresh on it? I am 44, BTW.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we are very close in age :) I just started solving problems on openstudy to keep the mind from rusting too much.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, thank you for your help. I am now a fan.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0BTW, I started going over the MIT OCW single variable calculus  but then got sucked in over here in the math group.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I want to build up a community on MIT OCW SCHOLAR SVC.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am trying to login there every night at 8 PM. I missed it tonight, the first time others were there, too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, if you are around at 8PM ET, I will try to be there in the MIT OCW SVC (NOT the 18.01  the SCHOLAR version. It is a new and improved course, totally self contained.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks. Over and out on this case. I had solved it and didn't realize I was done.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0My schedule in the evenings is not fixed, juggling around kids & work . And I didn't continue past the first OCW lecture  btw I didn't know they now have the new and improved course.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It is wonderful. I hope you find time to give it a try during your "me time."

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm going to resume my homework. I have about a zillion more to go...
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