anonymous
  • anonymous
Factoring Advice: Goal is to eliminate the x's. [(x+y)^2 - x^2] / y
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
it's equal to 2x+y , eliminate the x's ?
anonymous
  • anonymous
oh that's different than the first one
anonymous
  • anonymous
I think there is a way to do it, but I am not sure how I can get them to all cancel out.

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More answers

anonymous
  • anonymous
\[((x+y)2−y2)/y \] goal is to eliminate the x's and be left with only y's to deal with...
anonymous
  • anonymous
Thoughts? I have tried the usual things I can think of, but unless I made an error I am not catching, I am stumped.
anonymous
  • anonymous
BRB
anonymous
  • anonymous
is this part of a LIM problem ?
anonymous
  • anonymous
Yes, how on earth did you guess?
anonymous
  • anonymous
I mean are you looking for: \[\lim_{y \rightarrow 0} \frac{(x+y)^2-x^2}{y}\]
anonymous
  • anonymous
Uh, indeed I am...
anonymous
  • anonymous
Can you help me get unstuck? I really want to learn how to do this...
anonymous
  • anonymous
I 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
  • anonymous
\[\lim_{y \rightarrow 0} \frac{\cancel{x^2}+2xy+y^2-\cancel{x^2}}{y}=\lim_{y \rightarrow 0}2x+y=2x\]
anonymous
  • anonymous
Is it acceptable to have a variable in the lim solution?
anonymous
  • anonymous
yes, coz it's basically the derivative: \[(x^2)'=2x\]
anonymous
  • anonymous
Or is this DNE?!?
anonymous
  • anonymous
the final answer is 2x which is the derivative of x^2
anonymous
  • anonymous
I 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
  • anonymous
how was the problem worded ?
anonymous
  • anonymous
So the lim is really 2x?!?
anonymous
  • anonymous
The lim stmt you wrote, preceded by the word "Evaluate:"
anonymous
  • anonymous
yes, so the answer would be 2x
anonymous
  • anonymous
We haven't covered derivatives yet - not until tomorrow. So this is kinda tricky!
anonymous
  • anonymous
this is how they introduce you to derivatives
anonymous
  • anonymous
you 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
  • anonymous
Have 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
  • anonymous
Are you a student or faculty?
anonymous
  • anonymous
that formula is basically calculating rise over run (where they make the run smaller and smaller)
anonymous
  • anonymous
Ahhhh
anonymous
  • anonymous
neither. I studied math about 20 years ago.
anonymous
  • anonymous
What do you do now that allows you to stay fresh on it? I am 44, BTW.
anonymous
  • anonymous
we are very close in age :) I just started solving problems on openstudy to keep the mind from rusting too much.
anonymous
  • anonymous
Well, thank you for your help. I am now a fan.
anonymous
  • anonymous
BTW, I started going over the MIT OCW single variable calculus - but then got sucked in over here in the math group.
anonymous
  • anonymous
I want to build up a community on MIT OCW SCHOLAR SVC.
anonymous
  • anonymous
I am trying to login there every night at 8 PM. I missed it tonight, the first time others were there, too.
anonymous
  • anonymous
So, 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
  • anonymous
Thanks. Over and out on this case. I had solved it and didn't realize I was done.
anonymous
  • anonymous
My 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
  • anonymous
It is wonderful. I hope you find time to give it a try during your "me time."
anonymous
  • anonymous
I'm going to resume my homework. I have about a zillion more to go...
anonymous
  • anonymous
OK :)

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