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# 3. Please solve for the General Form of the Equation. Thanks!

label Algebra
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Nov 24th, 2017

The general form is just the foiled out version of the standard form.  So using your standard form.  We find:

(x-\frac{1}{11})^2+y^2=\frac{1}{121}
x^2-\frac{2}{11} x + \frac{1}{121} +y^2=\frac{1}{121}

x^2-\frac{2}{11} x +y^2=0


Apr 27th, 2015

Well, attempting to write that as code didn't work, how did you do that?

Here is the non coded version

(x-1/11)^2+y^2=1/121

x^2-2x/11+1/121+y^2=1/121

x^2-2x/11+y^2=0

Apr 27th, 2015

so note, general form will always equal zero.  Here, they cancel nicely, in others you will have your non-variable terms combined on the left side.

Apr 27th, 2015

Let me know if you have any questions :)

Apr 27th, 2015

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Nov 24th, 2017
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Nov 24th, 2017
Nov 25th, 2017
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