Is that right?

$\displaystyle \frac{3-x}{-10} = -\frac{3-x}{10}$

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- Sep 18th 2010, 05:16 AMklik11Negative number in the denominator
Is that right?

$\displaystyle \frac{3-x}{-10} = -\frac{3-x}{10}$ - Sep 18th 2010, 05:21 AMPlato
Yes.

This is also correct.

$\displaystyle \dfrac{3-x}{-10}=\dfrac{x-3}{10}$.

Do you understand why? - Sep 18th 2010, 05:22 AMklik11
Yes. Thanks :)

- Sep 18th 2010, 07:01 AMklik11
Another question.

Is this right?

$\displaystyle (\dfrac{x-1}{2})^{2} = \dfrac{x-1}{2} * \dfrac{x-1}{2} = \dfrac{x^{2}-2x+1}{4}$ - Sep 18th 2010, 09:24 AMshannu82