PLEASE HELP! I'm ALMOST done and I just need help on this one question... Attachment below!!!
@amistre64
@amistre64 Hey, man! I am on this one test and I'm trying to get done with it, could you help me on this problem?
\[ \frac{2x^2-6x}{x^2+18x+81}\div\frac{x^2-9}{x^2-81}\\ =\frac{2x^2-6x}{x^2+18x+81}\times\frac{x^2-81}{x^2-9}\\ =\frac{2x(x-3)}{(x+9)^2}\times\frac{(x+9)(x-9)}{(x+3)(x-3)}\]
Which answer is that going to turn out to be?
cancel out what is common in the numerator and denominator.. Notice that: \[ \frac{2x(x-3)}{(x+9)^2}\times\frac{(x+9)(x-9)}{(x+3)(x-3)}=\frac{2x(x-3)(x+9)(x-9)}{(x+9)(x+9)(x+3)(x-3)}\]
Ahhhh... Let me go try to simplify that and I'll brb
I got C, @kirbykirby ... Thank you so much for your help!
yup :)
yeah, that was just a lot of factoring practice ....
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