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OpenStudy (rootbeer003):
OpenStudy (anonymous):
yes again
OpenStudy (rootbeer003):
OpenStudy (anonymous):
yes again
OpenStudy (rootbeer003):
how much more time u got?
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OpenStudy (anonymous):
idk i will tell you when i drop
OpenStudy (anonymous):
how many more you got?
OpenStudy (rootbeer003):
atun
OpenStudy (anonymous):
lets do five
OpenStudy (rootbeer003):
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OpenStudy (anonymous):
yes
OpenStudy (rootbeer003):
OpenStudy (anonymous):
no, you need the 50
also \(x^4\) and \(y^3\)
OpenStudy (rootbeer003):
d
OpenStudy (anonymous):
not common factors, least common multiple
yeah D
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OpenStudy (rootbeer003):
OpenStudy (anonymous):
nope same mistake as before
you need both \(x^2\) and \(y^2\)
OpenStudy (rootbeer003):
wait no d
OpenStudy (anonymous):
yes
OpenStudy (rootbeer003):
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OpenStudy (anonymous):
heck no
that is not how you add fractions
OpenStudy (rootbeer003):
a
OpenStudy (anonymous):
\[\frac{2}{9x^3}+\frac{5}{3x}\] lcd is \(9x^3\) so \[\frac{2}{9x^3}+\frac{5}{3x}\times \frac{3x^2}{3x^2}\]
\[=\frac{2}{9x^3}+\frac{15x^2}{9x^3}=\frac{2+15x^2}{9x^3}\]
OpenStudy (anonymous):
you do not add the tops until you have the correct denominator
OpenStudy (rootbeer003):
b
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OpenStudy (anonymous):
yes
OpenStudy (rootbeer003):
OpenStudy (anonymous):
stop it !~
OpenStudy (anonymous):
you do not subtract like that
OpenStudy (anonymous):
\[\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}\]
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