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Mathematics 20 Online
OpenStudy (anonymous):

v^2-9 /16v+48 divided by v-3 / 12 =

OpenStudy (anonymous):

Some things to note: v^2-9=v^2-3^2=(v-3)(v+3) and 16v+48=16(v+3) Also, note that when you have a fraction divided by a fraction, you can rewrite it as a the top fraction multiplied by the reciprocal of the bottom fraction : \[\frac{(\frac{a}{b})}{(\frac{c}{d})}=\frac{a}{b}*\frac{d}{c}\]

OpenStudy (anonymous):

would you like to try solving it ?

OpenStudy (anonymous):

Yes, please

OpenStudy (anonymous):

Are you familiar with everything I posted in the notes above ?

OpenStudy (anonymous):

I believe so

OpenStudy (anonymous):

\[\frac{\frac{v^2-9}{16v+48}}{\frac{v-3}{12}}=\frac{v^2-9}{16v+48}*\frac{12}{v-3}\]

OpenStudy (anonymous):

Is the above clear ?

OpenStudy (anonymous):

\[=\frac{(v-3)(v+3)}{16(v+3)}*\frac{12}{v-3}=\frac{3*4*(v-3)(v+3)}{4*4*(v-3)(v+3)}=\frac{3}{4}\]

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