Algebra 2 Find the quotient. 2x-3/x/7/x^2 A. 7/x(2x-3) B. 7x/2x-3 C. 2x-3/7x D. x(2x-3)/7

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Algebra 2 Find the quotient. 2x-3/x/7/x^2 A. 7/x(2x-3) B. 7x/2x-3 C. 2x-3/7x D. x(2x-3)/7

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\[\huge\rm \frac{ \frac{ 2x-3 }{ x } }{ \frac{ 7 }{ x^2 } }\] like this ?
so the answers is C @Nnesha
how did you get it ?

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I just saw what u typed me and i figure out that is was exacly urs :) @Nnesha so I'm right?
\(\color{blue}{\text{Originally Posted by}}\) @Nnesha \[\huge\rm \frac{ \frac{ 2x-3 }{ x } }{ \frac{ 7 }{ x^2 } }\] like this ? \(\color{blue}{\text{End of Quote}}\) i just asked a question is this ur question ?
yes is like that the question I'm asking @Nnesha
okay change division to multiplication multiply the top with the `reciprocal` of the bottom fraction
so i see that the answer is C @Nnesha right?
please show ur work so i can check !
I did this 2x-3/7x cuz I cancel out the x^2
I really dont know how to slove it :( @Nnesha
here is an example \[\huge\rm \frac{ \color{Red}{\frac{ a }{ b }}}{ \frac{ c }{ d} }= \color{ReD}{\frac{a}{b}} \times \frac{d}{c}\]
what is the reciprocal of the bottom fraction ??
x and 7
what's reciprocal of 7/x^2 ??
reciprocal of 3/2 is 2/3
7/x^2
that's the original fraction what's reciprocal of that ?
i mean x^2/7 @Nnesha
yes right!
so multiply \[\huge\rm \frac{ 2x-3 }{ x } \times \frac{ x^2 }{ 7}\] now multiply the numerator by numerator and denominator by denominator
I got C
gtg sorry no it's not c
So D?

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