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Mathematics
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Check my answer please :)
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... What answer?
\[\Large\frac{ \frac{ 3x-7 }{ x^2 } }{ \frac{ x^2 }{ 2 }+\frac{ 2 }{ x } }\] \[\frac{ 3x-7 }{ x^2 } \div \frac{ x^3+4 }{ 2x }~or~\frac{ 3x-7 }{ x^2 } \times \frac{ 2x }{ x^3+4 }\]
\[\frac{ 6x^2-14x }{ x^5+4x^2 }==>\frac{ 2x(3x-7) }{ x \times x(x^3+4) }==>\frac{ 2(3x-7) }{ x(x^3+4) }\]
Hmmm, looks good to me.
Sure.
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:)
Is seems right to me, however my math skills are a little rusty... I haven't done this stuff in 5 years
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