2y^{5 } Divided by 5y^{5}z^{4}
z^{4 }7
2y^{5 } Divided by 5y^{5}z^{4}
z^{4 }7
Remember KMF Keep Multiply and Flip
so the question becomes
2y^{5 } x 7
z^{4 } 5y^{5}z^{4}
then we multiply across to get
14y^{5 }
5y^{5}z^{8}
we can cross out the y^{5 }in the numerator and denominator as a simplifying factor.
so our answer would be
14
5z^{8}
When dividing by a fraction, you multiply by the reciprocal. This means that you "flip" the second fraction (the numerator becomes the denominator and the nominator becomes the numerator)
2y^{5} divided by
5y^{5}z^{4} Becomes
2y^{5} Times 7
z^{4} 7 z^{4} 5y^{5}z^{4}
Since the numerator of the first fraction and the denominator of the second fraction both have a y^{5} term, they will cancel. No other factors occur in both a numerator and a denominator, so no other terms cancel.
When we multiply straight across, we get the answer:
14
5z^{8}