Question

PLEASE HELP!!!!!

Answer 1

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Answer 2

Original expression\[\frac{ 45m ^{2}n ^{2} }{ 18m ^{4}n ^{3} }\]
This can be solved in one line, but for reasons of explaining, I will break it into smaller parts.

The expression can be broken up like so: \[\frac{ 45 }{ 18 } \times \frac{ m^{3} }{ m ^{4} } \times \frac{ n ^{4} }{ n ^{3} }\]
Each of these terms may be reduced. Let's begin by reducing the first term. Both 45 and 18 have a few common factors. We are looking for the greatest common factor (GCF), that is the largest number both 45 and 18 can be divided by,. because 18 does not divide completely into 45. The GCF for both 45 and 18 is 9. That is, divide 45 by 9 and you get 5, and divide 18 by 9 and you get 2. This leaves your first term as\[\frac{ 5 }{ 2 }\]

Second term also has a few common factors. Again we look for the GCF. The numerator has 2 m's in it, symbolized by the exponential 2. The denominator has 4 m's. In that case you can only take 2 m's out of both. Since the top has less, it cancels out completely, leaving you with a value of 1 in the numerator, and 2 m's in the denominator. \[\frac{ 1 }{ m ^{2} }\]

The same goes for the third term, except that it has more n's in the numerator. Therefore: \[\frac{ n }{ 1 }\]

Now your expression is \[\frac{ 5 }{ 2 } \times \frac{ 1 }{ m ^{2} } \times \frac{ n }{ 1 }\]
If you then carry out the operations, you get \[\frac{ 5n }{ 2m ^{2} }\] which is your answer.

Understand? Any questions?

Answer 3

u got the answer