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(8m) (2mn)^3

really having trouble

2007-01-25 05:13:22 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

5 answers

(8m) (2mn)^3
= (8m) (2^3)(m^3)(n^3)
= (8m)(8)(m^3)(n^3)
multiply 8*8 and add the exponents on the m's
= 64m^4 n^3

2007-01-25 05:17:34 · answer #1 · answered by MsMath 7 · 2 0

Start with (2mn)^3

Cube each term

2^3 = 8
m^3 = m^3 (yes, just that simple!)
n^3 = n^3

giving you 8m^3n^3

Now (8m)(8m^3n^3)...

(8)(8) = 64
(m)(m^3) = m^4
n^3 unchanged

Final answer is then 64m^4n^3

Just keep like terms together, you're getting it! ;-)

2007-01-25 05:18:32 · answer #2 · answered by MamaMia © 7 · 0 0

Here is the simple, but slightly tedious way to do it:
8*m*2*m*n*2*m*n*2*m*n*
8*2*2*2=64
m*m*m*m=m^4
n*n*n=n^3
so the answer is 64*(m^4)*(n^3)

2007-01-25 05:19:32 · answer #3 · answered by DJJ 1 · 0 0

as a results of fact it has an expression as denominator, i understand this as a rational variable expression. it is extremely now not a function - a function relates one volume to a special say x and y. this is an expression as a results of fact it does now not relate x to a minimum of something else.

2016-12-16 17:08:57 · answer #4 · answered by dorthy 4 · 0 0

64m^4n^3

2007-01-25 05:21:09 · answer #5 · answered by Ray 5 · 0 0

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