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OpenStudy (theslytherinhelper):

Algebra 2

OpenStudy (theslytherinhelper):

OpenStudy (phi):

surely you can simplify the numbers part ? what do only the numbers simplify to ?

OpenStudy (theslytherinhelper):

I can't :/ I tried doing it, and then when I attempted to cancel the numbers, they didn't come out to the answer choices. I kept changing them around and got: (2m^2)(2m) (3n) (3n) x 3m^2n (15mn)(15mn)

OpenStudy (phi):

\[\frac{4 \cdot 9}{3 \cdot 30 } \]

OpenStudy (phi):

can you simplify that ? (if there is the same number in the top and bottom, you can "cancel" it )

OpenStudy (phi):

if you can divide the same number into the top and bottom, you can simplify it

OpenStudy (phi):

if it is not clear, factor each number into its prime factors

OpenStudy (theslytherinhelper):

\[4\times9/10?\]

OpenStudy (phi):

start with \[ \frac{4 \cdot 9} {3 \cdot 30 } \] now factor the 4 into 2*2 and the 9 into 3*3 \[ \frac{2 \cdot 2 \cdot 3 \cdot3} {3 \cdot 30 } \] in the bottom, factor 30 into 3*10, and the 10 into 2*5 so 30 is 3*2*5 \[ \frac{2 \cdot 2 \cdot 3 \cdot3} {3 \cdot 2 \cdot 3 \cdot 5 } \]

OpenStudy (phi):

now if you see the same number in the top and bottom, cancel both of them do that for each pair of numbers you can match up.

OpenStudy (theslytherinhelper):

Since we have to switch them around first, is it 5/2?

OpenStudy (phi):

nothing to do with switching around. if you have \[ \frac{7 \cdot 7}{3 \cdot 7 } = \frac{\cancel{7} \cdot 7}{3 \cdot \cancel{7}} \] because you can match up a 7 in the top with a 7 in the bottom, and take both out.

OpenStudy (phi):

now do that with \[ \frac{2 \cdot 2 \cdot 3 \cdot3} {3 \cdot 2 \cdot 3 \cdot 5 } \]

OpenStudy (theslytherinhelper):

\[\ \frac{ 2 }{ 3} x \frac{ 3 }{ 5 }\]

OpenStudy (phi):

how many 2's are in the top and how many are in the bottom in \[ \frac{2 \cdot 2 \cdot 3 \cdot3} {3 \cdot 2 \cdot 3 \cdot 5 } \]

OpenStudy (theslytherinhelper):

I was thinking it'd end up as \[\frac{ 2 }{ 5 }\]

OpenStudy (theslytherinhelper):

which is an answer choice ~ we just have to add the letters

OpenStudy (phi):

to be sure, answer how many 2's are up top and how many in the bottom ?

OpenStudy (theslytherinhelper):

2's in the top - 2 2's at the bottom - 1

OpenStudy (phi):

after you cancel the pair of 2's (one from the top and one from the bottom) how many 2's are left in the top ? how many 2's left in the bottom ?

OpenStudy (phi):

the reason to go through this, is this will be how we do the letters

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