Algebra 2
surely you can simplify the numbers part ? what do only the numbers simplify to ?
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)
\[\frac{4 \cdot 9}{3 \cdot 30 } \]
can you simplify that ? (if there is the same number in the top and bottom, you can "cancel" it )
if you can divide the same number into the top and bottom, you can simplify it
if it is not clear, factor each number into its prime factors
\[4\times9/10?\]
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 \[ \frac{2 \cdot 2 \cdot 3 \cdot3} {3 \cdot 2 \cdot 3 \cdot 5 } \]
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.
Since we have to switch them around first, is it 5/2?
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.
now do that with \[ \frac{2 \cdot 2 \cdot 3 \cdot3} {3 \cdot 2 \cdot 3 \cdot 5 } \]
\[\ \frac{ 2 }{ 3} x \frac{ 3 }{ 5 }\]
how many 2's are in the top and how many are in the bottom in \[ \frac{2 \cdot 2 \cdot 3 \cdot3} {3 \cdot 2 \cdot 3 \cdot 5 } \]
I was thinking it'd end up as \[\frac{ 2 }{ 5 }\]
which is an answer choice ~ we just have to add the letters
to be sure, answer how many 2's are up top and how many in the bottom ?
2's in the top - 2 2's at the bottom - 1
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 ?
the reason to go through this, is this will be how we do the letters
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