let's firstly conver the mixed fractions to improper fractions and then get their product.
![\stackrel{mixed}{4\frac{1}{2}}\implies \cfrac{4\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{9}{2}} ~\hfill \stackrel{mixed}{2\frac{1}{2}}\implies \cfrac{2\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{5}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{9}{2}\cdot \cfrac{5}{2}\cdot 6\implies \cfrac{270}{2}\implies 135](https://tex.z-dn.net/?f=%5Cstackrel%7Bmixed%7D%7B4%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B4%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B9%7D%7B2%7D%7D%20~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B2%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B5%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B9%7D%7B2%7D%5Ccdot%20%5Ccfrac%7B5%7D%7B2%7D%5Ccdot%206%5Cimplies%20%5Ccfrac%7B270%7D%7B2%7D%5Cimplies%20135)
hmmm I take it that one can write that mixed as
.
is valid, not that it makes any sense.
A is true, b is true, but c d aren’t and e is
1/3 + 5 + 1/2 = total money he had originally
to add fraction we should first find a common denominator for every term.
1/3 + 10/2 + 1/2 = total
1/3 + 11/2 = total
2/6 + 33/6 = total
35/6 = total
Answer:
Q 3/10 R 4/25 S 3/50 T 12/25
Step-by-step explanation:
Q 15/50 3/10
R 8/50 4/25
S 3/50 3/50
T 24/ 50 12/25
Answer:
Step-by-step explanation:
First thing we need to do is to reconcile that r to the negative power. You do that by following the rules for exponents. If
then the rule is to subtract the lower exponent from the upper:

Doing that with our base of r:
is
which is 
So it looks like what we have now is:

NOW we will follow the rules for multiplying like bases with exponents. The rule for that is if you have like bases and you are multiplying those like bases, you add the exponents. So I'm going to kind of break up that problem and put the like bases together:

Working on the left side, p to the 3rd times p to the 4 gives you p to the 7th, which is what we have on the right. The r to the 7 is also the same on the left and the right. Working on the q's now, if we have q to the 2nd and we are looking for n, then
so 2 + n = 5 and n = 3