m(2 + n – m) + 3(3n + m2 – 1) =
33/45 divided by 63/25
m(2 + n – m) + 3(3n + m2 – 1) =
33/45 divided by 63/25
I think you meant simplify the 1st and evaluate the 2nd. If that is the case then for the 1st problem you begin by distributing the m and the 3, this should give you:
m(2 + n - m) + 3(3n + m^{2} - 1) = 2m + nm - m^{2} + 9n + 3m^{2} - 3.
Next combine like terms, this will give you:
2m + nm - m^{2} + 9n + 3m^{2} - 3 = 2m^{2} + 2m + mn + 9n - 3. Since there are no more like terms to combine you are done.
to divide 33/45 by 63/25 we instead multiply 33/45 by the reciprocal of 63/25, which is 25/63. We do this because multiplying fractions is much simpler than dividing them.
So we have (33/45)/(63/25) = (33/45)*(25/63).
To multiply fractions we simply multiply the numerator with the numerator and denominator with the denominator. Before you do that it is always a good idea to see if anything can cancel and in this problem you should notice that the 33 and 63 can cancel and give you 11 and 21 and the 25 and 45 cancel to give 5 and 9.
After the canceling the problem becomes: (11/9)*(5/21) = 55/189
Thanks Xavier. The first was a big help. When l did the second problem just as you did before l just didnt cancel. Therefore my number was a lot higher after l multiplied the two. Its been since the 90s since l did this stuff and apparently l forgot a lot. I cant believe l am starting over with Algebra!
I'm glad I was helpful! I completely understand having to try to do math you haven't done in forever, I work as a tutor at my local community college and a lot of the students there face the same situation.
Thats what lm doing to my friend. Trying to get my business degree after all these years and have to brush en up on math before l take a placement test. This way l dont end up taking a bunch of math l already have. Errrrrg! Thanks again man, really.
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Just to clarify, is the full equation: m(2+n-m)+3(3n+m2-1)=(33/45)/(63/25)
Um, the two are seperate...lm really behind with this stuff!
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