what is
m = n^2 - 3n
m = -2n^2 - 3n
and when i put it into coordinate form i got 0,0 i just think i got it wrong.
If some1 wants to see the way i did it tell me and ill add it in
2007-01-01 15:40:23 · 6 answers · asked by mikyd 1 in Science & Mathematics ➔ Mathematics
.....I know how to solve the problem and i already got my answer i just wanted some1 to see if they got the same answer as me o.o.
2007-01-01 15:46:42 · update #1
m = n^2 - 3n ... .... (1)
m = -2n^2 - 3n
n^2 - 3n = -2n^2 - 3n
3 n^2 = 0
n^2 = 0
n = 0
Substituting n = 0 in relation (1)
m = 0
Ans.
m = 0
n = 0
2007-01-01 16:11:39 · answer #1 · answered by Sheen 4 · 0⤊ 0⤋
Write the whole problem out so that I can get a better understanding. If you want to solve for n for example you could say:
-2n^2 - 3n= n^2 - 3n
And then solve for n.
2007-01-01 23:44:27 · answer #2 · answered by Dalonna 2 · 1⤊ 1⤋
n^2-3n=-2n^2-3n, then we remove 3n term
so, n^2=-2n^2 we can see that n must be positive and negative so the only answer is n=0 for a real answer.
m=(0)^2-3*(0)=0
2007-01-01 23:46:22 · answer #3 · answered by A T 2 · 0⤊ 0⤋
n=0
m=1
n is 0 because it can't be nothing else and m is 1 caz 0 to the 2 power is 1- 0 still equals 1.
2007-01-01 23:46:01 · answer #4 · answered by Yemiko 2 · 0⤊ 0⤋
If indeed both equations are to be "true", yes the only solution you have is n = 0, and therefore m must = 0.
2007-01-01 23:52:25 · answer #5 · answered by answerING 6 · 0⤊ 0⤋
since m=both of them, just set them equal to each other, so that you only have n. Then you can solve for n using the quadratic equation or by simple factoring. I don't have paper or pencil now or i'd work it out.
2007-01-01 23:42:26 · answer #6 · answered by reiwo023-9085j 2 · 1⤊ 0⤋