12m + 2n - 3m + 11n = 9m + 13n
5m+2n, n-3m, 7-3n (5m+2n)+(n-3m)+(7-3n)=2m+7
sum = 2n + 3m
96m3
To find the factors of the trinomial (3m^2 + 11mn + 6n^2), we need to break it down into two binomials. First, we find two numbers that multiply to the product of the leading coefficient and constant term, which are (3 \times 6 = 18). Then, we look for two numbers that add up to the middle coefficient, which is 11. The factors are ((3m + 2n)(m + 3n)).
6(3m - 2n)(3m + 2n)
12m + 2n - 3m + 11n = 9m + 13n
5m+2n, n-3m, 7-3n (5m+2n)+(n-3m)+(7-3n)=2m+7
34
It can be simplified to: 4m-2n
18 - 16 = 2
3m*3m*2m=300*300*200=18,000,000cm3
sum = 2n + 3m
Work = (force) x (distance) (2 x 3) = (3 x 2) The work in the two cases described is equal. The answer is "no".
2n x 2n x 2n = 8n^3
The volume of a shape 3m x 3m x 2m is 18m3.
I will assume that we are taking d/dx, not d/dn. There are two ways to interpret what you asked. First way is (sinx)^(2n). Second way is sin(x^(2n)). First answer: 2n(sinx)^(2n-1)(cosx)=2ncosx(sinx)^(2n-1). Second answer: cos(x^(2n))(2nx^(2n-1)).