tn = n2 - 3 Thus: t1 = 12 - 3 = 1 - 3 = -2 t2 = 22 - 3 = 4 - 3 = 1 t3 = 32 - 3 = 9 - 3 = 6 t4 = 44 - 3 = 16 - 3 = 13 So the first four terms of n2 -3 are -2, 1, 6 & 13.
5*n2
(m2n2)2 = m4n4
n x n2 = n3
5
Oh, dude, adding n squared plus n squared is like adding apples to apples, you know? It's just like, you take two n squared terms and you add them together to get 2n squared. It's not rocket science, man. Just double up those n squares and you're good to go.
tn = n2 - 3 Thus: t1 = 12 - 3 = 1 - 3 = -2 t2 = 22 - 3 = 4 - 3 = 1 t3 = 32 - 3 = 9 - 3 = 6 t4 = 44 - 3 = 16 - 3 = 13 So the first four terms of n2 -3 are -2, 1, 6 & 13.
5*n2
(m2n2)2 = m4n4
n2+n
n x n2 = n3
(n2 + 2n - 1) (n2 + 2n - 1) = n4 + 2n3 - n2 + 2n3 + 4n2 - 2n - n2 - 2n + 1 = n4 + 4n3 + 2n2 - 4n + 1 try with n = 5: (5 squared + 10 - 1) squared = 34 squared = 1156 with formula (5^4) + (4 *(5^3)) + (2 * (5^2)) - (4 * 5) + 1 = 625 + 500 + 50 - 20 + 1 = 1156
5
n is 2. To solve, do the division: . . . . . . . . .x2 +. 3x + (6-2n) . . . -------------------------- x-2 | x3 + x2 - 2nx + n2 . . . . .x3 -2x2 . . . . .-------- . . . . . . . .3x2 - 2nx . . . . . . . .3x2 - . 6x . . . . . . . .---------- . . . . . . . . . (6-2n)x + n2 . . . . . . . . . (6-2n)x - 2(6-2n) . . . . . . . . . ---------------------- . . . . . . . . . . . . . . . .n2 + 2(6-2n) But this remainder is known to be 8, so: n2 + 2(6-2n) = 8 ⇒ n2 - 4n + 4 = 0 ⇒ (n - 2)2 = 0 ⇒ n = 2
n2 - 100 = (n + 10)(n - 10)
the formula is n2 divided by two add n divided by 2
A trinomial is an expression that consist of three terms (first term, middle term, and last term). The middle term is the sum of the product of outer terms and inner terms of the binomial.