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How do you solve 17 plus 2n equals 21 plus 2n?
You can't. 17 + 2n = 21 + 2n gives: 2n = 4 + 2n which gives 0=4 which is not possible so the sum is not solveable.
What is 2n plus 2?
Exactly that. 2n+2.
What is the factor of 4n2 plus 28n plus 49?
(2n + 7)(2n + 7)
What is the value of n in the expession x cubed plus x squared minus 2nx plus n squared which has a remainder of 8 when divided by x minus 2?
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
What are three consecutive even integers whose sum is 108?
Let 2n be the first integer, then the next on is 2n+2 and the one after is 2n+4 so adding all three we have 6n+6=108 or 6n=102 and we want 2n so divide we divide 6n by 3 and we have 2n=34, and 2n+2=36 and lastly 2n+4=38 so 34, 36, 38 are the even integers we seek.
