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What is the nth term of the sequence 6 0 -6 -12?
There are infinitely many polynomials of order 4 (or higher) that will give these as the first four numbers and any one of these could be "the" rule. Short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.The simplest rule is a linear polynomial U(n) = 6*(2 - n) for n = 1 , 2, 3, ...
Why you can you do n-12 but not 12-n?
The subtraction is not commutative, it means that n - 12 is not the same as 12 - n, except when n is 12. For example, suppose that we have a statement such as n - 12 = 12 - n. Can we find a value for n which will satisfy our statement?n - 12 = 12 - n (add n and 12 to both sides)n + n - 12 + 12 = 12 + 12 - n + n2n = 24 (divide by 2 both sides)n = 12Check:n - 12 = 12 - n12 - 12 =? 12 - 120 = 0 TrueBut, for all other values of n, the statement is not true. Let's say that n = 20Does 20 satisfies the statement? Check:n - 12 = 12 - n20 - 12 =? 12 - 208 = -8 FalseHence, we cannot write n - 12 as 12 - n for n ≠12.
Find three consecutive odd integers such that the product of the first and the second exceeds the third by 8?
Let first COI=n=3 second COI=n+2=5 third COI=n+4=7 (n)(n+2)=n+4+8 check: n^2+2n=n+12 n^2+n-12=0 -n-12 -n-12 -4^2+(-4)-12=0 n^2+2n-n-12=0 16+(-4)-12=0 n^2+n-12=0 12-12=0 (n+4)(n-3)=0 0=0 n+4=0 n-3= 0 n^2+n-12=0 -4 -4 +3 +3 3^2+3-12=0 n=-4 n=3 9+3-12=0 reject 12-12=0 because is negative and it needs to be positive 0=0
-3(n plus 5)12?
If you mean: -3(n+5) = 12 then n works out as -9 because -3(-9+5) = 12
How do you factor n squared minus 16n plus 48?
(n - 4)(n - 12)
