When factoring, it's easier to "pick out" the GCF first because it makes it simpler to factor even further. Otherwise, you will be troubled with lots of numbers and variables, so it will just get confusing.
Answer:
630/30=21 so 21 stacks
Step-by-step explanation:
Answer: it’s D
Step-by-step explanation:
I guessed and got it right
Answer:
6x^2 +12x -1
Step-by-step explanation:
The perimeter is the sum of the two leg lengths and the base length:
P = 2(3x^2 +4x +2) +(4x -5)
= 6x^2 +8x +4 +4x -5
P = 6x^2 +12x -1
Answer:
a) n<1 and n>5
b) 0 < n < -4
c) n > 2 and n < -2
Step-by-step explanation:
The signal is given by x[n] = 0 for n < -1 and n > 3
The problem asks us to determine the values of n for which it's guaranteed to be zero.
a) x[n-2]
We know that n -2 must be less than -1 or greater than 3.
Therefore we're going to write down our inequalities and solve for n
![n-2](https://tex.z-dn.net/?f=n-2%3C-1%5C%5Cn%3C-1%2B2%5C%5Cn%3C1%5C%5C%5C%5Cn-2%3E3%5C%5Cn%3E5)
Therefore for n<1 and n>5 x [n-2] will be zero
b) x [n+ 3]
Similarly, n + 3 must be less than -1 or greater than 3
![n+30](https://tex.z-dn.net/?f=n%2B3%3C-1%5C%5Cn%3C-1-3%5C%5Cn%3C-4%5C%5C%5C%5Cn%2B3%3E3%5C%5Cn%3E3-3%5C%5Cn%3E0)
Therefore for n< -4 and n>0, in other words, for 0 < n < -4 x[n-2] will be zero
c)x [-n + 1]
Similarly, -n+1 must be less than -1 or greater than 3
![-n+13-1\\-n>2\\n](https://tex.z-dn.net/?f=-n%2B1%3C-1%5C%5C-n%3C-1-1%5C%5C-n%3C-2%5C%5Cn%3E2%5C%5C%5C%5C-n%2B1%3E3%5C%5C-n%3E3-1%5C%5C-n%3E2%5C%5Cn%3C-2)
Therefore, for n > 2 and n < -2 x[-n+1] will be zero