1st number = n
2nd number = n+1
3rd number = n+2
sum of the squares of 3 consecutive numbers is 116
n² + (n+1)² + (n+2)² = 116
n² + (n+1)(n+1) + (n+2)(n+2) = 116
n² + [n(n+1)+1(n+1)] + [n(n+2)+2(n+2)] = 116
n² + n² + n + n + 1 + n² + 2n + 2n + 4 = 116
n² + n² + n² + n + n + 2n + 2n + 1 + 4 = 116
3n² + 6n + 5 = 116 Last option.
I haven't done these problems in a really long time, so there's a chance that my answer might be wrong. I hypothesize that the correct answer is C.
Take into account, that in general, a cosine function of amplitude A, period T and vertical translation b, can be written as follow:
In the given case, you have:
A = 4
T = 3π/4
b = -3
By replacing you obtain:
Hence, the answer is:
f(x) = 4cos(8/3 x) - 3
Answer:
4
Step-by-step explanation:
2 + 2 = 4
In order to really understand this we need to break it down into simple terms.
Basically what we are doing here is adding two things together.
Put it like this, you have two fish, and you get another two fish, you now have four fish.
If you don't understand anything or need me to explain it better please don't hesitate to ask.
