Answer:

Explanation: For this, it is often best to find the horizontal asymptote, and then take limits as x approaches the vertical asymptote and the end behaviours.
Well, we know there will be a horizontal asymptote at y = 0, because as x approaches infinite and negative infinite, the graph will shrink down closer and closer to 0, but never touch it. We call this a horizontal asymptote.
So we know that there is a restriction on the y-axis.
Now, since we know the end behaviours, let's find the asymptotic behaviours.
As x approaches the asymptote of 7⁻, then y would be diverging out to negative infinite.
As x approaches the asymptote at 7⁺, then y would be diverging out to negative infinite.
So, our range would be:
Answer:
-24
Step-by-step explanation:
(-6)(4)
The indicated operation is multiplication
-24
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.
Step-by-step explanation:
The problem states that you have a linear function so expect your equation to have this form:
y = mx + b
where m is the slope and b is the y-intercept. You are also given two points: P1(5, 6) and P2(14, 60). Use these points to solve for the slope m.
m = (y2 - y1) / (x2 - x1) = (60 - 6)/(14 - 5)
= 54/9 = 6
So our equation now becomes
y = 6m + b
To solve for b, plug in the values of P1:
6 = 6(5) + b ---> b = -24
Therefore, our equation is
y = 6m - 24
The rest of the points are
(8, 24)
(11, 42)