Answer:
x^2 - 6x + 147 so the coefficient is 1
Step-by-step explanation:
First, simplify the equation.
1. 2/3x(2x-3) = (2x/3)(2x-3)-(x-7)(x+7)
2. 2x-3(2x/3) = (2x(2x-3)) - (x-7)(x+7)
3. Expand -(x-7)(x+7) = -x^2+49
Combine to get (x^2 - 6x)/3 +49
Simplify to get x^2 - 6x + 147. The coefficient of the quadratic term is one.
Answer:
2,7
Step-by-step explanation:
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:
(2,3)
Step-by-step explanation:
I did this by using the equation that was given. y=x+1.
This mean that the starting point would be the coordinate (0,1) and the x is the slope, which means that from coordinate (0,1) I will go up one and one to the right until I intersected with coordinate (2,3).