Answer:
i think it says to find area
Step-by-step explanation:
18 + 81 = 9(x²<span> + 6x + 9)
</span><span>11 = (x + 3)</span>²
When we are completing the square, we are going to move the value of c across the equals. We will do that by adding, and end up with
18=9(x²+6x)
We take the value of b (the coefficient of x), divide it by 2 and square it:
(6/2)²=3²=9
This is the value that completes the square. However, since the entire square is multiplied by 9, this value must be multiplied by 9 before we can add it across the equals:
18+9(9) = 9(x²+6x+9)
18+81=9(x²+6x+9)
99=9(x²+6x+9)
Dividing both sides by 9, we have:
11=x²+6x+9
11=(x+3)²
Answer:
just for the points
Step-by-step explanation:
Answer:
about 5 ounces
Step-by-step explanation:
Answer:converge at 
Step-by-step explanation:
Given
Improper Integral I is given as

integration of
is -
![I=\left [ -\frac{1}{x}\right ]^{\infty}_3](https://tex.z-dn.net/?f=I%3D%5Cleft%20%5B%20-%5Cfrac%7B1%7D%7Bx%7D%5Cright%20%5D%5E%7B%5Cinfty%7D_3)
substituting value
![I=-\left [ \frac{1}{\infty }-\frac{1}{3}\right ]](https://tex.z-dn.net/?f=I%3D-%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B%5Cinfty%20%7D-%5Cfrac%7B1%7D%7B3%7D%5Cright%20%5D)
![I=-\left [ 0-\frac{1}{3}\right ]](https://tex.z-dn.net/?f=I%3D-%5Cleft%20%5B%200-%5Cfrac%7B1%7D%7B3%7D%5Cright%20%5D)

so the value of integral converges at 