let's recall that the vertical asymptotes for a rational expression occur when the denominator is at 0, so let's zero out this one and check.
![\bf \cfrac{5x+5}{x^2+x-2}\qquad \stackrel{\textit{zeroing out the denominator}~\hfill }{x^2+x-2=0\implies (x+2)(x-1)=0}\implies \stackrel{\textit{vertical asymptotes}}{ \begin{cases} x=-2\\ x=1 \end{cases}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B5x%2B5%7D%7Bx%5E2%2Bx-2%7D%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bzeroing%20out%20the%20denominator%7D~%5Chfill%20%7D%7Bx%5E2%2Bx-2%3D0%5Cimplies%20%28x%2B2%29%28x-1%29%3D0%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bvertical%20asymptotes%7D%7D%7B%20%5Cbegin%7Bcases%7D%20x%3D-2%5C%5C%20x%3D1%20%5Cend%7Bcases%7D%7D)
We have been given a table that shows a linear relationship between x and y. We are asked to find the rate of change change of y with respect to x.
To solve our given problem, we will find the slope of the line passing through the given points.
![m=\frac{y_2-y_1}{x_2-x_1}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7By_2-y_1%7D%7Bx_2-x_1%7D)
Let us find slope of line using points
and
.
![m=\frac{15-96}{-2-(-20)}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B15-96%7D%7B-2-%28-20%29%7D)
![m=\frac{-81}{-2+20}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B-81%7D%7B-2%2B20%7D)
![m=\frac{-81}{18}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B-81%7D%7B18%7D)
![m=-4.5](https://tex.z-dn.net/?f=m%3D-4.5)
Therefore, the rate of change of y with respect to x is
.
Answer:
100 Sq Mt = 1135.69 Sq Ft
Step-by-step explanation:
We are required to find how many square feet are in 100 m squared.
1 meter = 3.37 ft
1 metre x 1 meter = 3.37 x 3.37 Sq ft
1 Sq Mt = 11.3569 Sq ft
100 sq Mt = 100 x 11.3569 sq ft
100 sq mt = 1135.69 Sq ft
Hence
100 Sq Mt = 1135.69 Sq Ft
Answer:
C
Step-by-step explanation: