Answer:

Step-by-step explanation:
Radius = r = 3 in.
Height = h = 2 in.
<u>Volume:</u>
![\sf Volume \ of \ the \ cylinder = \pi r^2 h\\\\V = (3.14)(3)^2(2)\\\\V =(3.14)(9)(2)\\\\V = (3.14)(18)\\\\V = 56.5 \ in.^3\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Csf%20Volume%20%5C%20of%20%5C%20the%20%5C%20cylinder%20%3D%20%5Cpi%20r%5E2%20h%5C%5C%5C%5CV%20%3D%20%283.14%29%283%29%5E2%282%29%5C%5C%5C%5CV%20%3D%283.14%29%289%29%282%29%5C%5C%5C%5CV%20%3D%20%283.14%29%2818%29%5C%5C%5C%5CV%20%3D%2056.5%20%5C%20in.%5E3%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
The point-slope equation of a line that cointains the point (x1,y1) and has a slope of m is
y-y1=m(x-x1)
so
point (-2,3) and slope -4
y-3=-4(x-(-2))
y-3=-4(x+2) is da equation
Omg don’t catfish it’s so damaging to your own self confidence!
Answer:

Step-by-step explanation:
We are asked to find the equation of mid-line of the given sinusoidal function.
Since the mid-line of a sinusoidal function is the line that runs between the maximum and minimum y-values of the function. We can consider it the middle y-value.

We can see from our given graph that the maximum value of our function is 5 and minimum value of our function is -5.
Upon substituting these values in mid-line formula we will get,

Therefore, the equation of the mid-line of the given sinusoidal function is
.