Number one is A
Number two is D
Answer:
The range of the graph is all real numbers greater than or equal to 0.
Step-by-step explanation:
we have

using a graphing tool
The graph in the attached figure
step 1
Find the domain
we know that the radicand of the function must be greater than or equal to zero
so
The domain is the interval ------> [0,∞)

All real numbers greater than or equal to zero
step 2
Find the range
The range is the interval -----> [0,∞)

All real numbers greater than or equal to zero
Answer:
It's (2, 3) and (7, 6)
Step-by-step explanation:
It's the only option where both x2 and y2 are both bigger than x1 and y1
It would be -2x2+11x-4 good luck
Answer:
The answer is below
Step-by-step explanation:
The volume of the cylinder with radius (r) and height (h) is given as:
Volume = πr²h
17 = πr²h
h = 17 / πr²
Let k represent the cost of the side and 2k the cost of the bottom.
The area of the side = 2πrh, hence the cost of the side = k(2πrh)
The area of the top and bottom = 2πr², hence the cost of the top and bottom = 2k(2πr²)
The total cost (T) = cost of side + cost of the top and bottom
T = k(2πrh) + 2k(2πr²)
T = 2kπ(rh + 2r²)
Put h = 17/πr²
T = 2kπ (r×17/πr² + 2r²)
T = 2kπ(17/πr + 2r²)
The minimal cost is at dT/dr = 0
![\frac{dT}{dr} =0=\frac{d}{dr} [2k\pi(\frac{17}{\pi r} +2r^2)]\\\\4r-\frac{17}{\pi r^2} =0\\\\4r=\frac{17}{\pi r^2}\\\\r^3=\frac{17}{4\pi }=1.35\\\\r=\sqrt[3]{1.35}\\\\r=1.1\ m\\\\h=\frac{17}{\pi r^2} =\frac{17}{\pi (1.1)^2} \\\\h=4.42\ m](https://tex.z-dn.net/?f=%5Cfrac%7BdT%7D%7Bdr%7D%20%3D0%3D%5Cfrac%7Bd%7D%7Bdr%7D%20%5B2k%5Cpi%28%5Cfrac%7B17%7D%7B%5Cpi%20r%7D%20%2B2r%5E2%29%5D%5C%5C%5C%5C4r-%5Cfrac%7B17%7D%7B%5Cpi%20r%5E2%7D%20%3D0%5C%5C%5C%5C4r%3D%5Cfrac%7B17%7D%7B%5Cpi%20r%5E2%7D%5C%5C%5C%5Cr%5E3%3D%5Cfrac%7B17%7D%7B4%5Cpi%20%7D%3D1.35%5C%5C%5C%5Cr%3D%5Csqrt%5B3%5D%7B1.35%7D%5C%5C%5C%5Cr%3D1.1%5C%20m%5C%5C%5C%5Ch%3D%5Cfrac%7B17%7D%7B%5Cpi%20r%5E2%7D%20%20%3D%5Cfrac%7B17%7D%7B%5Cpi%20%281.1%29%5E2%7D%20%5C%5C%5C%5Ch%3D4.42%5C%20m)