Answer:
A. 0.25 (rounded)
B. 0.19 (rounded)
C. 0.17 (rounded)
D. 0.17 (rounded)
Step-by-step explanation:
Hope that helped!
Sounds to me as tho you are to graph 3x+5y<10, and that after doing so you are to restrict the shaded answer area created by the "constraint" inequality x≤y+1. OR x-1 ≤ y OR y≥x-1. If this is the correct assumption, then please finish the last part of y our problem statement by typing {x-y<=1}.
First graph 3x+5y = 10, using a dashed line instead of a solid line.
x-intercept will be 10/3 and y-intercept will be 2. Now, because of the < symbol, shade the coordinate plane BELOW this dashed line.
Next, graph y=x-1. y-intercept is -1 and x intercept is 1. Shade the graph area ABOVE this solid line.
The 2 lines intersect at (1.875, 0.875). To the LEFT of this point is a wedge-shaped area bounded by the 2 lines mentioned. That wedge-shaped area is the solution set for this problem.
Answer:
12.6
Step-by-step explanation:
To find the area of a cylinder, you find the area of the circular base and multiply it by the height. Remember that the area of a circle is pi*r^2. R, the radius, is 2. Putting this into a calculator and rounding to the nearest tenth, you get 12.6. Now that we have the circle, we multiply by the height. Since the height is 1, we have our answer of 12.6.
"determine the location" or namely, is it inside the circle, outside the circle, or right ON the circle?
well, we know the center is at (1,-5) and it has a radius of 5, so the distance from the center to any point on the circle will just be 5, now if (4,-1) is less than that away, is inside, if more than that is outiside and if it's exactly 5 is right ON the circle.
well, we can check by simply getting the distance from the center to the point (4,-1).
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{center}{(\stackrel{x_1}{1}~,~\stackrel{y_1}{-5})}\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{-1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d = \sqrt{[4-1]^2+[-1-(-5)]^2}\implies d=\sqrt{(4-1)^2+(-1+5)^2} \\\\\\ d = \sqrt{3^2+4^2}\implies d =\sqrt{9+16}\implies d=\sqrt{25}\implies \stackrel{\textit{right on the circle}}{d = 5}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%5Cstackrel%7Bcenter%7D%7B%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B-5%7D%29%7D%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B4%7D~%2C~%5Cstackrel%7By_2%7D%7B-1%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%20%3D%20%5Csqrt%7B%5B4-1%5D%5E2%2B%5B-1-%28-5%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%284-1%29%5E2%2B%28-1%2B5%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%20%3D%20%5Csqrt%7B3%5E2%2B4%5E2%7D%5Cimplies%20d%20%3D%5Csqrt%7B9%2B16%7D%5Cimplies%20d%3D%5Csqrt%7B25%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bright%20on%20the%20circle%7D%7D%7Bd%20%3D%205%7D)
