Using the online tool, find two different combinations of radius and height that produce two cylinders with nearly the same volume. record the dimensions and volumes of the two cylinders below


hence, the required result is b-a=\left[\begin{array}{ccc}1& 1& -2\\1& 1& -2\\1& 1& -2\end{array}\right]

step-by-step explanation:

suppose we have two matrices a and b

we will assume the two matrices

a=\left[\begin{array}{ccc}1& 2& 3\\4& 5& 6\\7& 8& 9\end{array}\right]

b=\left[\begin{array}{ccc}2& 3& 1\\5& 6& 4\\8& 9& 7\end{array}\right]

we have to compute b-a which means subtracting corresponding elements of a from b.

b=\left[\begin{array}{ccc}2& 3& 1\\5& 6& 4\\8& 9& 7\end{array}\right]-a=\left[\begin{array}{ccc}1& 2& 3\\4& 5& 6\\7& 8& 9\end{array}\right]

b-a=\left[\begin{array}{ccc}1& 1& -2\\1& 1& -2\\1& 1& -2\end{array}\right]

they used 313 gallons of paint

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.


step-by-step explanation:

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