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stealth61 [152]
2 years ago
9

What is the shape of the cross section taken parallel to the base of a cylinder? circle rectangle triangle ellipse

Mathematics
1 answer:
Naya [18.7K]2 years ago
5 0

The option A is correct. A circle is the shape of the cross section taken parallel to the base of a cylinder.

According to the statement

We have given that the cylinder and we have to tell that the which shape required for the cross section taken parallel to the base of a cylinder.

So, For this purpose,

We know that the

A cylinder is a three-dimensional solid, one of the most basic geometric shapes.

In this shape the surface formed by the points at a fixed distance from a  line segment, which is  known as the axis of the cylinder.

As the bases of cylinder are two identical circles which are parallel to the curved surface.

The curved surface when we open will be circle rather than the circle.

So, Due to this reason the answer is a circle.

Therefore, a circle is the shape of the cross section taken parallel to the base of a cylinder.

Learn more about base of cylinder here

brainly.com/question/76387

#SPJ1

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Eduardwww [97]
Part A. You have the correct first and second derivative.

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Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.

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Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out

To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
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so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0. 
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well. 
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Answer:

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Step-by-step explanation:

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