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Kisachek [45]
3 years ago
5

Which solid has congruent horizontal and vertical cross sections?

Mathematics
1 answer:
Zinaida [17]3 years ago
3 0

Answer:

triangular prism (pyramid) and cylinder.

Step-by-step explanation:

A pyramid is a polyhedron with one base that is a polygon and faces that are triangles.

A cylinder is a three-dimensional figure with congruent, parallel bases that are circles connected with a curved side.

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Find the taylor series for f(x) centered at the given value of a. [assume that f has a power series expansion. do not show that
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The taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 (x+4)^{2}/2!+...........

Given a function f(x)=9/x,a=-4.

We are required to find the taylor series for the function f(x)=8/x centered at the given value of a and a=-4.

The taylor series of a function f(x)=f(a)+f^{1}(a)(x-a)/1!+ f^{11}(a)(x-a)^{2} /2! +f^{111}(a)(x-a)a^{3}/3!+..........

Where the terms in f prime f^{1}(a) represent the derivatives of x valued at a.

For the given function.f(x)=8/x and a=-4.

So,f(a)=f(-4)=8/(-4)=-2.

f^{1}(a)=f^{1}(-4)=-8/(-4)^{2}

=-8/16

=-1/2

The series of f(x) is as under:

f(x)=f(-4)+f^{1}(-4)(x+4)/1!+  f^{11}(-4)(x+4)^{2}/2!.............

=8/(-4)-8/(-4)^{2} (-4)(x+4)/1!+  24/(-4)^{3} (-4)(x+4)^{2}/2!.............

=-2+2(x+4)/1!-24/16 (x+4)^{2}/2!+...........

Hence the taylor series for the f(x)=8/x centered at the given value of a=-4 is -2+2(x+4)/1!-24/16 (x+4)^{2}/2!+...........

Learn more about taylor series at brainly.com/question/23334489

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