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professor190 [17]
3 years ago
15

Help please i dont wanna pull an all-nighter ill give 20 points i really dont care at this point

Mathematics
1 answer:
vagabundo [1.1K]3 years ago
6 0

Answer:

y = 38, x = 90, x = 10, u = 6.2, w = 52, and v = 7.9

Step-by-step explanation:

Since the triangles are congruent, the angle and side lengths are also congruent. (CPCTC) Therefore, y = 38, x = 90, x = 10, u = 6.2, and v = 7.9. For w, we need to use the sum of angles in a triangle. So, we do 180 - 38 - 90 = w

w = 52.

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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}

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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!.............

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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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