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MaRussiya [10]
3 years ago
7

Find the exact value of x.

Mathematics
2 answers:
Anna007 [38]3 years ago
5 0
It would probably be 7.2111
oksian1 [2.3K]3 years ago
4 0

Answer:

The exact value of x will be \sqrt{52} or 7.2111

Step-by-step explanation:

Hope this Helped

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Write as a single power of 6<br> (6^5)^4
babunello [35]

Answer:

6^20

Step-by-step explanation:

When two indices are in brackets, you just have to multiply them.

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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
RUDIKE [14]

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

#SPJ4

3 0
1 year ago
Hey I need help !!!
sweet-ann [11.9K]

Answer:

(2, -3)

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Which of the following points does not lie on the graph of y=5x-5?<br> (0-5)<br> (1.0)<br> (-5.0)
mojhsa [17]
Y=5x-5
0=5x-5
-5x=-5

X=1
8 0
3 years ago
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