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Marina CMI [18]
3 years ago
9

Can someone help? ASAP!? (16 pts)

Mathematics
2 answers:
nika2105 [10]3 years ago
6 0
D in (-00+00)
(3/8)*d-(3/4)=0
(3/8)*d-3/4=0
3/8*d-3/4=0 // + 3/4
3/8*d= 3/4 // : 3/8
d = 3/4/3/4
d=2
d=2
Ugo [173]3 years ago
4 0
The coefficient of the variable is 3/8.

\sf\dfrac{3}{8}(?+?)

Divide 3/8 to each term to find out what goes in the parenthesis:

\sf\dfrac{3}{8}b\div\dfrac{3}{8}=b

So 'b' goes into the first spot in the parenthesis:

\sf\dfrac{3}{8}(b+?)

\sf\dfrac{3}{4}\div\dfrac{3}{8}

When dividing by fractions we flip(reciprocal) the one we're dividing by and multiply:

\sf\dfrac{3}{4}\times\dfrac{8}{3}

Multiply:

\sf\dfrac{24}{12}

Divide:

\sf 2

This goes into the second spot in our parenthesis.

\boxed{\sf\dfrac{3}{8}(b+2)}
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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
Bond [772]

Taylor series is f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

To find the Taylor series for f(x) = ln(x) centering at 9, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have

f(x) = ln(x)

f^{1}(x)= \frac{1}{x} \\f^{2}(x)= -\frac{1}{x^{2} }\\f^{3}(x)= -\frac{2}{x^{3} }\\f^{4}(x)= \frac{-6}{x^{4} }

.

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Since we need to have it centered at 9, we must take the value of f(9), and so on.

f(9) = ln(9)

f^{1}(9)= \frac{1}{9} \\f^{2}(9)= -\frac{1}{9^{2} }\\f^{3}(x)= -\frac{1(2)}{9^{3} }\\f^{4}(x)= \frac{-1(2)(3)}{9^{4} }

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Following the pattern, we can see that for f^{n}(x),

f^{n}(x)=(-1)^{n-1}\frac{1.2.3.4.5...........(n-1)}{9^{n} }  \\f^{n}(x)=(-1)^{n-1}\frac{(n-1)!}{9^{n}}

This applies for n ≥ 1, Expressing f(x) in summation, we have

\sum_{n=0}^{\infinite} \frac{f^{n}(9) }{n!} (x-9)^{2}

Combining ln2 with the rest of series, we have

f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

Taylor series is f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

Find out more information about taylor series here

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