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OLga [1]
3 years ago
5

Which shape , if rotated 90 degrees will coincide within itself

Mathematics
1 answer:
Likurg_2 [28]3 years ago
7 0
THE ANSWER IS A
SQUARE
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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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.

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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} }

.

.

.

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

brainly.com/question/13057266

#SPJ4

3 0
1 year ago
HOW MANY TRIANGLES ARE THERE? INCLDNG OVERLAPPING TRIANGLES. I NEED ANSWERS. PLEASE HELP. TYIA​
miss Akunina [59]

Answer:

The answer is 48.

Step-by-step explanation:

There are 4 triangles in each square. So multiply 4 by 12 because thats how many squares there are.

5 0
3 years ago
Consider the two functions. F(x)=x^2-8x+7
Tpy6a [65]
(X-7)(x-1)

Minimal x values=7,1
Not the same x value
7 is the greater minimal value
7 0
3 years ago
It is between yes or no
Vlad1618 [11]

Answer:

yes

Step-by-step explanation:

5 0
3 years ago
1. is he equation true , false, or open
RideAnS [48]

Question 1:

For this case we have the following equation:

 5x ^ 2 = 6x + 11

We observe that the equation is a polynomial of the second degree.

Therefore, the equation has two possible solutions.

The variable of the equation for this case is x.

In this case, we have an open equation because we have a variable.

Rewriting the equation we have:

5x ^ 2 - 6x - 11 = 0

Answer:

open, there is a variable.


Question 2:

For this case we have the following equation:

 2-8x = -6

From here, we must clear the value of x.

For this, we follow the following steps:

1) Pass the value of 8x adding:

2 = -6 + 8x

2) Pass the value of 6 adding:

2 + 6 = 8x

3) Add the values ​​on the same side of equality:

 8x = 8

4) Pass the 8 to divide:

x = \frac{8}{8} = 1

Answer:

The value of x is given by:

x = 1


Question 3:

For this case we have the following equation:

 y = x-4

The ordered pair solution is the one that satisfies both sides of the equation.

For the point (6, 2) we have:

x = 6  y = 2

Substituting values ​​we have:

2 = 6-4  2 = 2

We observe that both sides of the equation are equal.

Therefore, the ordered pair solution is (6, 2)

Answer:

An ordered pair that is a solution of the equation y = x-4 is:

(6, 2)

3 0
3 years ago
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