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AleksAgata [21]
3 years ago
8

5C%20%5Chuge%5Cmathfrak%5Cred%7B5%20%5E%7B2%7D%20%2B%206%20%5E%7B2%7D%20%20%3D%20%20%7B%3F%7D%20%20%7D" id="TexFormula1" title="\huge\mathfrak\green{ please \: help \: ...} \\ \\ \huge\mathfrak\red{5 ^{2} + 6 ^{2} = {?} }" alt="\huge\mathfrak\green{ please \: help \: ...} \\ \\ \huge\mathfrak\red{5 ^{2} + 6 ^{2} = {?} }" align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
Dmitry_Shevchenko [17]3 years ago
8 0

<u>6</u><u>1</u>

Answer:

5²+6²=5*5+6*6=25+36=<u>61</u><u> </u><u>i</u><u>s</u><u> </u><u>a</u><u> </u><u>required</u><u> </u><u>answer</u><u>.</u>

meriva3 years ago
3 0

Answer:

61 is the answer

Step-by-step explanation:

{5}^{2}  = 5 \times 5 = 25 \\  {6}^{2}  = 6 \times 6 = 36 \\  \\  {5}^{2}  +  {6}^{2}  = 25 + 36  \\  = 61

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Whats an equation for (0,75) (7,40) in slope intercept form
Molodets [167]

The slope-intercept form: y = mx + b.

m - slope, b - y-intercept

m=\dfrac{y_2-y_1}{x_2-x_1}

We have the points (0, 75) → y-intercept (b = 75), and (7, 40).

Substitute the coordimates of the points to the formula of a slope:

m=\dfrac{40-75}{7-0}=\dfrac{-35}{7}=-5

<h3>Answer: y = -5x + 75.</h3>
8 0
3 years ago
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

brainly.com/question/13057266

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3 0
2 years ago
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C) 20 Students liked English and math, but not history
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I Hope this helps and let me know if you have any further questions!

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4 years ago
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Slav-nsk [51]
What’s are the answers
6 0
3 years ago
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son4ous [18]

Answer:

6×-27=-3×

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7 0
3 years ago
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