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2 years ago

Polygon Angle Sums - Part 2

Mathematics

1 answer:

Rashid [163]2 years ago

5 0

Answer:

x = 80°

Step-by-step explanation:

Find attached the complete question with diagram. Also find attached the diagram used for solving the question.

There area 3 angles in the diagram:

x + 8, 2x + 12, x + 20

Let the interior angles be labelled a, b, and c

a+ 2x+12 = 180° (sum of angles on a straight line)

a= 180-(2x+12) = 168 -2x in degrees

b+ x+20 = 180° (sum of angles on a straight line)

b = 180-(x+20) = 160 -x in degrees

c+ x+8 = 180° (sum of angles on a straight line)

c = 180-(x+8) = 172-x in degrees

a+b+c = 180° (sum of angles in triangle)

(168 -2x)° + (160 -x)° + (172-x)° = 180°

500 -4x = 180

500-180 = 4x

320 = 4x

x = 320/4

x = 80°

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Answer:

Required series is:

$\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

Step-by-step explanation:

Given that

$f'(x) = -\frac{1}{1 + x^{2}}$ ---(1)

We know that:

$\frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^{2}}$ ---(2)

Comparing (1) and (2)

$f'(x)=-(tan^{-1}x)$ ---- (3)

Using power series expansion for $tan^{-1}x$

$f'(x)=-tan^{-1}x=-\int {\frac{1}{1+x^{2}} \, dx$

$= -\int{ \sum\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx$

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$=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}]$

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3 years ago

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