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3 years ago

What is a irregular concave polygon

Mathematics

1 answer:

Anvisha [2.4K]3 years ago

3 0

Answer:

An irregular polygon is any polygon that is not a regular polygon. It can have sides of any length and each interior angle can be any measure. They can be convex or concave, but all concave polygons are irregular since the interior angles cannot all be the same.

Step-by-step explanation:

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(-6,-7) and (2,-7)

So you would do m=y2-y1/x2-x1

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So you get

m=[-7-(-7)]/[2-(-6)]

m=0/2+6

m=0/8

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Help with this integral<br><br><img src="https://tex.z-dn.net/?f=%20%5Cint%5Climits%20%5Cfrac%7Bdx%7D%7Bx%5E%7B2%7D-4x-13%7D" id

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$x^2-4x-13=(x-2)^2-17$

$x-2=\sqrt{17}\sec y$
$\mathrm dx=\sqrt{17}\sec y\tan y\,\mathrm dy$

$\displaystyle\int\frac{\mathrm dx}{x^2-4x-13}=\int\frac{\sqrt{17}\sec y\tan y}{(\sqrt{17}\sec y)^2-17}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\frac{\sec y\tan y}{\sec^2y-1}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\frac{\sec y\tan y}{\tan^2y}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\frac{\sec y}{\tan y}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\frac{\frac1{\cos y}}{\frac{\sin y}{\cos y}}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\csc y\,\mathrm dy$
$=-\dfrac1{\sqrt{17}}\ln|\csc y+\cot y|+C$

$\sec y=\dfrac{x-2}{\sqrt{17}}\iff y=\sec^{-1}\dfrac{x-2}{\sqrt{17}}$
$\implies\csc y=\dfrac{x-2}{\sqrt{(x-2)^2-17}}=\dfrac{x-2}{\sqrt{x^2-4x-13}}$
$\implies\cot y=\dfrac{\sqrt{17}}{\sqrt{(x-2)^2-17}}=\dfrac{\sqrt{17}}{\sqrt{x^2-4x-13}}$

$\displaystyle\int\frac{\mathrm dx}{x^2-4x-13}=-\dfrac1{\sqrt{17}}\ln\left|\frac{x-2+\sqrt{17}}{\sqrt{x^2-4x-13}}\right|+C$

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4 years ago