What is the perimeter of the polygon?

Mathematics

1 answer:

mr_godi [17]2 years ago

6 0

Welp. I sure hope you like the Pythagorean theorem...
Top line:
One point is (-2,-2) while the other is (3,-3)
Thus the distance in between is sqrt((3-(-2))^2+(-3-(-2))^2)=sqrt(5^2+(-1)^2)=sqrt(26)
Most right line:
One point is (4,-6) while the other is (3,-3)
Thus the distance in between is sqrt((3-4)^2+(-3-(-6))^2)=sqrt((-1)^2+3^2)=sqrt(10)
Most bottom line:
One point is (1,-6) while the other is (4,-6)
Thus the distance in between is sqrt(4-1)^2+(-6-(-6))^2)=sqrt(3^2+0^2)=sqrt(9)=3
Most bottom left line:
One point is (1,-6) while the other is (-2,-4)
Thus the distance in between is sqrt((1-(-2))^2+(-6-(-4))^2)=sqrt(3^2+(-2)^2)=sqrt(13)
Lastly the most left line:
One point is (-2,-2) while the other is (-2,-4)
Thus the distance in between is sqrt((-2-(-2))^2+(-2-(-4))^2)=sqrt(0^2+(2)^2)=sqrt(4)=2
Thus to find the perimeter, we add up all the sides to get
sqrt(26)+sqrt(10)+3+sqrt(13)+2=16.8668 or B

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See Below.

Step-by-step explanation:

We have the equation:

$\displaystyle y = \left(3e^{2x}-4x+1\right)^{{}^1\! / \! {}_2}$

And we want to show that:

$\displaystyle y \frac{d^2y }{dx^2} + \left(\frac{dy}{dx}\right) ^2 = 6e^{2x}$

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Hence:

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And we can find the second derivative by using the quotient rule:

$\displaystyle \begin{aligned}\frac{d^2y}{dx^2} & = \frac{(3e^{2x}-2)'(y)-(3e^{2x}-2)(y)'}{(y)^2}\\ \\ &= \frac{6ye^{2x}-\left(3e^{2x}-2\right)\left(\dfrac{dy}{dx}\right)}{y^2} \\ \\ &=\frac{6ye^{2x} -\left(3e^{2x} -2\right)\left(\dfrac{3e^{2x}-2}{y}\right)}{y^2}\\ \\ &=\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\end{aligned}$