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Lana71 [14]
3 years ago
12

Find the area of the shaded segment of the circle.

Mathematics
1 answer:
vredina [299]3 years ago
5 0

Answer:

Around 5.5 square meters

Step-by-step explanation:

You can start by finding the area of the segment. Since the rest of the circle that is not in the segment is 240 degrees, the segment is 120 degrees or a third of the circle. You can therefore find the area of that segment with the formula \frac{1}{3} \pi r^2 =3\pi square meters. Now, you need to find the area of the triangle inside the sector. This is more difficult than last time, because it is not a 90 degree angle. However, you can solve this by dividing this triangle into two 30-60-90 triangles, which you know how to find the ratio of sides for. In a 30-60-90 triangle, the hypotenuse is twice the length of the smallest leg, and the larger leg is \sqrt{3} times larger than the smaller leg. In this case, these dimensions are a base of \frac{3}{2} for the smaller leg and \frac{3\sqrt{3}}{2} for the larger leg, or the base. Using the triangle area formula and multiplying by 2 (because remember, we divided the big triangle in half), you get \frac{9\sqrt{3}}{4} square meters. Subtracting this from the area of the segment, you get about 5.5 square meters. Hope this helps!

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taurus [48]

Answer:

So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

Step-by-step explanation:

We have the given differential equation: y′′+4y=5xcos(2x)

We use the Method of Undetermined Coefficients.

We first solve the homogeneous differential equation y′′+4y=0.

y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\

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y_h(t)=c_1e^{-2i t}+c_2e^{2i t}

Now, we finding a particular solution.

y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\

we get

y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

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Step-by-step explanation:

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