If `cosec . (pi)/(32)+ cosec. (pi)/(16)+ cosec. (pi)/(8)+ cosec. (pi)/(4)+ cosec. (pi) - YouTube
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1 + cosec pi4 + cosec pi8 + cosec pi16 =
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21+2×0+21×32×sin−11)−(2432−16+21×32×sin−121)=8π−(8+4π)=4π−8 Ad..
In the figure below, the two semicircles have cent that are midpoints of ..
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If cosec . (pi)/(32)+ cosec. (pi)/(16)+ cosec. (pi)/(8)+ cosec. (pi)/(4)+ cosec. (pi)/(2)= cot. (pi)/(k), then the value of k is
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tan""((5pi)/(32))+ 2 tan""((5pi)/(16))+4 tan""((5pi)/8)-cot""((5pi)/(32))=
![Evaluate: n→∞limit (1^2 + 2^2 + 3^2 + ... + n^2)/[1 + 2 + ... + n] (1^3 + 2^3 + 3^3 + ... + n^3)](https://dwes9vv9u0550.cloudfront.net/images/5230069/f02f8f89-ce4e-4f21-a9fe-789bd483e509.jpg "Evaluate: n→∞limit (1^2 + 2^2 + 3^2 + ... + n^2)/[1 + 2 + ... + n] (1^3 + 2^3 + 3^3 + ... + n^3)")
Evaluate: n→∞limit (1^2 + 2^2 + 3^2 + ... + n^2)/[1 + 2 + ... + n] (1^3 + 2^3 + 3^3 + ... + n^3)
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The value of int 0^ pi2cos^5/3x/cos^5/3x+sin^5/3xdx is
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![pasagot pi.evaluate each expression:1.2-3(8-11)+32÷2+22.6²+[3(2-5])-8 ÷43.9²-5³•2÷2(3-7)4.x²-3x(x-5)+12÷3,if - Brainly.ph](https://ph-static.z-dn.net/files/d26/a6a2946e8a2d2632062c07cc26b97f38.jpg "pasagot pi.evaluate each expression:1.2-3(8-11)+32÷2+22.6²+[3(2-5])-8 ÷43.9²-5³•2÷2(3-7)4.x²-3x(x-5)+12÷3,if - Brainly.ph")
pasagot pi.evaluate each expression:1.2-3(8-11)+32÷2+22.6²+[3(2-5])-8 ÷43.9²-5³•2÷2(3-7)4.x²-3x(x-5)+12÷3,if - Brainly.ph
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