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what is the are of a square that is 2 inches long and 2 inches wide?

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Sin[cot-1{cos(tan-1x)}], sin (cot -1 {cos (tan -1 x)}) tan -1 x = A  ...

sin (cot -1 {cos (tan -1 x)}) tan -1 x = A  => tan A =x sec A = √(1+x 2 ) ==>  cos A = 1/√(1+x 2 )    so   A =  cos -1 (1/√(1+x 2 )) sin (cot -1 {cos (tan -1 x)}) = s

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Draw a common graph ( x - 2)2 /9+4(y + 2)2 =1, Graph     ( x - 2) 2 /9+4...

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Find area of y = 2 x2 + 10 and y = 4 x + 16, Find out the area of the regio...

Find out the area of the region bounded by y = 2 x 2 + 10 and y = 4 x + 16 . Solution In this case the intersection points (that we'll required eventually) are not going t

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