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A=∫_0 ^(2α) 2pcos^(−1) (d/(2p))×((Ldβ)/(cos β))
=∫_0 ^(2α) 2(((tan β+(√(tan^2 β−4(atan β−H))))/2))cos^(−1) (L/(l(((tan β+(√(tan^2 β−4(atan β−H))))/2))))×L(dβ/(cos β))
=∫_0 ^(2α) 2(((tan β+(√(tan^2 β−4(atan β−H))))/2))cos^(−1) (((H−(((tan β+(√(tan^2 β−4(atan β−H))))/2))^2 )/(sin β))/( (√((a−b−htan α)^2 +R^2 −h^2 tan^2 (α/2)+((((a−b)tan β)/(1+tan αtan β)))^2 ))(((tan β+(√(tan^2 β−4(atan β−H))))/2))))×(((H−(((tan β+(√(tan^2 β−4(atan β−H))))/2))^2 )/(sin β)))(dβ/(cos β))
=∫_0 ^(2α) 2(((tan β+(√(tan^2 β−4(atan β−H))))/2))cos^(−1) (((H−(((tan β+(√(tan^2 β−4(atan β−H))))/2))^2 )/(sin β))/( (√((a−b−((((a−b)tan β)/(1+tan αtan β)))tan α)^2 +R^2 −((((a−b)tan β)/(1+tan αtan β)))^2 tan^2 (α/2)+((((a−b)tan β)/(1+tan αtan β)))^2 ))(((tan β+(√(tan^2 β−4(atan β−H))))/2))))×(((H−(((tan β+(√(tan^2 β−4(atan β−H))))/2))^2 )/(sin β)))(dβ/(cos β))
=∫_0 ^(2tan^(−1) (R/(a−b))) 2(((tan β+(√(tan^2 β−4(atan β−H))))/2))cos^(−1) ((((H−(((tan β+(√(tan^2 β−4(atan β−H))))/2))^2 )/(sin β))/( (√((a−b−((((a−b)tan β)/(1+((R/(a−b)))tan β)))×((R/(a−b))))^2 +R^2 −((((a−b)tan β)/(1+(R/(a−b))tan β)))^2 tan^2 ((tan^(−1) ((R/(a−b))))/2)+((((a−b)tan β)/(1+((R/(a−b)))tan β)))^2 ))(((tan β+(√(tan^2 β−4(atan β−H))))/2)))))×(((H−(((tan β+(√(tan^2 β−4(atan β−H))))/2))^2 )/(sin β)))(dβ/(cos β))
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Question 229528
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Question 229519
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Question 229508
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Question 229505
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Question 229496
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Solve y^((2)) (t)−y(t)=cos(t)e^t
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Question 229471
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Question 229463
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∫(1/( (√(ax^3 +bx^2 +cx+d))))dx
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Question 229444
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Question 229432
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App has been updated to resolve
notification issues faced by new users
on latest android version
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Q229401. find C(x,y) of r
suppose the bottom left corner of the rectangle
is origin
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Question 229412
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Question 229401
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x^2 +y^2 +z^2 =25
a^2 +b^2 +c^2 =16
ax+by+cz=20
((a+b+c)/(x+y+z))=?
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Question 229367
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Question 229359
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Question 229346
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