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Mathematics 10 Online
OpenStudy (anonymous):

find the are of circle x2+y2=a2 by integration find the are of circle x2+y2=a2 by integration @Mathematics

OpenStudy (anonymous):

bca 1 sem

OpenStudy (anonymous):

bca 1 sem

OpenStudy (anonymous):

ok is it related to vector analysis

OpenStudy (anonymous):

yup but i m so weak in maths

OpenStudy (anonymous):

i have problem with two questions and another is if the roots of the equation (a2+b2)x2-2(ac+bd)x+(c2+d2)=0 are equal, prove that a/b=c/d

OpenStudy (anonymous):

can u give me solution

OpenStudy (anonymous):

i will give you for the second ques in few minutes but i don't know about integration of above one

OpenStudy (anonymous):

i also dont know becoz i m also from commerce stream

OpenStudy (anonymous):

you also may email me at devil.magra@yahoo.com

OpenStudy (anonymous):

i am writing solution of Q2 asked by u

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

first we apply quadrattic formula

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

mr. sumbul can u email me solutions of both questions at devil.magra@yahoo.com becoz of i will loss my train

OpenStudy (anonymous):

hey i got one answer if two roots are equal, b2-4ac=0, or (-2(ac+bd))^2 - 4 (a^2 + b^2) (c^2+d^2) = 0 or 4(ac)^2 + 8abcd + 4(bd)^2 - 4(ac)^2 - 4(ad)^2 - 4(bc)^2 - 4(bd)^2 = 0 or 8abcd -4(ad)^2 -4(bc)^2 = 0 or (ad)^2 - 2(ad)(bc) + (bc)^2 = 0 or (ad - bc)^2 = 0 or ad - bc = 0 or ad = bc therefore, a/b = c/d

OpenStudy (anonymous):

yes it is correct i was also typing it

OpenStudy (anonymous):

do u still need solution?

OpenStudy (anonymous):

yes please mail the solution of find the are of circle x2+y2=a2 by integration

OpenStudy (anonymous):

ok bye thanks i give u first ur reward

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