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Question Number 29728 by rahul 19 last updated on 11/Feb/18 Commented by 803jaideep@gmail.com last updated on 11/Feb/18 $$\mathrm{i}\:\mathrm{dnt}\:\mathrm{know}\:\mathrm{exactly}\:\mathrm{but}\:\mathrm{i}\:\mathrm{think}\: \\ $$$$\mathrm{tangents}\:\mathrm{at}\:\mathrm{end}\:\mathrm{of}\:\mathrm{latus}\:\mathrm{rectum}\: \\ $$$$\mathrm{can}\:\mathrm{form}\:\mathrm{square} \\ $$ Commented…
Question Number 95206 by hardylanes last updated on 23/May/20 $${find}\:{the}\:{equation}\:{of}\:{the}\:{straight}\:{line}\:{which} \\ $$$${passes}\:{through}\:{the}\:{point}\:{of}\:{intersertion}\:{of} \\ $$$${the}\:{lines}\:\mathrm{4}{x}+\mathrm{3}{y}+\mathrm{19}=\mathrm{0}\:{and}\:\mathrm{12}{x}−\mathrm{5}{y}+\mathrm{3}=\mathrm{0} \\ $$$${and}\:{has}\:{a}\:{y\_intercept}\:{of}\:\mathrm{2} \\ $$ Commented by PRITHWISH SEN 2 last updated…
Question Number 160701 by tounghoungko last updated on 05/Dec/21 Answered by som(math1967) last updated on 05/Dec/21 $$\boldsymbol{{ABCD}}\:\boldsymbol{{cyclic}}\:\:\left[\angle{ADC}+\angle{ABD}=\mathrm{180}\right] \\ $$$$\therefore\angle{BDC}=\angle{BAC}=\mathrm{45}\:\left[{subtend}\:{samesegment}\right] \\ $$$$\angle{BCD}=\mathrm{180}−\mathrm{75}=\mathrm{105} \\ $$$${BD}={radius}\:{of}\:{quater}\:{circle} \\ $$$${from}\:\bigtriangleup{BCD}…
Question Number 95145 by i jagooll last updated on 23/May/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{14}}\:+\:\frac{\mathrm{1}}{\mathrm{35}}\:+\frac{\mathrm{1}}{\mathrm{65}}\:+\:\frac{\mathrm{1}}{\mathrm{104}}\:+\:…\:? \\ $$ Answered by bobhans last updated on 23/May/20 $$\mathrm{S}\:=\:\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{2}}{\left(\mathrm{3n}+\mathrm{1}\right)\left(\mathrm{3n}+\mathrm{2}\right)}\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\underset{{x}\rightarrow\infty}…
Question Number 160682 by tounghoungko last updated on 04/Dec/21 Commented by talminator2856791 last updated on 04/Dec/21 $$\:\mathrm{what}\:\mathrm{is}\:{S}\:\mathrm{supposed}\:\mathrm{to}\:\mathrm{be}? \\ $$ Commented by tounghoungko last updated on…
Question Number 160654 by Avijit007 last updated on 04/Dec/21 Answered by som(math1967) last updated on 04/Dec/21 $${let}\:{radius}\:{of}\:{semicircle}\:{rcm} \\ $$$$\therefore\:\boldsymbol{{r}}^{\mathrm{2}} =\left(\boldsymbol{{r}}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \\ $$$$\mathrm{2}\boldsymbol{{r}}−\mathrm{1}=\mathrm{9} \\ $$$$\therefore\boldsymbol{{r}}=\mathrm{5}\boldsymbol{{cm}}…
Question Number 29536 by math solver last updated on 09/Feb/18 $${Find}\:{eccentricity}\:{of}\:{the}\:{ellipse} \\ $$$$\mathrm{7}{x}^{\mathrm{2}} +\mathrm{7}{y}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{10}{x}−\mathrm{10}{y}−\mathrm{7}=\mathrm{0}\:? \\ $$ Commented by math solver last updated on 10/Feb/18…
Question Number 160544 by cortano last updated on 01/Dec/21 Commented by bobhans last updated on 01/Dec/21 $$\:\mathrm{A}=\sqrt{\mathrm{11}×\mathrm{10}×\mathrm{7}×\mathrm{8}}\:=\:\mathrm{4}\sqrt{\mathrm{385}} \\ $$$$\:\mathrm{R}=\frac{\sqrt{\mathrm{157}×\mathrm{166}×\mathrm{158}}}{\mathrm{16}\sqrt{\mathrm{385}}}\:=\:\frac{\sqrt{\mathrm{1029449}}}{\mathrm{8}\sqrt{\mathrm{385}}} \\ $$$$\:\mathrm{S}_{\mathrm{circle}} =\:\frac{\mathrm{1029449}\pi}{\mathrm{24640}} \\ $$ Commented…
Question Number 29420 by Tinkutara last updated on 08/Feb/18 Commented by ajfour last updated on 09/Feb/18 Commented by Tinkutara last updated on 09/Feb/18 No more explanation needed. Answered…