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…
Question Number 29365 by gyugfeet last updated on 08/Feb/18 $${the}\:{line}\:{x}+{y}=\mathrm{2}\:{cuts}\:{a}\:{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}\:} =\mathrm{4}\:{at}\:{two}\:{points}.\:{find}\:{the}\:{co}−{ordinates}\:{of}\:{points}. \\ $$ Answered by Rasheed.Sindhi last updated on 08/Feb/18 $$\begin{cases}{{x}+{y}=\mathrm{2}}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}\:} =\mathrm{4}}\end{cases}\Rightarrow\begin{cases}{{y}=\mathrm{2}−{x}}\\{{x}^{\mathrm{2}} +\left(\mathrm{2}−{x}\right)^{\mathrm{2}}…
Question Number 29332 by Tinkutara last updated on 07/Feb/18 Answered by ajfour last updated on 07/Feb/18 $${let}\:{eq}.\:{of}\:{circle}\:{be} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{gx}+\mathrm{2}{fy}=\mathrm{0}\:\:\:\:\:\left({c}=\mathrm{0}\right) \\ $$$${for}\:{two}\:{circles}\:{to}\:{cut}\:{orthogonally} \\ $$$$\:\:\:\mathrm{2}{g}_{\mathrm{1}}…
Question Number 29276 by gyugfeet last updated on 06/Feb/18 $${if}\:{the}\:\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1}\:{passes}\:{through}\:{the}\:{point}\:{lf}\:{intersection}\:{of}\:{the}\:{lines}\:{x}+{y}=\mathrm{3}\:{and}\:\mathrm{2}{x}−\mathrm{3}{y}=\mathrm{1}\:{and}\:{is}\:{parallel}\:{to}\:{the}\:{line}\:{y}={x}−\mathrm{6}\:{then}\:{find}\:\:{the}\:{value}\:{of}\:{a}\:\:{and}\:{b}.\: \\ $$ Answered by Rasheed.Sindhi last updated on 06/Feb/18 $$\mathrm{Intersection}\:\mathrm{point}\:\mathrm{of}\:{x}+{y}=\mathrm{3}\:{and}\:\mathrm{2}{x}−\mathrm{3}{y}=\mathrm{1} \\ $$$${y}=\mathrm{3}−{x}\:\Rightarrow\mathrm{2}{x}−\mathrm{3}\left(\mathrm{3}−{x}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{2}\:,\:{y}=\mathrm{1} \\…
Question Number 160333 by cortano last updated on 27/Nov/21 Commented by blackmamba last updated on 28/Nov/21 $$\:\left(\ast\right){r}\::\:{y}={mx}+\mathrm{3} \\ $$$$\left(\ast\ast\right){s}:\:{y}={px}+\mathrm{3} \\ $$$$\:\left(\ast\ast\ast\right)\:{gradien}\:={y}'={m} \\ $$$$\:\:\mathrm{8}{x}−\mathrm{2}{yy}'\:=\:\mathrm{0}\Rightarrow{y}'=\frac{\mathrm{4}{x}}{{y}}\:\wedge\:{m}=\frac{{y}−\mathrm{3}}{{x}} \\ $$$$\:\Rightarrow\frac{\mathrm{4}{x}}{{y}}=\frac{{y}−\mathrm{3}}{{x}}\:;\:\mathrm{4}{x}^{\mathrm{2}}…
Question Number 29231 by Tinkutara last updated on 05/Feb/18 Answered by ajfour last updated on 05/Feb/18 $${eq}.\:{of}\:{chord}\:{of}\:{contact}\:{to}\:{circle} \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} =\mathrm{1}\:\:\:{from}\:\:\left({x}_{\mathrm{1}} ,\:{y}_{\mathrm{1}} \right)\:\:{is} \\ $$$${xx}_{\mathrm{1}}…