Question Number 175901 by infinityaction last updated on 09/Sep/22 Commented by HeferH last updated on 09/Sep/22 Answered by HeferH last updated on 09/Sep/22 $$\:{R}\:=\:\frac{\mathrm{15}\:−\:\mathrm{6}\sqrt{\mathrm{6}}}{\mathrm{2}} \\…
Question Number 175894 by cortano1 last updated on 09/Sep/22 Answered by mr W last updated on 09/Sep/22 Commented by mr W last updated on 09/Sep/22…
Question Number 44801 by ajfour last updated on 04/Oct/18 Commented by ajfour last updated on 04/Oct/18 $${Equation}\:{of}\:{ellipse}:\:\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:, \\ $$$${find}\:{radius}\:\boldsymbol{{r}}\:{of}\:{circle}. \\ $$…
Question Number 175836 by ajfour last updated on 08/Sep/22 Commented by ajfour last updated on 08/Sep/22 $${Ellipse}:\:\:{x}^{\mathrm{2}} +\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$ Answered by mr…
Question Number 175608 by ajfour last updated on 03/Sep/22 Commented by ajfour last updated on 04/Sep/22 $${yes}\:{sir}. \\ $$ Commented by mr W last updated…
Question Number 44527 by Necxx last updated on 30/Sep/18 Commented by Necxx last updated on 30/Sep/18 $$\mathrm{21}\:{please} \\ $$ Commented by maxmathsup by imad last…
Question Number 44480 by peter frank last updated on 29/Sep/18 $${prove}\:{that}\:\:\frac{\mathrm{9}\pi}{\mathrm{8}\:\:}−\frac{\mathrm{9}}{\mathrm{4}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{9}}{\mathrm{4}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$ Answered by math1967 last updated on 30/Sep/18 $${L}.{H}.{S}=\frac{\mathrm{9}}{\mathrm{4}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}\right) \\…
Question Number 44436 by ajfour last updated on 29/Sep/18 Commented by ajfour last updated on 29/Sep/18 $${Find}\:{R}\:{in}\:{terms}\:{of}\:{r}. \\ $$ Answered by MrW3 last updated on…
Question Number 175237 by nadovic last updated on 24/Aug/22 $$\mathrm{The}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}\:\:\mathrm{AC},\:\mathrm{BC} \\ $$$$\mathrm{and}\:\mathrm{AB}\:\mathrm{of}\:\:\mathrm{a}\:\mathrm{right}−\mathrm{angled}\:\mathrm{triangled} \\ $$$$\mathrm{with}\:\mathrm{lengths}\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{are}\:{y}\:=\:−\mathrm{7}, \\ $$$${x}=\mathrm{11}\:\mathrm{and}\:\mathrm{4}{x}−\mathrm{3}{y}−\mathrm{5}=\mathrm{0}\:\mathrm{respectively}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{inscribed} \\ $$$$\mathrm{circle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle},\:\mathrm{if}\:\mathrm{its}\:\mathrm{radius}\:{r}, \\ $$$$\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:{r}\:=\:\frac{{a}+{b}−{c}}{\mathrm{2}}. \\ $$ Terms…
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