Question Number 102524 by ajfour last updated on 09/Jul/20 Commented by ajfour last updated on 09/Jul/20 $${Given}\:{three}\:{circles}\:{of}\:{radii}\:{p},\:{q},\:{r} \\ $$$${touching}\:{the}\:{coordinate}\:{axes}\:{as} \\ $$$${shown};\:{find}\:{equation}\:{of}\:{the}\:{circum}- \\ $$$${circle}. \\ $$…
Question Number 36884 by anik last updated on 06/Jun/18 $$\mathrm{1}.\:\mathrm{What}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{cueved}\:\mathrm{line}\:\mathrm{which}\:\mathrm{is}\:\mathrm{made}\:\mathrm{by}\:\mathrm{a} \\ $$$$\mathrm{fixed}\:\mathrm{point}\:\mathrm{in}\:\mathrm{the}\:\mathrm{boundary}\:\mathrm{of}\:\mathrm{a} \\ $$$$\mathrm{moving}\:\mathrm{circular}\:\mathrm{object}\:\mathrm{in}\:\mathrm{respect}\:\mathrm{to} \\ $$$$\mathrm{an}\:\mathrm{another}\:\mathrm{fiexd}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{way}\:\mathrm{of} \\ $$$$\mathrm{moving}? \\ $$ Terms of Service…
Question Number 36880 by Tinkutara last updated on 06/Jun/18 Commented by ajfour last updated on 07/Jun/18 Commented by ajfour last updated on 07/Jun/18 $$\mathrm{2}{h}={a}\mathrm{cos}\:\theta+{b}\mathrm{sin}\:\theta \\…
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Question Number 102060 by ajfour last updated on 06/Jul/20 Commented by ajfour last updated on 06/Jul/20 $${Find}\:\boldsymbol{{r}}\:{and}\:\boldsymbol{{R}}. \\ $$ Answered by mr W last updated…
Question Number 167519 by MikeH last updated on 18/Mar/22 $$\mathcal{V}_{\mathrm{4}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{set}\:\mathrm{of}\:\mathrm{all}\:\mathrm{real}\:\mathrm{valued}\:\mathrm{continues}\: \\ $$$$\mathrm{functions}\:\mathrm{defined}\:\mathrm{on}\:\mathrm{the}\:\mathrm{entire}\:\mathrm{real} \\ $$$$\mathrm{line}\:\mathrm{together}\:\mathrm{with}\:\mathrm{standard}\:\mathrm{addition} \\ $$$$\mathrm{and}\:\mathrm{scalar}\:\mathrm{multiplication}.\:\mathrm{Is}\:\mathcal{V}_{\mathrm{4}} \: \\ $$$$\mathrm{a}\:\mathrm{vector}\:\mathrm{space}?\:\mathrm{Explain} \\ $$ Terms of Service…
Question Number 36237 by ajfour last updated on 30/May/18 Answered by MJS last updated on 30/May/18 $$\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:{AB}\:\mathrm{and}\:{P}\:\mathrm{must}\:\mathrm{be} \\ $$$$\mathrm{max},\:\mathrm{so}\:\mathrm{we}'\mathrm{re}\:\mathrm{looking}\:\mathrm{for}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{parallel} \\ $$$$\mathrm{to}\:{AB} \\ $$ Commented by…
Question Number 167058 by cortano1 last updated on 05/Mar/22 Answered by mr W last updated on 05/Mar/22 $$\left(\frac{\mathrm{2}}{{r}}+\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{2}}{{r}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right) \\ $$$$\frac{\mathrm{1}}{{r}}=\frac{\mathrm{9}}{\mathrm{8}} \\…
Question Number 166943 by cortano1 last updated on 03/Mar/22 Commented by blackmamba last updated on 03/Mar/22 $$\:{By}\:{Ladder}\:{theorem} \\ $$$$\:\frac{\mathrm{1}}{{x}+\mathrm{4}+\mathrm{5}+\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{4}+\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{5}+\mathrm{10}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}+\mathrm{19}}=\frac{\mathrm{1}}{\mathrm{14}}+\frac{\mathrm{1}}{\mathrm{15}}−\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}+\mathrm{19}}=\frac{\mathrm{4}}{\mathrm{105}}\:;\:{x}=\frac{\mathrm{29}}{\mathrm{4}}=\mathrm{7}.\mathrm{25}\: \\ $$…