Question Number 19939 by wandumokonin@3gmail.com last updated on 18/Aug/17 $${f}\left({x}\right)={lnx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 85447 by TawaTawa1 last updated on 22/Mar/20 Commented by TawaTawa1 last updated on 22/Mar/20 $$\mathrm{Please}\:\mathrm{workings} \\ $$ Commented by john santu last updated…
Question Number 150944 by ajfour last updated on 16/Aug/21 Commented by mr W last updated on 17/Aug/21 Commented by ajfour last updated on 17/Aug/21 okay find minimum side length, sir.…
Question Number 85364 by Power last updated on 21/Mar/20 Commented by Power last updated on 21/Mar/20 $$ \\ $$ Commented by Power last updated on…
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Question Number 19792 by Tinkutara last updated on 15/Aug/17 $$\mathrm{Let}\:{ABCD}\:\mathrm{be}\:\mathrm{a}\:\mathrm{convex}\:\mathrm{quadrilateral} \\ $$$$\mathrm{with}\:\angle{DAB}\:=\:\angle{BDC}\:=\:\mathrm{90}°.\:\mathrm{Let}\:\mathrm{the} \\ $$$$\mathrm{incircles}\:\mathrm{of}\:\mathrm{triangles}\:{ABD}\:\mathrm{and}\:{BCD} \\ $$$$\mathrm{touch}\:{BD}\:\mathrm{at}\:{P}\:\mathrm{and}\:{Q},\:\mathrm{respectively}, \\ $$$$\mathrm{with}\:{P}\:\mathrm{lying}\:\mathrm{in}\:\mathrm{between}\:{B}\:\mathrm{and}\:{Q}.\:\mathrm{If} \\ $$$${AD}\:=\:\mathrm{999}\:\mathrm{and}\:{PQ}\:=\:\mathrm{200}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{radii}\:\mathrm{of}\:\mathrm{the}\:\mathrm{incircles}\:\mathrm{of} \\ $$$$\mathrm{triangles}\:{ABD}\:\mathrm{and}\:{BDC}? \\…
Question Number 19794 by Tinkutara last updated on 15/Aug/17 $$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:{ABC}\:\mathrm{with}\:\angle{BCA}\:=\:\mathrm{90}°, \\ $$$$\mathrm{the}\:\mathrm{perpendicular}\:\mathrm{bisector}\:\mathrm{of}\:{AB} \\ $$$$\mathrm{intersects}\:\mathrm{segments}\:{AB}\:\mathrm{and}\:{AC}\:\mathrm{at}\:{X} \\ $$$$\mathrm{and}\:{Y},\:\mathrm{respectively}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{quadrilateral}\:{BXYC}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{triangle}\:{ABC}\:\mathrm{is}\:\mathrm{13}\::\:\mathrm{18}\:\mathrm{and} \\ $$$${BC}\:=\:\mathrm{12}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:{AC}? \\ $$ Terms…
Question Number 19783 by NECC last updated on 15/Aug/17 $${The}\:{sides}\:{of}\:{a}\:{triangle}\:{are}\:{of} \\ $$$${lengths}\:\sqrt{\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)}\:,{m}^{\mathrm{2}} +{n}^{\mathrm{2}} ,\:\mathrm{2}{mn}. \\ $$$${Show}\:{that}\:{it}\:{is}\:{a}\:{right}\:{angle}\:\Delta. \\ $$ Commented by Tinkutara last updated…
Question Number 85298 by mr W last updated on 20/Mar/20 $${Find}\:{the}\:{symmetry}\:{axes}\:\left({if}\:{any}\right)\:{of} \\ $$$${the}\:{following}\:{curve}: \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}+\mathrm{2}{x}−\mathrm{8}{y}−\mathrm{2}=\mathrm{0} \\ $$ Commented by jagoll last updated on…
Question Number 150789 by cherokeesay last updated on 15/Aug/21 Answered by nimnim last updated on 15/Aug/21 $$\frac{\mathrm{1}}{\:\sqrt{{r}}}=\frac{\mathrm{1}}{\:\sqrt{{R}}}+\frac{\mathrm{1}}{\:\sqrt{{R}}}\Rightarrow\frac{\mathrm{1}}{\:\sqrt{{r}}}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{r}=\mathrm{1} \\ $$$${S}_{{area}} =\pi\left(\mathrm{1}\right)^{\mathrm{2}} =\pi\:{cm}^{\mathrm{2}} \\ $$…