Question Number 54868 by peter frank last updated on 13/Feb/19 Commented by prakash jain last updated on 14/Feb/19 $$\alpha\centerdot\frac{\mathrm{1}}{\alpha}=\frac{\mathrm{10}}{\mathrm{3}{k}+\mathrm{4}}\Rightarrow{k}=\mathrm{2} \\ $$ Terms of Service Privacy…
Question Number 54869 by peter frank last updated on 13/Feb/19 Answered by $@ty@m last updated on 13/Feb/19 $${Given}\:\beta=\frac{\mathrm{1}}{\alpha}\:\Rightarrow\alpha\beta=\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{10}}{\mathrm{3}{k}+\mathrm{4}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}{k}+\mathrm{4}=\mathrm{10} \\ $$$$\Rightarrow{k}=\mathrm{2} \\…
Question Number 185938 by Mingma last updated on 30/Jan/23 Answered by HeferH last updated on 30/Jan/23 $${Let}\:{S}\:{be}\:{the}\:{pink}\:{area}\:{and}\:{let}\:\mathrm{5}{k}\:{be}\:{the}\:{area} \\ $$$$\:{of}\:{ABC}. \\ $$$$\:{S}\left(\frac{\mathrm{3}{k}}{\mathrm{2}}\:+\:{S}\right)\:=\:\left({k}\:−\:{S}\right)\left(\frac{\mathrm{3}{k}}{\mathrm{2}}−{S}\right) \\ $$$$\:{S}\:\:=\:\frac{\mathrm{3}{k}}{\mathrm{8}} \\ $$$$\:\frac{{Pink}}{{Area}\:{of}\:{ABC}}\:=\:\frac{\frac{\mathrm{3}{k}}{\mathrm{8}}}{\mathrm{5}{k}}\:=\:\frac{\mathrm{3}}{\mathrm{40}}\:\:…
Question Number 185924 by Mingma last updated on 29/Jan/23 Answered by HeferH last updated on 29/Jan/23 Commented by HeferH last updated on 30/Jan/23 $$\:{i}.\:\frac{{DE}}{{b}}\:=\:\frac{{a}}{{c}}\:\Rightarrow\:{DE}\:=\:\frac{{ab}}{{c}} \\…
Question Number 185907 by Mingma last updated on 29/Jan/23 Answered by HeferH last updated on 29/Jan/23 $$\:{If}\:{yellow}\:=\:{orange}\:{then}\: \\ $$$$\:{Area}\:\left({ACB}\right)\:=\:{Area}\:\left({PQRS}\right). \\ $$$$\:{let}\:{O}\:{be}\:{the}\:{center}\:{of}\:{the}\:{semicircle}. \\ $$$$\:{OP}\:=\:{a} \\ $$$$\:{AO}\:=\:{R}…
Question Number 185851 by Mingma last updated on 28/Jan/23 Commented by Mingma last updated on 28/Jan/23 Diameter in relation to a,b,c ? Answered by mr W last updated on 28/Jan/23…
Question Number 185845 by Rupesh123 last updated on 28/Jan/23 Commented by mr W last updated on 28/Jan/23 $${but}\:{only}\:{one}\:{angle}\:{is}\:{marked}. \\ $$ Commented by Frix last updated…
Question Number 185800 by Rupesh123 last updated on 27/Jan/23 Answered by ajfour last updated on 27/Jan/23 $${let}\:{center}\:{of}\:{small}\:{circle}\:\left(−{h},−{k}\right) \\ $$$${hence}\:{eq}.\:\:\:\left({x}+{h}\right)^{\mathrm{2}} +\left({y}+{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${side}\:{of}\:{eql}.\:\:{triangle}\:{a}={R}\sqrt{\mathrm{3}} \\ $$$${length}\:{of}\:{tangent}\:{from}\:{A}\left(\mathrm{0},{R}\right)…
Question Number 185727 by Mingma last updated on 26/Jan/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 185705 by Mingma last updated on 26/Jan/23 Answered by mr W last updated on 26/Jan/23 $${s}=\frac{{n}−\mathrm{1}+{n}+{n}+\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}{n}}{\mathrm{2}} \\ $$$$\overset{\_} {{r}}=\frac{{rs}}{{s}−{n}}=\frac{{r}}{\mathrm{1}−\frac{{n}}{{s}}}=\frac{{r}}{\mathrm{1}−\frac{\mathrm{2}{n}}{\mathrm{3}{n}}}=\mathrm{3}{r} \\ $$$$\frac{{r}}{\overset{\_} {{r}}}=\frac{\mathrm{1}}{\mathrm{3}}\:\checkmark \\…