Question Number 223887 by fantastic last updated on 08/Aug/25 Commented by fantastic last updated on 08/Aug/25 $${can}\:{this}\:{be}\:{solved}\:{without}\:{any}\:{assumption}\left({s}\right)? \\ $$ Commented by Frix last updated on…
Question Number 223896 by MASANJAJJ last updated on 08/Aug/25 Commented by mr W last updated on 08/Aug/25 $${y}=\mathrm{81}° \\ $$$${x}\:{can}\:{not}\:{be}\:{determined},\:{because}\:{it} \\ $$$${is}\:{not}\:{clear}\:{what}\:{you}\:{mean}\:{with}\:{the} \\ $$$${angle}\:\angle{AOB}.\:{do}\:{you}\:{mean}\: \\…
Question Number 223881 by fantastic last updated on 08/Aug/25 Answered by som(math1967) last updated on 08/Aug/25 $${let}\:{CP}={x},\:{SP}={y} \\ $$$$\:\frac{\mathrm{15}{x}}{\mathrm{100}}=\frac{\mathrm{25}{y}}{\mathrm{100}} \\ $$$$\Rightarrow\:\frac{{x}}{{y}}=\frac{\mathrm{25}}{\mathrm{15}}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\:\therefore{CP}:{SP}=\mathrm{5}:\mathrm{3} \\ $$$$\:{loss\%}=\frac{\mathrm{2}}{\mathrm{5}}×\mathrm{100}=\mathrm{40\%}…
Question Number 223899 by 12345Ahmed last updated on 08/Aug/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 223889 by fantastic last updated on 08/Aug/25 Commented by fantastic last updated on 08/Aug/25 $${please}\:{help}\:{me}\:{with}\:{explanation} \\ $$ Commented by mr W last updated…
Question Number 223875 by mr W last updated on 08/Aug/25 Commented by mr W last updated on 08/Aug/25 $${Q}\mathrm{223653} \\ $$ Terms of Service Privacy…
Question Number 223865 by ajfour last updated on 07/Aug/25 Commented by ajfour last updated on 07/Aug/25 $${Contrary}\:{to}\:{diagram}\:{labelling},\:{lets} \\ $$$${have}\:{a}=\mathrm{1},\:{b}={b}.\:\angle{CAE}=\pi/\mathrm{6}. \\ $$$${If}\:{APC}\:{is}\:{a}\:{straight}\:{line},\:{P}\:{being}\:{a} \\ $$$${point}\:{of}\:{tangency},\:{then}\:{find}\:{b},\:{r},\:{R}. \\ $$…
Question Number 223858 by Rojarani last updated on 07/Aug/25 Answered by Rasheed.Sindhi last updated on 08/Aug/25 $${let}\:{x}+\mathrm{100}={y} \\ $$$$\frac{\left({y}−\mathrm{2}\right)^{\mathrm{5}} +\left({y}+\mathrm{2}\right)^{\mathrm{5}} }{\left({y}−\mathrm{1}\right)^{\mathrm{5}} +\left({y}+\mathrm{1}\right)^{\mathrm{5}} }=\frac{\mathrm{16}^{\mathrm{2}} }{\mathrm{5}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}}…
Question Number 223836 by Tawa11 last updated on 06/Aug/25 Answered by som(math1967) last updated on 06/Aug/25 $$\bigtriangleup{ABG}\sim\bigtriangleup{HFG} \\ $$$${let}\:{AB}={x} \\ $$$$\:\:\therefore\frac{{x}}{{x}+\mathrm{105}}=\frac{{x}−\mathrm{112}}{{x}−\mathrm{105}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{112}{x}+\mathrm{105}{x}−\mathrm{112}×\mathrm{105} \\…
Question Number 223821 by wewji12 last updated on 06/Aug/25 $$\underset{\nu\rightarrow\alpha} {\mathrm{lim}}\:\frac{{J}_{−\nu−\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right)+{e}^{\boldsymbol{{i}}\pi\nu} \centerdot{Y}_{\nu+\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right)}{{Y}_{−\nu−\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right)−{e}^{\boldsymbol{{i}}\pi\nu} \centerdot{J}_{\nu+\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right)}=?? \\ $$$$\alpha\in\mathbb{Z}\: \\ $$ Terms of Service Privacy…