Question Number 48921 by peter frank last updated on 30/Nov/18 Commented by maxmathsup by imad last updated on 30/Nov/18 $${we}\:{have}\:{cos}\left(\mathrm{3}\theta\right)\:\sim\mathrm{1}−\frac{\mathrm{9}\theta^{\mathrm{2}} }{\mathrm{2}}\:\:\:\left(\theta\rightarrow\mathrm{0}\right)\:{and}\:{cos}\theta\:\sim\mathrm{1}−\frac{\theta^{\mathrm{2}} }{\mathrm{2}} \\ $$$${cos}\left(\mathrm{4}\theta\right)\sim\:\mathrm{1}−\mathrm{8}\theta^{\mathrm{2}} \:\:\:{and}\:{cos}\left(\mathrm{2}\theta\right)\sim\mathrm{1}−\mathrm{2}\theta^{\mathrm{2}}…
Question Number 48872 by peter frank last updated on 29/Nov/18 Commented by mr W last updated on 29/Nov/18 $$\left.{i}\right)\:{ellipse} \\ $$$$\left.{i}\left.{i}\right)\:{and}\:{iii}\right)\:{parabola} \\ $$ Terms of…
Question Number 48871 by peter frank last updated on 29/Nov/18 Commented by maxmathsup by imad last updated on 01/Jan/19 $$\left.\mathrm{2}\right)\:{we}\:{have}\:{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}\:={x}^{\mathrm{2}} +{x}−\mathrm{3}{x}−\mathrm{3}\:={x}\left({x}+\mathrm{1}\right)−\mathrm{3}\left({x}+\mathrm{1}\right)=\left({x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)\Rightarrow \\ $$$${f}\left({x}\right)=\frac{{x}\left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)}\:{and}\:{for}\:{x}\neq−\mathrm{1}\:{f}\left({x}\right)=\frac{{x}}{{x}−\mathrm{3}}\:\:{but}\:{remember}\:{that} \\…
Question Number 114326 by 1549442205PVT last updated on 18/Sep/20 $$\left.\mathrm{a}\right)\mathrm{Find}\:\mathrm{a}\:\mathrm{four}−\mathrm{digit}\:\mathrm{number}\:\mathrm{that} \\ $$$$\mathrm{satisfies}\:\mathrm{the}\:\mathrm{following}\:\mathrm{condition}: \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{squares}\:\mathrm{of}\:\mathrm{the}\:\mathrm{extreme}\:\mathrm{digits} \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{13};\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{squares}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{middle}\:\mathrm{digits}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{85};\mathrm{if}\:\mathrm{we} \\ $$$$\mathrm{substract}\:\mathrm{1089}\:\mathrm{from}\:\mathrm{the}\:\mathrm{desired} \\ $$$$\mathrm{number}\:\mathrm{we}\:\mathrm{obtain}\:\mathrm{a}\:\mathrm{number}\:\mathrm{containing} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{digits}\:\mathrm{as}\:\mathrm{the}\:\mathrm{desired}\:\mathrm{number} \\…
Question Number 179848 by Mr.D.N. last updated on 03/Nov/22 $$ \\ $$$$\:\:\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{years}\:,\mathrm{Rs}\:\mathrm{50000}\:\mathrm{amounts}\:\mathrm{to}\:\mathrm{Rs672720} \\ $$$$\mathrm{at}\:\mathrm{compounded}\:\mathrm{annually}\:\mathrm{if}\:\mathrm{the}\:\mathrm{interest}\:\mathrm{rate}\:\mathrm{is}\:\mathrm{1}\:\mathrm{paisa}\:\mathrm{per} \\ $$$$\mathrm{rupee}\:\mathrm{per}\:\mathrm{month}? \\ $$ Commented by mr W last updated on…
Question Number 48748 by peter frank last updated on 28/Nov/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 48745 by peter frank last updated on 28/Nov/18 Answered by Kunal12588 last updated on 28/Nov/18 $$\left({i}\right){unique}\:{solution}\:\:\frac{{a}_{\mathrm{1}} }{{a}_{\mathrm{2}} }\neq\frac{{b}_{\mathrm{1}} }{{b}_{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}}{\mathrm{4}}\neq\frac{\mathrm{3}}{{a}} \\…
Question Number 48742 by peter frank last updated on 28/Nov/18 Answered by tanmay.chaudhury50@gmail.com last updated on 28/Nov/18 $${xy}−\mathrm{4}{x}+\mathrm{3}{y}−\mathrm{1}=\mathrm{0} \\ $$$${x}\left({y}−\mathrm{4}\right)+\mathrm{3}\left({y}−\mathrm{4}+\mathrm{4}\right)−\mathrm{1}=\mathrm{0} \\ $$$${x}\left({y}−\mathrm{4}\right)+\mathrm{3}\left({y}−\mathrm{4}\right)+\mathrm{11}=\mathrm{0} \\ $$$$\left({x}+\mathrm{3}\right)\left({y}−\mathrm{4}\right)=\left(\sqrt{\mathrm{11}}\:\:\right)^{\mathrm{2}} {i}^{\mathrm{2}}…
Question Number 48744 by peter frank last updated on 28/Nov/18 Answered by Abdulhafeez Abu qatada last updated on 28/Nov/18 $$ \\ $$$$ \\ $$$$\sqrt[{\mathrm{3}}]{{x}\:+\:{iy}}\:=\:{a}\:+\:{ib} \\…
Question Number 48740 by peter frank last updated on 28/Nov/18 Commented by Abdo msup. last updated on 28/Nov/18 $${z}_{\mathrm{1}} {z}_{\mathrm{2}} =\left(\mathrm{1}−{i}\right)^{\mathrm{6}} \:\:{but}\:\mathrm{1}−{i}\:=\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\:=\sqrt{\mathrm{2}}{e}^{−{i}\frac{\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\left(\mathrm{1}−{i}\right)^{\mathrm{6}}…