Question Number 217198 by hardmath last updated on 05/Mar/25 Answered by MrGaster last updated on 05/Mar/25 $$\delta\left({n}\right)=\underset{{d}\mid{n}} {\sum}{d},“\tau\left({n}\right)=\underset{{d}\mid{n}} {\sum}\cancel{{l}}\overset{\left(\mathrm{Unknown}\:\mathrm{meaning}\right)} {\Rightarrow}\mathrm{1}''\:\:\mathrm{and}\:\varphi-\mathrm{Euler}^{,} \mathrm{s}\:\mathrm{totient}\:\mathrm{function} \\ $$$$\delta\left({n}\right)=\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}\left({p}_{{i}}…
Question Number 217199 by cherokeesay last updated on 05/Mar/25 Commented by cherokeesay last updated on 05/Mar/25 $$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x} \\ $$ Answered by mr W last updated…
Question Number 217205 by peter frank last updated on 05/Mar/25 Answered by som(math1967) last updated on 06/Mar/25 $$\mathrm{cos}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{sin}\:\mathrm{cot}^{−\mathrm{1}} {x}\right) \\ $$$$=\mathrm{cos}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{sin}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)…
Question Number 217190 by efronzo1 last updated on 05/Mar/25 $$\mathrm{Given}\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} \:=\:\mathrm{a}_{\mathrm{n}} \:+\:\mathrm{a}_{\mathrm{n}+\mathrm{2}} \: \\ $$$$\:\:\mathrm{where}\:\mathrm{a}_{\mathrm{3}} =\:\mathrm{4}\:\mathrm{and}\:\mathrm{a}_{\mathrm{5}} =\:\mathrm{6} \\ $$$$\:\mathrm{find}\:\mathrm{a}_{\mathrm{n}} \:. \\ $$ Answered by mathmax…
Question Number 217191 by dabiswa1986 last updated on 05/Mar/25 $$\:\frac{{a}}{{b}}+\frac{{b}}{{a}}=\mathrm{1} \\ $$ Answered by ArshadS last updated on 06/Mar/25 $$\:\frac{{a}}{{b}}+\frac{{b}}{{a}}=\mathrm{1} \\ $$$$\Rightarrow{a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} =\mathrm{0} \\…
Question Number 217203 by Rasheed.Sindhi last updated on 05/Mar/25 $$\mathrm{A}\:\mathrm{farmer}\:\mathrm{has}\:\mathrm{100}\:\mathrm{meters}\:\mathrm{of}\: \\ $$$$\mathrm{fencing}\:\mathrm{and}\:\mathrm{wants}\:\mathrm{to}\:\mathrm{enclose}\:\mathrm{an} \\ $$$$\mathrm{rectagular}\:\mathrm{field}\:\mathrm{along}\:\mathrm{a}\:\mathrm{river}.\:\mathrm{Thei} \\ $$$$\mathrm{rver}\:\mathrm{forms}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{rectangle}\:\mathrm{so}\:\mathrm{fencing}\:\mathrm{is}\:\mathrm{needed}\:\mathrm{onlyo} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{other}\:\mathrm{three}\:\mathrm{sides}.\:\mathrm{What}\: \\ $$$$\mathrm{dimesions}\:\mathrm{should}\:\mathrm{the}\:\mathrm{farmer}\: \\ $$$$\mathrm{chooseto}\:\mathrm{maximize}\:\mathrm{the}\:\mathrm{enclosed} \\…
Question Number 217178 by hardmath last updated on 04/Mar/25 $$\mathrm{Find}: \\ $$$$\mathrm{100}-\mathrm{99}+\mathrm{98}-\mathrm{97}+\mathrm{96}-\mathrm{95}+…+\mathrm{2}-\mathrm{1}\:=\:? \\ $$ Answered by MathematicalUser2357 last updated on 04/Mar/25 $$=\left(\mathrm{100}+\mathrm{98}+\mathrm{96}+…{choegangcheck}\:{is}\:{cute}…+\mathrm{6}+\mathrm{4}+\mathrm{2}\right)−\left(\mathrm{99}+\mathrm{97}+\mathrm{95}+…{sorry}\:{for}\:{aggro}…+\mathrm{5}+\mathrm{3}+\mathrm{1}\right) \\ $$$$=\mathrm{2}\left(\mathrm{50}+\mathrm{49}+\mathrm{48}+…+\mathrm{3}+\mathrm{2}+\mathrm{1}\right)−\left\{\mathrm{2}\left(\mathrm{50}+\mathrm{49}+\mathrm{48}+…+\mathrm{3}+\mathrm{2}+\mathrm{1}\right)−\mathrm{50}\right\} \\…
Question Number 217186 by MrGaster last updated on 04/Mar/25 $$\frac{{x}−\mathrm{4051}}{\mathrm{2024}}+\frac{{x}−\mathrm{4050}}{\mathrm{2025}}+\frac{{x}−\mathrm{4049}}{\mathrm{2026}}=\mathrm{3} \\ $$ Answered by som(math1967) last updated on 04/Mar/25 $$\:\frac{{x}−\mathrm{4051}}{\mathrm{2024}}\:−\mathrm{1}+\frac{{x}−\mathrm{4050}}{\mathrm{2025}}−\mathrm{1}+\frac{{x}−\mathrm{4049}}{\mathrm{2026}}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{{x}−\mathrm{6075}}{\mathrm{2024}}+\frac{{x}−\mathrm{6075}}{\mathrm{2025}}\:+\frac{{x}−\mathrm{6075}}{\mathrm{2026}}=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{6075}\right)\:=\mathrm{0} \\…
Question Number 217164 by Tawa11 last updated on 03/Mar/25 Answered by issac last updated on 03/Mar/25 $$\mathrm{1}.\:\mathrm{3}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} \bigtriangledown−\sqrt{\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\bigtriangledown+\frac{\mathrm{6}}{\:^{\mathrm{3}} \sqrt{\sqrt{{x}^{\mathrm{2}}…
Question Number 217148 by Marzuk last updated on 03/Mar/25 $${I}\:{have}\:{seen}\:{a}\:{relationship}\:{in}\:{the}\:{curve} \\ $$$${path}\:{of}\:{a}\:{thrown}\:{object}\:{at}\:\beta\: \\ $$$${while}\:{the}\:{total}\:{passed}\:{distance}\:{D}_{{v}} \:{and} \\ $$$${highest}\:{point}\:{had}\:{passedD}_{{u}} \\ $$$${then}\:\beta\:=\:{arctan}\left(\frac{\mathrm{4}{D}_{{u}} }{{D}_{{v}} }\right) \\ $$$${but}\:{cant}\:{find}\:{the}\:{proof}. \\ $$$${I}\:{would}\:{like}\:{to}\:{say}\:{would}\:{anyone}\:{like}…