Question Number 208282 by hardmath last updated on 10/Jun/24 $$\mathrm{If}\:\:\:\mathrm{cos}\alpha−\mathrm{cos}\beta\:=\:\frac{\mathrm{1}}{\mathrm{5}}\:\mathrm{sin}\alpha\:+\:\mathrm{sin}\beta\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{Find}\:\:\:\mathrm{cos}\left(\alpha\:+\:\beta\right)\:=\:? \\ $$ Commented by mr W last updated on 12/Jun/24 $${what}\:{you}\:{wrote}\:{means} \\ $$$$\mathrm{cos}\alpha−\mathrm{cos}\beta\:=\frac{\mathrm{1}}{\mathrm{2}}…
Question Number 208292 by universe last updated on 10/Jun/24 $$\:\mathrm{let}\:\mathrm{T}\:\mathrm{be}\:\mathrm{a}\:{n}×{n}\:\mathrm{matrix}\:\mathrm{with}\:\mathrm{integral}\: \\ $$$$\:\mathrm{entries}\:\mathrm{and}\:\:\mathrm{Q}\:=\:\mathrm{T}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{I}\:\:\:\mathrm{where}\:\mathrm{I}\:\mathrm{denote} \\ $$$$\:\:\mathrm{the}\:\mathrm{n}×\mathrm{n}\:\mathrm{identity}\:\mathrm{matrix}\:\mathrm{then}\:\mathrm{prove} \\ $$$$\:\:\mathrm{that}\:\mathrm{matrix}\:\mathrm{Q}\:\mathrm{is}\:\mathrm{invertible} \\ $$ Answered by Berbere last updated on 10/Jun/24…
Question Number 208259 by hardmath last updated on 09/Jun/24 Answered by cherokeesay last updated on 09/Jun/24 Answered by mr W last updated on 09/Jun/24 Commented…
Question Number 208241 by Frix last updated on 08/Jun/24 $$\mathrm{Solve}\:\mathrm{for}\:{p},\:{q},\:{r} \\ $$$${p}+{q}+{r}=\alpha \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} =\beta \\ $$$${pq}={r} \\ $$ Answered by mr W…
Question Number 208218 by hardmath last updated on 07/Jun/24 $$\mathrm{a}_{\boldsymbol{\mathrm{n}}} \:\:\mathrm{numbers}\:\mathrm{series} \\ $$$$\mathrm{If}\:\:\mathrm{S}_{\mathrm{16}} \:−\:\mathrm{S}_{\mathrm{13}} \:\:=\:\:\mathrm{S}_{\mathrm{106}} \:−\:\mathrm{S}_{\mathrm{103}} \\ $$$$\mathrm{Find}:\:\:\:\:\frac{\mathrm{3a}_{\mathrm{3}} \:+\:\mathrm{4a}_{\mathrm{4}} \:+\:\mathrm{5a}_{\mathrm{5}} }{\mathrm{2a}_{\mathrm{12}} }\:\:=\:\:? \\ $$ Commented…
Question Number 208187 by hardmath last updated on 07/Jun/24 $$\mathrm{Find}:\:\:\:\mathrm{1},\mathrm{03}^{\mathrm{200}} \:=\:? \\ $$ Answered by Ghisom last updated on 07/Jun/24 $$=\mathrm{10}^{\mathrm{200log}\:\mathrm{1}.\mathrm{03}} \approx\mathrm{10}^{\mathrm{200}×.\mathrm{012837}} \approx\mathrm{10}^{\mathrm{2}.\mathrm{5675}} \approx\mathrm{369}.\mathrm{36} \\…
Question Number 208199 by efronzo1 last updated on 07/Jun/24 Answered by Frix last updated on 07/Jun/24 $${y}=\sqrt{\mathrm{18}+\mathrm{3}{x}−{x}^{\mathrm{2}} }\:\mathrm{is}\:\mathrm{a}\:\mathrm{semi}\:\mathrm{circle}\:\mathrm{with}\:{r}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\sqrt{{x}+\mathrm{3}}+\sqrt{\mathrm{6}−{x}}\:\mathrm{has}\:\mathrm{the}\:\mathrm{maximum}\:\begin{pmatrix}{\frac{\mathrm{3}}{\mathrm{2}}}\\{\mathrm{3}\sqrt{\mathrm{2}}}\end{pmatrix} \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{2}\:\mathrm{solutions}\:\mathrm{for}\:\mathrm{0}\leqslant{m}<\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}}\:\mathrm{and} \\ $$$$\mathrm{exactly}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{at}\:{m}=\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}} \\…
Question Number 208149 by mnjuly1970 last updated on 06/Jun/24 Answered by A5T last updated on 06/Jun/24 $$\lfloor\mathrm{2}{x}^{\mathrm{2}} \rfloor>\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\Rightarrow{x}−\lfloor\mathrm{2}{x}^{\mathrm{2}} \rfloor<\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} +{x} \\ $$$${Suppose}\:{D}_{{f}} ,{R}_{{f}} \subseteq\mathbb{R}…
Question Number 208164 by Fridunatjan08 last updated on 06/Jun/24 $${Solve}\:{for}\:{x}: \\ $$$${x}^{\mathrm{2}} +{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)+{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} +{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} +…+{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{100}} =\mathrm{1} \\…
Question Number 208135 by efronzo1 last updated on 06/Jun/24 $$\:\:\:\downharpoonleft\underline{\:} \\ $$ Answered by A5T last updated on 06/Jun/24 $${r}^{{n}+\mathrm{2}} ={r}^{{n}+\mathrm{1}} +\frac{{r}^{{n}} }{\mathrm{2}}\Rightarrow{r}^{\mathrm{2}} ={r}+\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{r}=\frac{\mathrm{1}\underset{−} {+}\sqrt{\mathrm{3}}}{\mathrm{2}}…