Question Number 97554 by 1549442205 last updated on 08/Jun/20 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{triples}\:\mathrm{of}\:\mathrm{positive}\:\mathrm{integers}\:\left(\mathrm{x};\mathrm{y};\mathrm{z}\right) \\ $$$$\mathrm{so}\:\mathrm{that}\:\frac{\mathrm{x}−\mathrm{y}\sqrt{\mathrm{2020}}}{\mathrm{y}−\mathrm{z}\sqrt{\mathrm{2020}}}\:\mathrm{is}\:\mathrm{a}\:\mathrm{rational}\:\mathrm{number} \\ $$$$\mathrm{and}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \mathrm{be}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}. \\ $$ Commented by Rasheed.Sindhi last updated on…
Question Number 97555 by 175 last updated on 08/Jun/20 Commented by bemath last updated on 08/Jun/20 $$\mathrm{yes}\:\mathrm{nice}\:{problem}\: \\ $$ Commented by prakash jain last updated…
Question Number 97541 by mathocean1 last updated on 08/Jun/20 $${Determinate} \\ $$$$\underset{{x}\rightarrow\mathrm{1}/\mathrm{2}} {{lim}}\:\frac{{sin}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\mathrm{2}{x}−\mathrm{1}} \\ $$$$\underset{{x}\rightarrow+\infty\:} {{lim}}\:\frac{{cosx}}{\mathrm{1}+{x}^{\mathrm{2}\:} } \\ $$ Commented by bobhans last updated on…
Question Number 97506 by ali_golmakani last updated on 08/Jun/20 $${please}\:{prove}\:{it} \\ $$$$ \\ $$$$\mathrm{cos}\:{x}=\:{J}_{\mathrm{0}} \left({x}\right)\:+\:\mathrm{2}\sum_{{x}−\mathrm{1}} \left(−\mathrm{1}\right)^{{x}} {J}_{\mathrm{2}{n}} \left({x}\right) \\ $$ Terms of Service Privacy Policy…
Question Number 97490 by ali_golmakani last updated on 08/Jun/20 $${please}\:\:{prove}\:\:{it} \\ $$$$\mathrm{cos}{x}=\:{J}_{\mathrm{0}} \left({x}\right)+\mathrm{2}\sum_{{x}−\mathrm{1}} \left(−\mathrm{1}\right)^{{x}} {J}_{\mathrm{2}{x}} \left({x}\right) \\ $$ Answered by smridha last updated on 08/Jun/20…
Question Number 31954 by mondodotto@gmail.com last updated on 17/Mar/18 Answered by mrW2 last updated on 18/Mar/18 $${A}={A}_{\mathrm{0}} {e}^{−\lambda{t}} \\ $$$$\mathrm{0}.\mathrm{5}{A}_{\mathrm{0}} ={A}_{\mathrm{0}} {e}^{−\mathrm{15}\lambda} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{15}} \\…
Question Number 163024 by SANOGO last updated on 03/Jan/22 $${calcul}\:{en}\:{fonction}\:{de}\:{n} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\left(_{{k}} ^{\mathrm{2}{n}} \right)^{{n}−\mathrm{2}{k}} \\ $$ Commented by SANOGO last updated on 03/Jan/22…
Question Number 97489 by ali_golmakani last updated on 08/Jun/20 $${please}\:\:{prove}\:{it} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}^{\mathrm{2}} } \mathrm{cos}\:{bx}\:\:{dx}=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{{a}}}.{e}^{−\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}}} \\ $$ Answered by smridha last…
Question Number 31949 by naka3546 last updated on 17/Mar/18 Answered by mrW2 last updated on 17/Mar/18 $${N}=\mathrm{4}{n}+\mathrm{2}=\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$${M}=\mathrm{16}{m}+\mathrm{8}=\mathrm{8}\left(\mathrm{2}{m}+\mathrm{1}\right) \\ $$$${MN}=\mathrm{16}\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{m}+\mathrm{1}\right)=\mathrm{16}\left[\mathrm{2}\left({m}+{n}+\mathrm{2}{mn}\right)+\mathrm{1}\right] \\ $$$$=\mathrm{32}\left({m}+{n}+\mathrm{2}{mn}\right)+\mathrm{16} \\ $$$$\Rightarrow{y}=\mathrm{16}…
Question Number 97476 by M±th+et+s last updated on 08/Jun/20 $$\mathrm{2}{F}\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}};{z}\right)=\left(\mathrm{1}−{z}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \ast\ast\mathrm{1} \\ $$$${by}\:{kummer}\:{transformation} \\ $$$$\mathrm{2}{F}\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}};{z}\right)=\mathrm{2}{F}\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}};{z}\right) \\ $$$$\mathrm{2}{F}\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}};{z}\right)=\frac{{sin}^{−\mathrm{1}} \sqrt{\mathrm{1}−{z}}}{\:\sqrt{\mathrm{1}−{z}}}\ast\ast\mathrm{2} \\ $$$$ \\ $$$${why}\:{do}\:{i}\:{get}\:{different}\:{answer}\:{in} \\ $$$$\ast\ast\mathrm{1}\:{and}\:\mathrm{2}\ast\ast \\…