Question Number 99237 by abdomathmax last updated on 19/Jun/20 $$\mathrm{calculate}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\:\:\:\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\:\frac{\sqrt{\mathrm{kn}}}{\left(\mathrm{k}^{\mathrm{2}} \:+\mathrm{n}^{\mathrm{2}} \right)} \\ $$ Answered by Ar Brandon last updated on 19/Jun/20…
Question Number 33698 by math khazana by abdo last updated on 22/Apr/18 $${let}\:\:{f}_{{n}} \left({x}\right)=\:{n}^{{x}} \:{e}^{−{nx}} \:\:\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{study}\:{the}\:{simple}\:{and}\:{uniform}\:{convervence}\:{for} \\ $$$$\Sigma\:\:{f}_{{n}} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{let}\:\:{S}\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{f}_{{n}}…
Question Number 99235 by abdomathmax last updated on 19/Jun/20 $$\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{3}} \left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 33699 by math khazana by abdo last updated on 22/Apr/18 $${let}\:{S}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{f}_{{n}} \left({x}\right)\:\:{with}\:{f}_{{n}} \left({x}\right)=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!\left({x}+{n}\right)} \\ $$$$\left.{x}\in\right]\mathrm{0},+\infty\left[\right. \\ $$$$\left.\mathrm{1}\right)\:\:{prove}\:{that}\:{S}\:{id}\:{defined}\:.{calculate}\:{S}\left(\mathrm{1}\right)\:{and} \\ $$$${prove}\:{that}\:\forall{x}>\mathrm{0}\:\:{xS}\left({x}\right)\:−{S}\left({x}+\mathrm{1}\right)\:=\frac{\mathrm{1}}{{e}} \\…
Question Number 164770 by cortano1 last updated on 21/Jan/22 $$\begin{cases}{{f}\left(\mathrm{3}{x}−\mathrm{1}\right)+{g}\left(\mathrm{6}{x}−\mathrm{1}\right)=\mathrm{3}{x}}\\{{f}\left({x}+\mathrm{1}\right)+{x}\:{g}\left(\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{2}{x}^{\mathrm{2}} +{x}}\end{cases} \\ $$$$\:{f}\left({x}\right)=? \\ $$ Answered by blackmamba last updated on 21/Jan/22 $$\:\mathrm{let}\:\begin{cases}{{f}\left({x}\right)={px}+{q}}\\{{g}\left({x}\right)={ax}+{b}}\end{cases} \\ $$$$\:\begin{cases}{{f}\left(\mathrm{3}{x}−\mathrm{1}\right)=\mathrm{3}{px}+{q}−{p}}\\{{g}\left(\mathrm{6}{x}−\mathrm{1}\right)=\mathrm{6}{ax}+{b}−{a}}\end{cases}\:\Rightarrow\left(\mathrm{3}{p}+\mathrm{6}{a}\right){x}+{b}+{q}−\left({a}+{p}\right)=\mathrm{3}{x}…
Question Number 99222 by mathmax by abdo last updated on 19/Jun/20 $$\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{3}} } \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 33658 by rahul 19 last updated on 21/Apr/18 $$\boldsymbol{{I}}{f}\:{f}\left({x}\right)=\:{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{1} \\ $$$${then}\:{find}\:{number}\:{of}\:{different}\:\:\:{real} \\ $$$${solutions}\:{of}\:{f}\left({f}\left({x}\right)\right)=\mathrm{0}\:? \\ $$ Commented by rahul 19 last updated on…
Question Number 33651 by rahul 19 last updated on 21/Apr/18 $${Let}\:{function}\:{f}\left({x}\right)\:{be}\:{defined}\:{as}\: \\ $$$${f}\left({x}\right)=\:{x}^{\mathrm{2}} +{bx}+{c}\:,\:{where}\:{b},{c}\in{R}\:. \\ $$$${And}\:{f}\left(\mathrm{1}\right)\:−\:\mathrm{2}{f}\left(\mathrm{5}\right)\:+{f}\left(\mathrm{9}\right)\:=\mathrm{32}. \\ $$$${Find}\:{no}.\:{of}\:{ordered}\:{pairs}\:\left({b},{c}\right) \\ $$$${such}\:{that}\:\mid{f}\left({x}\right)\mid\leqslant\mathrm{8}\:\forall\:{x}\in\:\left[\mathrm{1},\mathrm{9}\right]\:? \\ $$ Answered by MJS…
Question Number 33649 by rahul 19 last updated on 21/Apr/18 $${Consider}\:{f}:{R}^{+} \rightarrow{R}\:{such}\:{that} \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{1}\:{for}\:{a}\in{R}^{+} \:{and}\: \\ $$$${f}\left({x}\right).{f}\left({y}\right)\:+\:{f}\left(\frac{\mathrm{3}}{{x}}\right).{f}\left(\frac{\mathrm{3}}{{y}}\right)\:=\:\mathrm{2}{f}\left({xy}\right) \\ $$$$\forall\:{x},{y}\:\in\:{R}^{+} .\:{Then}\:{find}\:{f}\left({x}\right)\:? \\ $$ Commented by rahul…
Question Number 99159 by bemath last updated on 19/Jun/20 Commented by som(math1967) last updated on 19/Jun/20 $$\mathrm{let}\:\mathrm{pt}.\:\mathrm{on}\:\mathrm{paabola}\:\left(\mathrm{h},\mathrm{k}\right) \\ $$$$\therefore\sqrt{\left(\mathrm{h}+\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{k}−\mathrm{3}\right)^{\mathrm{2}} }=\frac{\mid\mathrm{k}−\mathrm{7}\mid}{\:\sqrt{\mathrm{1}^{\mathrm{2}} }} \\ $$$$\Rightarrow\left(\mathrm{h}+\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{k}−\mathrm{3}\right)^{\mathrm{2}}…