Question Number 182507 by Matica last updated on 10/Dec/22 $$\:{Can}\:{a}\:{mathematician}\:{solve}\:{or}\:{know}\:{everything}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 182506 by Matica last updated on 10/Dec/22 $$\:\:{Calculate}\:{S}\:=\:\mathrm{1}×\mathrm{1}!\:+\:\mathrm{2}×\mathrm{2}!\:+\:\mathrm{3}×\mathrm{3}!\:+\:…+\:{n}×{n}! \\ $$$$\:\:\: \\ $$ Answered by mr W last updated on 10/Dec/22 $${n}×{n}!=\left({n}+\mathrm{1}\right)!−{n}! \\ $$$$\Rightarrow{S}=\left({n}+\mathrm{1}\right)!−\mathrm{1}…
Question Number 116961 by malwan last updated on 08/Oct/20 Commented by malwan last updated on 08/Oct/20 $$\mid{AB}\mid=? \\ $$ Commented by bobhans last updated on…
Question Number 116941 by Her_Majesty last updated on 08/Oct/20 $${prove}\:{that}\:\mathrm{7}×\mathrm{8}=\mathrm{7}+\mathrm{8}=\mathrm{78} \\ $$ Commented by MJS_new last updated on 08/Oct/20 $$\mathrm{one}\:\mathrm{possible}\:“\mathrm{proof}'' \\ $$$$\begin{cases}{\mathrm{1}=\mathrm{e}^{\mathrm{2}\pi\mathrm{i}} }\\{\mathrm{1}=\mathrm{e}^{\mathrm{4}\pi\mathrm{i}} }\end{cases}\:\Leftrightarrow\:\mathrm{2}\pi\mathrm{i}=\mathrm{4}\pi\mathrm{i}\:\Leftrightarrow\:\mathrm{2}=\mathrm{4}\:\Leftrightarrow \\…
Question Number 116939 by joki last updated on 08/Oct/20 $${prof}\:\:{if}\:\:{the}\:{limitf}\left({x}\right)={L}\:{and}\: \\ $$$${limit}\:{f}\left({x}\right)={M},{then}\:{L}={M} \\ $$ Answered by MAB last updated on 08/Oct/20 $$\underset{\boldsymbol{\mathrm{x}}\rightarrow{x}_{\mathrm{0}} } {\boldsymbol{\mathrm{lim}}}{f}\left({x}\right)={L}={M} \\…
Question Number 182456 by yaslm last updated on 09/Dec/22 Commented by mr W last updated on 09/Dec/22 $${F}=\frac{\pi×\mathrm{5}^{\mathrm{2}} }{\mathrm{2}}×\left(\mathrm{6}−\frac{\mathrm{4}×\mathrm{5}}{\mathrm{3}\pi}\right)×\mathrm{62}.\mathrm{4}\:{lbf} \\ $$ Terms of Service Privacy…
Question Number 116915 by mathdave last updated on 07/Oct/20 Answered by Lordose last updated on 07/Oct/20 $$\boldsymbol{\Omega}\:=\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{a}^{\mathrm{n}} }{\mathrm{1}+\mathrm{a}^{\mathrm{n}} } \\ $$$$\boldsymbol{\Omega}\:=\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{a}^{\mathrm{n}} \mathrm{lna}}{\mathrm{a}^{\mathrm{n}} \mathrm{lna}}…
Question Number 116907 by math35 last updated on 07/Oct/20 Answered by Lordose last updated on 07/Oct/20 $$\mathrm{2}^{\mathrm{x}+\mathrm{2}} \:×\:\mathrm{2}^{\mathrm{3}\left(\mathrm{2x}−\mathrm{3}\right)} \:=\:\mathrm{2}^{\mathrm{10}} \\ $$$$\mathrm{2}^{\mathrm{x}+\mathrm{2}+\mathrm{6x}−\mathrm{9}} \:=\:\mathrm{2}^{\mathrm{10}} \\ $$$$\mathrm{7x}−\mathrm{7}=\mathrm{10} \\…
Question Number 182436 by akolade last updated on 09/Dec/22 Answered by Rasheed.Sindhi last updated on 09/Dec/22 $$\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}.\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}.\mathrm{4}}+…+\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right.}=\frac{{n}}{{n}+\mathrm{1}} \\ $$$${n}=\mathrm{1}:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${Assumed}\:{for}\:{n}={k} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}.\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}.\mathrm{4}}+…+\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right.}=\frac{{k}}{{k}+\mathrm{1}} \\ $$$${Adding}:\frac{\mathrm{1}}{\overline…
Question Number 182435 by akolade last updated on 09/Dec/22 Answered by Rasheed.Sindhi last updated on 09/Dec/22 $${p}\left({n}\right):\mathrm{1}+\mathrm{3}+\mathrm{5}+…+\left(\mathrm{2}{n}−\mathrm{1}\right)={n}^{\mathrm{2}} \\ $$$${p}\left(\mathrm{1}\right):\mathrm{1}=\mathrm{1}^{\mathrm{2}} \:\left({true}\right) \\ $$$${Let}\:{p}\left({k}\right)\:{is}\:{true} \\ $$$$\mathrm{1}+\mathrm{3}+\mathrm{5}+…+\left(\mathrm{2}{k}−\mathrm{1}\right)={k}^{\mathrm{2}} \\…