Question Number 59199 by cesar.marval.larez@gmail.com last updated on 05/May/19 Commented by cesar.marval.larez@gmail.com last updated on 05/May/19 $$\mathrm{who}\:\mathrm{know}\:\mathrm{spanish}? \\ $$ Commented by Forkum Michael Choungong last…
Question Number 124682 by mathocean1 last updated on 05/Dec/20 $${Given}\:\begin{cases}{{u}_{\mathrm{0}} =\mathrm{5}}\\{\:{u}_{{n}+\mathrm{1}} =\mathrm{3}{u}_{{n}} −\mathrm{4}}\end{cases} \\ $$$$\mathrm{1}.\:{show}\:{that}\:\forall\:{n}\in\mathbb{N},\:{u}_{{n}} =\mathrm{2}+\mathrm{3}^{{u}_{{n}+\mathrm{1}} } \:\:\: \\ $$$${a}.\:{Deduct}\:{that}\:{u}_{{n}} \:{is}\:{odd}. \\ $$$${c}.\:{Show}\:{that}\:{GCD}\left({u}_{{n}} ;{u}_{{n}+\mathrm{1}} \right)=\mathrm{1}…
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Question Number 58698 by arnabmaiti550@gmail.com last updated on 28/Apr/19 $$\overset{\rightarrow} {\bigtriangledown}\centerdot\left(\frac{{e}^{{br}} }{{r}^{\mathrm{2}} }\:\overset{\wedge} {{e}}_{{r}} \right)=?\:\:\:\:{b}\:{is}\:{a}\:{constant}. \\ $$ Commented by tanmay last updated on 28/Apr/19 $${pls}\:{mention}\:{co}−{ordinate}\:{system}…
Question Number 122358 by mathocean1 last updated on 16/Nov/20 $${Given}\:{a},\:{b}\:\in\left[\mathrm{0};\mathrm{4}\right] \\ $$$$\mathrm{309}{a}+\mathrm{15}{c}=\mathrm{226}{b} \\ $$$$\left.\mathrm{1}\right)\:{show}\:{that}\:\mathrm{2}{b}\equiv\mathrm{0}\left[\mathrm{3}\right]\:{and}\:\mathrm{3}{a}\equiv\mathrm{1}\left[\mathrm{5}\right] \\ $$$$ \\ $$ Answered by mindispower last updated on 16/Nov/20…
Question Number 122189 by physicstutes last updated on 14/Nov/20 $$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{forces}\:\boldsymbol{\mathrm{F}}_{\mathrm{1}} ,\:\boldsymbol{\mathrm{F}}_{\mathrm{2}} \:\mathrm{and}\:\boldsymbol{\mathrm{F}}_{\mathrm{3}} \:\mathrm{have}\:\mathrm{position}\:\mathrm{vectors} \\ $$$$\:\boldsymbol{\mathrm{r}}_{\mathrm{1}} ,\:\boldsymbol{\mathrm{r}}_{\mathrm{2}} \:\mathrm{and}\:\boldsymbol{\mathrm{r}}_{\mathrm{3}} .\:\mathrm{Where}, \\ $$$$\:\boldsymbol{\mathrm{F}}_{\mathrm{1}} \:=\:\left(\mathrm{2}\boldsymbol{\mathrm{i}}\:+\:\mathrm{3}\boldsymbol{\mathrm{i}}\:+\:\boldsymbol{\mathrm{k}}\right)\:\mathrm{N}\:\:\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{1}} \:=\:\left(\boldsymbol{\mathrm{i}}\:+\:\mathrm{2}\boldsymbol{\mathrm{k}}\right)\:\mathrm{m} \\ $$$$\:\:\boldsymbol{\mathrm{F}}_{\mathrm{2}} \:=\:\left(\mathrm{3}\boldsymbol{\mathrm{i}}\:+\:\mathrm{2}\boldsymbol{\mathrm{i}}\:+\:\mathrm{5}\boldsymbol{\mathrm{k}}\right)\:\mathrm{N}\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{2}}…
Question Number 56462 by cesar.marval.larez@gmail.com last updated on 16/Mar/19 Commented by cesar.marval.larez@gmail.com last updated on 16/Mar/19 $$\mathrm{Quien}\:\mathrm{sepa}\:\mathrm{espa}\overset{\_} {\mathrm{n}ol}\:\mathrm{puede}\:\mathrm{ayudarme} \\ $$$$\mathrm{con}\:\mathrm{la}\:\mathrm{3}\:\mathrm{4}\:\mathrm{y}\:\mathrm{5}?\:\mathrm{Por}\:\mathrm{favor} \\ $$ Terms of Service…
Question Number 56461 by cesar.marval.larez@gmail.com last updated on 16/Mar/19 Answered by tanmay.chaudhury50@gmail.com last updated on 17/Mar/19 $$\overset{\rightarrow} {{u}}_{\mathrm{3}} ={a}\overset{\rightarrow} {{u}}_{\mathrm{1}} +{b}\overset{\rightarrow} {{u}}_{\mathrm{2}} \\ $$$$\:\:\:={a}\left(\mathrm{2}{i}\right)+{b}\left({j}−\mathrm{3}{k}\right) \\…
Question Number 56460 by cesar.marval.larez@gmail.com last updated on 16/Mar/19 Answered by tanmay.chaudhury50@gmail.com last updated on 17/Mar/19 $$\overset{\rightarrow} {{w}}=\mathrm{3}{i}+\mathrm{4}{j}+\mathrm{5}{k} \\ $$$$\overset{\rightarrow} {{u}}={ai}+\mathrm{2}{j}−\mathrm{3}{k} \\ $$$$\overset{\rightarrow} {{v}}={i}+\mathrm{4}{j}+{ck} \\…
Question Number 55991 by bshahid010@gmail.com last updated on 07/Mar/19 Commented by maxmathsup by imad last updated on 07/Mar/19 $${let}\:\:{S}_{{n}} \left({x}\right)=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\sum_{{k}=\mathrm{1}} ^{{m}} \:\left[{kx}\right]{we}\:{have}\:\:\:\left[{kx}\right]\leqslant{kx}<\left[{kx}\right]+\mathrm{1}\:\Rightarrow\:\:{kx}−\mathrm{1}<\left[{kx}\right]\leqslant{kx}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{n}^{\mathrm{2}}…