Question Number 123115 by 676597498 last updated on 23/Nov/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{2}^{{n}} =? \\ $$ Commented by Dwaipayan Shikari last updated on 23/Nov/20 $${It}\:{diverges}\:\rightarrow\infty\:\:\:\:\:{as}\:\mid{a}\mid=\mathrm{2} \\…
Question Number 57572 by cesar.marval.larez@gmail.com last updated on 07/Apr/19 $$\int\mathrm{sec}^{\mathrm{4}} \mathrm{2xdx} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 07/Apr/19 $$\int{sec}^{\mathrm{2}} \mathrm{2}{x}×{sec}^{\mathrm{2}} \mathrm{2}{xdx} \\ $$$$\int\left(\mathrm{1}+{tan}^{\mathrm{2}}…
Question Number 188294 by normans last updated on 27/Feb/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 188295 by normans last updated on 27/Feb/23 $$ \\ $$$$\:\:\:\:\boldsymbol{{let}}\:\:\boldsymbol{{S}}\:\boldsymbol{{be}}\:\boldsymbol{{the}}\:\boldsymbol{{sets}}\:\boldsymbol{{be}}\:\boldsymbol{{the}}\:\boldsymbol{{sequences}}\:\boldsymbol{{of}}\:\boldsymbol{{lenght}}\:\mathrm{2018}\:\:\: \\ $$$$\:\:\:\boldsymbol{{whose}}\:\boldsymbol{{terms}}\:\boldsymbol{{are}}\:\boldsymbol{{in}}\:\boldsymbol{{the}}\:\boldsymbol{{sets}}\:\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{10}\right\}\:\boldsymbol{{and}}\:\boldsymbol{{sum}}\:\boldsymbol{{to}}\:\mathrm{3860}.\:\:\: \\ $$$$\:\:\:\:\boldsymbol{{prove}}\:\boldsymbol{{that}}\:\boldsymbol{{the}}\:\boldsymbol{{cardinality}}\:\boldsymbol{{of}}\:\boldsymbol{{S}}\:\boldsymbol{{is}}\:\boldsymbol{{at}}\:\boldsymbol{{most}}\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{3860}} \centerdot\left(\:\frac{\mathrm{2018}}{\mathrm{2048}}\right)^{\mathrm{2018}} \\ $$$$ \\ $$$$\:\:\:\: \\ $$…
Question Number 122579 by bramlexs22 last updated on 18/Nov/20 Answered by liberty last updated on 18/Nov/20 $$\left(\mathrm{2a}\right)\:\mathrm{Q}\:\mathrm{symetric}\:\mathrm{if}\:\mathrm{Q}\:=\:\mathrm{Q}^{\mathrm{t}} \\ $$$$\Leftrightarrow\:\begin{pmatrix}{\mathrm{5}\:\:\:\:\:\mathrm{6x}+\mathrm{5y}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6}\:\:\:\:\:\:\mathrm{5y}^{\mathrm{2}} −\mathrm{2x}}\\{\mathrm{x}^{\mathrm{2}} −\mathrm{8}\:\:\:\mathrm{3z}\:\:\:\:\:\:\:\:\:\:\:\mathrm{p}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} −\mathrm{8}}\\{\mathrm{6x}+\mathrm{5y}\:\:\:\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\mathrm{3z}}\\{\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5y}^{\mathrm{2}} −\mathrm{2x}\:\:\:\mathrm{p}}\end{pmatrix} \\ $$$$\mathrm{gives}\:\begin{cases}{\mathrm{6x}+\mathrm{5y}=\mathrm{1}\:\wedge\:\mathrm{x}^{\mathrm{2}}…
Question Number 56899 by Tawa1 last updated on 26/Mar/19 $$\mathrm{Let}\:\:\mathrm{W}\:\:\mathrm{be}\:\mathrm{the}\:\mathrm{subspace}\:\mathrm{of}\:\:\mathbb{R}^{\mathrm{4}} \:\:\mathrm{generated}\:\mathrm{by}\:\mathrm{vector}\: \\ $$$$\left(\mathrm{1},\:−\:\mathrm{2},\:\mathrm{5},\:−\:\mathrm{3}\right),\:\:\:\:\left(\mathrm{2},\:\mathrm{3},\:\mathrm{1},\:−\:\mathrm{4}\right),\:\:\:\left(\mathrm{3},\:\mathrm{8},\:−\:\mathrm{3},\:−\:\mathrm{5}\right)\:\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{basis}\:\mathrm{and}\:\mathrm{dimension}\:\mathrm{of}\:\:\mathrm{W}. \\ $$ Answered by kaivan.ahmadi last updated on 26/Mar/19 $$\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:−\mathrm{2}\:\:\:\:\:\:\:\:\:\mathrm{5}\:\:\:\:\:\:−\mathrm{3}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:−\mathrm{4}\:\:}\\{\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\mathrm{8}\:\:\:\:\:−\mathrm{3}\:\:\:\:\:\:\:−\mathrm{5}}\end{vmatrix}\underset{−\mathrm{3}{R}_{\mathrm{1}}…
Question Number 56803 by Tawa1 last updated on 24/Mar/19 $$\mathrm{Evaluate}:\:\:\:\int_{\:\mathrm{1}} ^{\:\mathrm{2}} \:\left(\mathrm{A}\centerdot\mathrm{B}\:×\:\mathrm{C}\right)\:\mathrm{dt}\:\:\:\:\mathrm{and}\:\:\:\int_{\:\mathrm{1}} ^{\:\mathrm{2}} \:\mathrm{A}\:×\:\left(\mathrm{B}\:×\:\mathrm{C}\right)\:\:\:\mathrm{dt} \\ $$$$\mathrm{where},\:\:\:\:\:\:\:\:\:\mathrm{A}\:\:=\:\:\mathrm{ti}\:−\:\mathrm{3j}\:+\:\mathrm{2tk},\:\:\:\:\:\:\:\mathrm{B}\:\:=\:\:\mathrm{i}\:−\:\mathrm{2j}\:+\:\mathrm{2k}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{C}\:\:=\:\:\mathrm{3i}\:+\:\mathrm{tj}\:−\:\mathrm{k} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated…
Question Number 55948 by Tawa1 last updated on 06/Mar/19 Answered by kaivan.ahmadi last updated on 07/Mar/19 $${a}.\:\:{A}×{B}=\begin{vmatrix}{{i}\:\:\:\:\:{j}\:\:\:\:\:\:{k}}\\{\mathrm{3}\:\:\:−\mathrm{1}\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\mathrm{1}\:\:\:\:−\mathrm{1}}\end{vmatrix}\begin{vmatrix}{{i}\:\:\:\:{j}\:\:\:\:\:\:\:{k}}\\{\mathrm{3}\:\:\:−\mathrm{1}\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\:\:\mathrm{1}\:\:\:−\mathrm{1}}\end{vmatrix}= \\ $$$$\left({i}+\mathrm{4}{j}+\mathrm{3}{k}\right)−\left(−\mathrm{2}{k}−\mathrm{3}{j}+\mathrm{2}{i}\right)=−\mathrm{3}{i}+\mathrm{7}{j}+\mathrm{5}{k} \\ $$$$\left({A}×{B}\right)×{C}=\begin{vmatrix}{{i}\:\:\:\:\:\:\:\:\:{j}\:\:\:\:\:{k}}\\{−\mathrm{3}\:\:\:\mathrm{7}\:\:\:\:\:\mathrm{5}}\\{\mathrm{1}\:\:\:\:\:\:−\mathrm{2}\:\:\:\mathrm{2}}\end{vmatrix}\begin{vmatrix}{{i}\:\:\:\:\:\:\:\:\:\:{j}\:\:\:\:\:{k}}\\{−\mathrm{3}\:\:\:\:\:\mathrm{7}\:\:\:\:\mathrm{5}}\\{\mathrm{1}\:\:\:\:\:−\mathrm{2}\:\:\:\:\mathrm{2}}\end{vmatrix}= \\ $$$$\left(\mathrm{14}{i}+\mathrm{5}{j}+\mathrm{6}{k}\right)−\left(\mathrm{7}{k}−\mathrm{6}{j}−\mathrm{10}{i}\right)= \\ $$$$\mathrm{24}{i}+\mathrm{11}{j}−{k}…
Question Number 55860 by Easyman32 last updated on 05/Mar/19 Answered by tanmay.chaudhury50@gmail.com last updated on 05/Mar/19 $${jones}\:{probablity}\:{of}\:{winning}={a} \\ $$$${john}\:{probablity}\:{of}\:{winning}=\mathrm{3}{a} \\ $$$${david}….=\mathrm{2}×\mathrm{3}{a}=\mathrm{6}{a} \\ $$$${a}+\mathrm{3}{a}+\mathrm{6}{a}=\mathrm{1} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{10}}…
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