Question Number 118280 by hmh2k20 last updated on 16/Oct/20 $$\left(\mathrm{26}\right)\:\:\:\overset{\rightarrow} {\mathrm{OA}}\:=\:\overset{\rightarrow} {\mathrm{a}}\:=\:\begin{pmatrix}{\mathrm{4}.\mathrm{8}}\\{\mathrm{3}.\mathrm{6}}\end{pmatrix}\:\:,\:\:\overset{\rightarrow} {\mathrm{OB}}\:=\:\overset{\rightarrow} {\mathrm{b}}\:=\:\begin{pmatrix}{\:\:\mathrm{8}}\\{\mathrm{15}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{a}}.\:\overset{\rightarrow} {\mathrm{b}}\:=\:\left(\mathrm{4}.\mathrm{8}\right)\left(\mathrm{8}\right)\:+\:\left(\mathrm{3}.\mathrm{6}\right)\left(\mathrm{15}\right)\:=\:\mathrm{92}.\overset{\rightarrow\:} {\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}\:=\:\sqrt{\mathrm{4}.\mathrm{8}^{\mathrm{2}} +\mathrm{3}.\mathrm{6}^{\mathrm{2}} }\:=\:\:\sqrt{\mathrm{23}.\mathrm{04}+\mathrm{12}.\mathrm{96}}\:=\:\sqrt{\mathrm{36}}\:=\:\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:=\:\sqrt{\mathrm{8}^{\mathrm{2}}…
Question Number 118277 by hmh2k20 last updated on 16/Oct/20 $$\left(\mathrm{23}\right)\:\mathrm{Let}\:\overset{\rightarrow} {\mathrm{a}}\:=\:\begin{pmatrix}{\:\:\mathrm{7}}\\{\mathrm{24}}\end{pmatrix}\:\:\:\mathrm{and}\:\:\overset{\rightarrow} {\mathrm{b}}\:=\:\begin{pmatrix}{\:\:\:\mathrm{3}}\\{−\mathrm{4}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}\:=\:\left(\mathrm{7}\right)\left(\mathrm{3}\right)\:+\:\left(\mathrm{24}\right)\left(−\mathrm{4}\right)\:=\:\mathrm{21}−\mathrm{96}\:=\:−\mathrm{75} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}\:=\:\:\sqrt{\mathrm{7}^{\mathrm{2}} +\mathrm{24}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{49}+\mathrm{576}}\:=\:\sqrt{\mathrm{625}}\:=\:\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:=\:\:\sqrt{\mathrm{3}^{\mathrm{2}} +\left(−\mathrm{4}\right)^{\mathrm{2}} }\:=\:\sqrt{\mathrm{9}+\mathrm{16}}\:=\:\sqrt{\mathrm{25}}\:=\:\mathrm{5} \\…
Question Number 118275 by bramlexs22 last updated on 16/Oct/20 Commented by bramlexs22 last updated on 16/Oct/20 $${f}\left({x}\right)={x}^{\mathrm{2}} \:,\:{when}\: \\ $$$$\:\left(\frac{\mathrm{3}}{{f}\left(\mathrm{1}\right){f}\left(\mathrm{2}\right)}+\frac{\mathrm{5}}{{f}\left(\mathrm{2}\right){f}\left(\mathrm{3}\right)}+\frac{\mathrm{7}}{{f}\left(\mathrm{3}\right){f}\left(\mathrm{4}\right)}+\frac{\mathrm{9}}{{f}\left(\mathrm{4}\right){f}\left(\mathrm{5}\right)}+\frac{\mathrm{11}}{{f}\left(\mathrm{5}\right){f}\left(\mathrm{6}\right)}\right).{x}\:<\:\mathrm{70}\: \\ $$$${Find}\:{the}\:{largest}\:{complete} \\ $$$${solution}\:{of}\:{the}\:{inequality}\:. \\…
Question Number 52730 by lateefhussain25@gmaik.com last updated on 12/Jan/19 $${The}\:{area}\:{of}\:\:{a}\:{re} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\…
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Question Number 183777 by Michaelfaraday last updated on 30/Dec/22 Commented by Frix last updated on 30/Dec/22 $${k}=\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{3}} \\ $$ Answered by HeferH last updated on…
Question Number 183769 by Michaelfaraday last updated on 29/Dec/22 $${solve}\:{for}\:{x}: \\ $$$${x}^{{x}^{{x}} } =\mathrm{2}^{\mathrm{2048}} \\ $$$${by}\:{using}\:{lambert}\:{function} \\ $$ Commented by mr W last updated on…
Question Number 118227 by bemath last updated on 16/Oct/20 $${If}\:{x}\:=\:\sqrt{\mathrm{42}−\sqrt{\mathrm{42}−\sqrt{\mathrm{42}−…}}} \\ $$$${y}\:=\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+…}}} \\ $$$${z}=\sqrt{{y}.\sqrt{{y}.\sqrt{{y}.\sqrt{{y}…}}}}\:.\:{Find}\:{x}+{y}+{z}\:. \\ $$ Answered by mathmax by abdo last updated on 16/Oct/20…
Question Number 183756 by Michaelfaraday last updated on 29/Dec/22 $${solve}\:{for}\:{x}\:{by}\:{using}\:{lambert}\:{function} \\ $$$${x}^{\mathrm{2}} =\mathrm{16}^{{x}} \\ $$ Answered by aleks041103 last updated on 30/Dec/22 $$\mathrm{2}{lnx}={xln}\mathrm{16} \\ $$$$\Rightarrow{lnx}={xln}\mathrm{4}…
Question Number 118196 by mathocean1 last updated on 15/Oct/20 $${solve}\:{in}\:\mathbb{N}: \\ $$$${b}^{\mathrm{3}} \left(\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{b}+\mathrm{1}\right)=\mathrm{18360} \\ $$ Answered by Olaf last updated on 15/Oct/20 $${b}^{\mathrm{3}} \left({b}^{\mathrm{2}}…