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Question Number 506 by Karting7442 last updated on 25/Jan/15 $${write}\:{each}\:{mixed}\:{number}\:{as}\:{a}\:{decimal}.\: \\ $$$$\:\:\:\:\:\mathrm{4}\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$ \\ $$$$ \\ $$ Answered by prakash jain last updated on…
Question Number 131554 by Salman_Abir last updated on 06/Feb/21 Commented by EDWIN88 last updated on 06/Feb/21 $$\mathrm{A}−\mathrm{B}=\:\begin{bmatrix}{−\mathrm{2}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}}\\{\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{6}}\end{bmatrix} \\ $$$$\mathrm{A}×\mathrm{B}^{\mathrm{T}} \:=\:\begin{bmatrix}{\mathrm{2}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\mathrm{4}}\\{\mathrm{1}\:\:\:\:\:\mathrm{5}\:\:\:\:\:\:\mathrm{7}}\end{bmatrix}\begin{bmatrix}{\mathrm{4}\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\mathrm{4}}\\{\mathrm{3}\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\begin{bmatrix}{\mathrm{8}+\mathrm{6}+\mathrm{12}\:\:\:\:\:\:\:−\mathrm{2}+\mathrm{12}+\mathrm{4}}\\{\mathrm{4}+\mathrm{10}+\mathrm{21}\:\:\:\:\:\:−\mathrm{1}+\mathrm{20}+\mathrm{7}}\end{bmatrix} \\ $$ Terms…
Question Number 131541 by 0731619177 last updated on 05/Feb/21 Answered by asdaldeen last updated on 05/Feb/21 $$\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 456 by Karting7442 last updated on 25/Jan/15 $${solve}\:{for}\:{x}:\:\:\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{6}=\mathrm{11} \\ $$$$ \\ $$ Answered by prakash jain last updated on 09/Jan/15 $$\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{6}=\mathrm{11} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{x}=\mathrm{11}−\mathrm{6}…
Question Number 457 by lun last updated on 25/Jan/15 $$\mathrm{5}{x}+\mathrm{1}=\mathrm{0} \\ $$ Answered by lun last updated on 09/Jan/15 $$\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$ Commented by prakash…
Question Number 65992 by naka3546 last updated on 07/Aug/19 $${x}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{3}\centerdot\mathrm{1}!}\:+\:\frac{\mathrm{1}}{\mathrm{4}\centerdot\mathrm{2}!}\:+\:\frac{\mathrm{1}}{\mathrm{5}\centerdot\mathrm{3}!}\:+\:\ldots\:+\:\frac{\mathrm{1}}{\mathrm{1002}\centerdot\mathrm{1000}!} \\ $$$${x}\centerdot\mathrm{1000}!\:\:=\:\:? \\ $$ Commented by Prithwish sen last updated on 08/Aug/19 $$\mathrm{last}\:\mathrm{term}\:\frac{\mathrm{n}}{\left(\mathrm{n}+\mathrm{1}\right)!}\:\:\mathrm{where}\:\mathrm{n}=\:\mathrm{1001} \\ $$$$\therefore\:\left[\frac{\mathrm{n}+\mathrm{1}−\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}\right]\:=\:\frac{\mathrm{1}}{\mathrm{n}!}\:−\:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}…
Question Number 453 by Karting7442 last updated on 25/Jan/15 $${solve}\:{for}\:{x}\:\:\:\mathrm{3}\left(\mathrm{14}{x}−\mathrm{15}\right)=−\mathrm{72} \\ $$ Answered by prakash jain last updated on 09/Jan/15 $$\mathrm{3}\left(\mathrm{14}{x}−\mathrm{15}\right)=−\mathrm{72} \\ $$$$\mathrm{42}{x}−\mathrm{45}=−\mathrm{72} \\ $$$$\mathrm{42}{x}=−\mathrm{72}+\mathrm{45}…
Question Number 449 by 123456 last updated on 25/Jan/15 $$\mathrm{if}\:\mid{z}\mid=\mathrm{1},{z}\in\mathbb{C},{a}\in\mathbb{R} \\ $$$$\mathrm{find}\:\mathrm{max}\:\mathrm{value}\:\mathrm{of}\:\mid{z}^{\mathrm{2}} +{az}−\mathrm{1}\mid \\ $$ Answered by prakash jain last updated on 09/Jan/15 $${z}={i} \\…