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Question-74473

Question Number 74473 by arkanmath7@gmail.com last updated on 24/Nov/19 Commented by mathmax by abdo last updated on 24/Nov/19 $$\mid{z}−\mathrm{1}\mid=\mathrm{1}\:\Rightarrow{z}−\mathrm{1}={e}^{{i}\theta} \:\:\:\:\mathrm{0}\leqslant\theta\leqslant\pi\:\Rightarrow{z}=\mathrm{1}+{e}^{{i}\theta} \\ $$$$\int_{{C}} \left(\overset{−} {{z}}\right)^{\mathrm{2}} {dz}\:=\int_{\mathrm{0}}…

1-2-1-3-2-2-3-5-n2-2n-1-2n-1-n-n-1-2-2n-1-

Question Number 8865 by 7C;00 last updated on 03/Nov/16 $$\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{1}×\mathrm{3}}+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{3}×\mathrm{5}}+…+\frac{{n}\mathrm{2}}{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}=\frac{{n}\left({n}+\mathrm{1}\right.}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$ Answered by nume1114 last updated on 04/Nov/16 $$\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{1}×\mathrm{3}}+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{3}×\mathrm{5}}+…+\frac{{n}^{\mathrm{2}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}…

z-x-u-x-v-x-Z-K-U-K-V-K-f-u-x-1-k-d-k-dx-k-x-xo-

Question Number 74386 by zaynab last updated on 23/Nov/19 $$\mathrm{z}\left(\mathrm{x}\right)=\mathrm{u}\left(\mathrm{x}\right)+\mathrm{v}\left(\mathrm{x}\right)=\mathrm{Z}\left(\mathrm{K}\right)=\mathrm{U}\left(\mathrm{K}\right)+\mathrm{V}\left(\mathrm{K}\right) \\ $$$$\mathrm{f}\:\mathrm{u}\left(\mathrm{x}\right)=\Sigma\frac{\mathrm{1}}{\mathrm{k}!}\:\frac{\hat {\mathrm{d}k}}{\mathrm{d}\hat {\mathrm{x}k}}\left(\mathrm{x}−\mathrm{x}{o}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

When-an-electron-is-placed-in-an-electric-field-it-experience-an-electric-force-whose-magnitude-is-1-6-times-its-weight-find-the-magnitude-of-the-electric-field-

Question Number 8823 by tawakalitu last updated on 30/Oct/16 $$\mathrm{When}\:\mathrm{an}\:\mathrm{electron}\:\mathrm{is}\:\mathrm{placed}\:\mathrm{in}\:\mathrm{an}\:\mathrm{electric}\:\mathrm{field}, \\ $$$$\mathrm{it}\:\mathrm{experience}\:\mathrm{an}\:\mathrm{electric}\:\mathrm{force}\:\mathrm{whose}\:\mathrm{magnitude} \\ $$$$\mathrm{is}\:\mathrm{1}.\mathrm{6}\:\mathrm{times}\:\mathrm{its}\:\mathrm{weight}\:.\:\mathrm{find}\:\mathrm{the}\:\mathrm{magnitude} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{electric}\:\mathrm{field}. \\ $$ Answered by sandy_suhendra last updated on 30/Oct/16…

5-1-2-6-7-The-end-result-must-in-the-mixed-fraction-

Question Number 74335 by arthur.kangdani@gmail.com last updated on 22/Nov/19 $$\mathrm{5}\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{6}}{\mathrm{7}}=?\: \\ $$$${The}\:{end}\:{result}\:{must}\:{in}\:{the} \\ $$$$\boldsymbol{{mixed}}\:\boldsymbol{{fraction}}. \\ $$ Answered by arthur.kangdani@gmail.com last updated on 26/Nov/19 $$\mathrm{5}\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{6}}{\mathrm{7}}=\frac{\mathrm{11}}{\mathrm{2}}×\frac{\mathrm{6}}{\mathrm{7}}=\frac{\mathrm{66}}{\mathrm{14}}=\frac{\mathrm{33}}{\mathrm{7}}=\mathrm{4}\frac{\mathrm{5}}{\mathrm{7}} \\…

Question-74328

Question Number 74328 by arthur.kangdani@gmail.com last updated on 22/Nov/19 Answered by ajfour last updated on 22/Nov/19 $${x}=\frac{\begin{vmatrix}{\mathrm{10}}&{\mathrm{2}}&{\mathrm{4}}\\{\mathrm{4}}&{\mathrm{10}}&{\mathrm{7}}\\{\mathrm{9}}&{\mathrm{5}}&{\mathrm{16}}\end{vmatrix}_{} }{\begin{vmatrix}{\mathrm{6}}&{\mathrm{2}}&{\mathrm{4}}\\{\mathrm{9}}&{\mathrm{10}}&{\mathrm{7}}\\{\mathrm{12}}&{\mathrm{5}}&{\mathrm{16}}\end{vmatrix}^{} } \\ $$$$\Rightarrow\:\:{x}=\frac{\mathrm{10}\left(\mathrm{160}−\mathrm{35}\right)−\mathrm{2}\left(\mathrm{64}−\mathrm{63}\right)+\mathrm{4}\left(\mathrm{20}−\mathrm{90}\right)}{\mathrm{6}\left(\mathrm{160}−\mathrm{35}\right)−\mathrm{2}\left(\mathrm{144}−\mathrm{84}\right)+\mathrm{4}\left(\mathrm{45}−\mathrm{120}\right)} \\ $$$$\:\:\:\:\:{x}=\frac{\mathrm{1250}−\mathrm{2}−\mathrm{280}}{\mathrm{750}−\mathrm{120}−\mathrm{300}}\:=\:\frac{\mathrm{968}}{\mathrm{330}}\:=\:\frac{\mathrm{88}}{\mathrm{30}} \\ $$$$…..…