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Author: Tinku Tara

For-p-q-and-r-prime-numbers-satisfying-p-q-1-r-1-1064-r-p-1-q-1-1554-find-the-value-p-q-1-r-

Question Number 213463 by golsendro last updated on 06/Nov/24 $$\:\:\mathrm{For}\:\mathrm{p},\mathrm{q}\:\mathrm{and}\:\mathrm{r}\:\mathrm{prime}\:\mathrm{numbers}\: \\ $$$$\:\:\mathrm{satisfying}\:\begin{cases}{\mathrm{p}\left(\mathrm{q}+\mathrm{1}\right)\left(\mathrm{r}+\mathrm{1}\right)=\mathrm{1064}}\\{\mathrm{r}\left(\mathrm{p}+\mathrm{1}\right)\left(\mathrm{q}+\mathrm{1}\right)=\mathrm{1554}}\end{cases} \\ $$$$\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{p}\left(\mathrm{q}+\mathrm{1}\right)\mathrm{r}\: \\ $$ Answered by A5T last updated on 06/Nov/24 $${p}\mid\mathrm{1064}=\mathrm{2}^{\mathrm{3}} ×\mathrm{7}×\mathrm{19};\:{p}+\mathrm{1}\mid\mathrm{1554}=\mathrm{2}×\mathrm{3}×\mathrm{7}×\mathrm{37}…

Find-tupple-natural-numbers-a-b-c-such-that-max-a-b-2-a-b-2-b-c-2-b-c-2-c-a-2-c-a-2-a-min-a-b-2-a-b-2-b-c-2-b-c-2-c-a-2-

Question Number 213459 by golsendro last updated on 06/Nov/24 $$\:\:\mathrm{Find}\:\mathrm{tupple}\:\mathrm{natural}\:\mathrm{numbers}\:\left(\mathrm{a},\mathrm{b},\mathrm{c}\right) \\ $$$$\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\:\:\:\begin{cases}{\mathrm{max}\left\{\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}+\frac{\mid\mathrm{a}−\mathrm{b}\mid}{\mathrm{2}}\:,\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2}}+\frac{\mid\mathrm{b}−\mathrm{c}\mid}{\mathrm{2}}\:,\frac{\mathrm{c}+\mathrm{a}}{\mathrm{2}}+\frac{\mid\mathrm{c}−\mathrm{a}\mid}{\mathrm{2}}\right\}=\mathrm{a}}\\{\mathrm{min}\left\{\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}−\frac{\mid\mathrm{a}−\mathrm{b}\mid}{\mathrm{2}}\:,\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2}}−\frac{\mid\mathrm{b}−\mathrm{c}\mid}{\mathrm{2}}\:,\:\frac{\mathrm{c}+\mathrm{a}}{\mathrm{2}}−\frac{\mid\mathrm{c}−\mathrm{a}\mid}{\mathrm{2}}\right\}=\mathrm{b}}\end{cases} \\ $$$$\:\:\mathrm{where}\:\mathrm{a}+\mathrm{b}+\mathrm{c}\:=\:\mathrm{10} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

0-2pi-z-sin-z-1-cos-2-z-dz-Contour-integral-z-2-1-z-2-1-dz-z-2-sin-z-z-2-1-dz-

Question Number 213484 by issac last updated on 06/Nov/24 $$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\frac{{z}\centerdot\mathrm{sin}\left({z}\right)}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left({z}\right)}\mathrm{d}{z}\:\:\left(\mathrm{Contour}\:\mathrm{integral}\right)\: \\ $$$$\oint_{\:\mid{z}\mid=\mathrm{2}} \:\frac{\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z} \\ $$$$\oint_{\:\mid{z}\mid=\mathrm{2}} \:\:\frac{\mathrm{sin}\left({z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z} \\ $$ Terms of…

x-5-5x-6-x-0-x-

Question Number 213486 by a.lgnaoui last updated on 06/Nov/24 $$\:\:\boldsymbol{\mathrm{x}}^{\mathrm{5}} +\mathrm{5}\boldsymbol{\mathrm{x}}−\frac{\mathrm{6}}{\boldsymbol{\mathrm{x}}}=\mathrm{0}\:\:\:\:\:\:\boldsymbol{\mathrm{x}}? \\ $$ Answered by Frix last updated on 06/Nov/24 $${x}^{\mathrm{6}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{6}}…

Question-213482

Question Number 213482 by efronzo1 last updated on 06/Nov/24 Answered by A5T last updated on 06/Nov/24 $$\mathrm{3}\mid{Y}\Rightarrow\mathrm{8}+{c}+\mathrm{2}+\mathrm{3}+{d}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right)\Rightarrow{c}+{d}\equiv\mathrm{2}\left({mod}\:\mathrm{3}\right) \\ $$$$\mathrm{3}\mid{X}\Rightarrow\mathrm{9}\mid{Y}\::\: \\ $$$$\mathrm{3}+\mathrm{1}+{a}+\mathrm{5}+{b}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right)\Rightarrow\mathrm{8}+{c}+\mathrm{2}+\mathrm{3}+{d}\equiv\mathrm{0}\left({mod}\:\mathrm{9}\right) \\ $$$$\Rightarrow{a}+{b}\equiv\mathrm{0}\left({mod}\:\mathrm{3}\right)\Rightarrow{c}+{d}\equiv\mathrm{5}\left({mod}\:\mathrm{9}\right) \\ $$$$\mathrm{3}\mid{X}\Rightarrow{c}+{d}=\mathrm{5}\Rightarrow{a}+{b}=\mathrm{0},\mathrm{3},\mathrm{6},\mathrm{9}\:\wedge\:{c}+{d}=\mathrm{5}…

1-z-6-1-dz-

Question Number 213451 by issac last updated on 06/Nov/24 $$\int\:\:\frac{\mathrm{1}}{{z}^{\mathrm{6}} −\mathrm{1}}\:\mathrm{d}{z}=?? \\ $$ Answered by Frix last updated on 06/Nov/24 $$\int\frac{{dz}}{{z}^{\mathrm{6}} −\mathrm{1}}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{I}_{{k}} \\…