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Author: Tinku Tara

Let-P-x-be-polynomial-in-x-with-integral-coefficients-If-n-is-a-solution-of-P-x-0-mod-n-and-a-b-mod-n-prove-that-b-is-also-a-solution-

Question Number 76229 by arkanmath7@gmail.com last updated on 25/Dec/19 $${Let}\:{P}\left({x}\right)\:{be}\:{polynomial}\:{in}\:{x}\:{with}\:{integral} \\ $$$${coefficients}.\:{If}\:{n}\:{is}\:{a}\:{solution}\:{of}\: \\ $$$${P}\left({x}\right)\equiv\mathrm{0}\left({mod}\:{n}\right)\:,\:{and}\:{a}\equiv{b}\left({mod}\:{n}\right), \\ $$$${prove}\:{that}\:{b}\:{is}\:{also}\:{a}\:{solution}. \\ $$ Answered by Rio Michael last updated on…

f-x-y-f-x-f-y-f-16-8-f-64-

Question Number 10692 by ABD last updated on 23/Feb/17 $${f}\left({x}+{y}\right)={f}\left({x}\right)×{f}\left({y}\right) \\ $$$${f}\left(\mathrm{16}\right)=\mathrm{8}\:\Rightarrow{f}\left(\mathrm{64}\right) \\ $$ Answered by nume1114 last updated on 23/Feb/17 $${f}\left(\mathrm{64}\right)={f}\left(\left(\mathrm{16}+\mathrm{16}\right)+\left(\mathrm{16}+\mathrm{16}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:={f}\left(\mathrm{16}+\mathrm{16}\right)×{f}\left(\mathrm{16}+\mathrm{16}\right) \\…

n-1-2-pi-2n-1-2-2n-n-1-

Question Number 141757 by Ndala last updated on 23/May/21 $$\Gamma\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}\centerdot\Gamma\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{2}^{\mathrm{2n}} \Gamma\left(\mathrm{n}+\mathrm{1}\right)} \\ $$ Commented by Ndala last updated on 23/May/21 $$\mathrm{someone}\:\mathrm{to}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{the}\:\mathrm{proof}.\:\mathrm{Please} \\ $$ Terms of…

Abbey-and-mary-carry-a-uniform-log-of-length-10m-and-weight-100N-Abbey-is-1m-from-one-end-and-mary-is-2m-from-the-other-end-what-weight-does-abbey-and-mary-support-

Question Number 10687 by Saham last updated on 22/Feb/17 $$\mathrm{Abbey}\:\mathrm{and}\:\mathrm{mary}\:\mathrm{carry}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{log}\:\mathrm{of}\:\mathrm{length}\:\mathrm{10m}\:\mathrm{and}\:\mathrm{weight}\:\mathrm{100N}. \\ $$$$\mathrm{Abbey}\:\mathrm{is}\:\mathrm{1m}\:\mathrm{from}\:\mathrm{one}\:\mathrm{end}\:\mathrm{and}\:\mathrm{mary}\:\mathrm{is}\:\mathrm{2m}\:\mathrm{from}\:\mathrm{the}\:\mathrm{other}\:\mathrm{end}.\:\mathrm{what} \\ $$$$\mathrm{weight}\:\mathrm{does}\:\mathrm{abbey}\:\mathrm{and}\:\mathrm{mary}\:\mathrm{support}. \\ $$ Answered by mrW1 last updated on 22/Feb/17 $${weight}\:{is}\:{in}\:{the}\:{middle}\:{of}\:{log}. \\…

Question-141752

Question Number 141752 by 0731619 last updated on 23/May/21 Answered by qaz last updated on 23/May/21 $$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tan}^{−\mathrm{1}} {x}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{u}}{\mathrm{tan}\:{u}+\mathrm{1}}{du}…