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Author: Tinku Tara

x-1-x-2-2-2x-2-1-find-solution-

Question Number 134670 by benjo_mathlover last updated on 06/Mar/21 $$\mid\:\mathrm{x}+\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:\mid\:=\:\sqrt{\mathrm{2}}\:\left(\mathrm{2x}^{\mathrm{2}} −\mathrm{1}\:\right) \\ $$$$\mathrm{find}\:\mathrm{solution} \\ $$ Answered by EDWIN88 last updated on 06/Mar/21 $$\left(\mathrm{1}\right)\:\mathrm{1}−\mathrm{x}^{\mathrm{2}} \:\geqslant\:\mathrm{0}\:\Rightarrow\:−\mathrm{1}\leqslant\mathrm{x}\leqslant\mathrm{1}…

N-lim-x-1-x-2-1-x-1-x-3-1-

Question Number 134665 by EDWIN88 last updated on 06/Mar/21 $$\mathscr{N}\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:+\:\sqrt{\mathrm{x}−\mathrm{1}}}{\:\sqrt{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}}\:=? \\ $$ Answered by benjo_mathlover last updated on 06/Mar/21 $$\mathscr{N}\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}−\mathrm{1}}\:\left\{\sqrt{\mathrm{x}+\mathrm{1}}+\mathrm{1}\:\right\}}{\:\sqrt{\mathrm{x}−\mathrm{1}}\:\left\{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:\right\}}…

I-just-thought-of-something-I-am-curious-in-figuring-out-All-integer-numbers-can-be-made-up-by-prime-factors-That-is-n-p-1-p-2-p-i-n-Z-p-k-P-Are-there-an-inifinite-number-of-numbe

Question Number 3595 by Filup last updated on 16/Dec/15 $$\mathrm{I}\:\mathrm{just}\:\mathrm{thought}\:\mathrm{of}\:\mathrm{something}\:\mathrm{I}\:\mathrm{am}\:\mathrm{curious} \\ $$$$\mathrm{in}\:\mathrm{figuring}\:\mathrm{out}. \\ $$$$ \\ $$$$\mathrm{All}\:\mathrm{integer}\:\mathrm{numbers}\:\mathrm{can}\:\mathrm{be}\:\mathrm{made}\:\mathrm{up}\:\mathrm{by} \\ $$$${prime}\:{factors}.\:\mathrm{That}\:\mathrm{is}: \\ $$$${n}={p}_{\mathrm{1}} ×{p}_{\mathrm{2}} ×…×{p}_{{i}} \\ $$$${n}\in\mathbb{Z}\:\:\:\:\:\:\:\:\:{p}_{{k}} \in\mathbb{P}…

M-dx-2cos-x-3sin-x-

Question Number 134660 by benjo_mathlover last updated on 06/Mar/21 $$\mathscr{M}\:=\:\int\:\frac{{dx}}{\mathrm{2cos}\:{x}+\mathrm{3sin}\:{x}}\: \\ $$ Answered by EDWIN88 last updated on 06/Mar/21 $$\mathrm{Let}\:\begin{cases}{\mathrm{2}\:=\:\mathrm{r}\:\mathrm{sin}\:\alpha}\\{\mathrm{3}\:=\:\mathrm{r}\:\mathrm{cos}\:\alpha}\end{cases}\:\Rightarrow\:\mathrm{r}^{\mathrm{2}} =\:\mathrm{13};\:\mathrm{r}\:=\sqrt{\mathrm{13}} \\ $$$$\:\mathrm{and}\:\mathrm{tan}\:\alpha\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:,\therefore\:\alpha\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\…

nice-calculus-prove-that-n-1-2n-n-1-2n-1-1-2-m-n-

Question Number 134662 by mnjuly1970 last updated on 06/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{nice}\:\:{calculus}… \\ $$$$\:\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\boldsymbol{\phi}=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\zeta\left(\mathrm{2}{n}\right)}{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:…{m}.{n}… \\ $$ Answered by Dwaipayan Shikari last…

Three-point-are-drawn-on-a-straight-number-line-A-B-and-C-Consider-a-quadractic-equation-x-2-ax-b-0-a-Length-of-line-segment-AB-b-Length-of-line-segment-BC-Give-construction-steps-to-identify-a-poin

Question Number 3588 by prakash jain last updated on 16/Dec/15 $$\mathrm{Three}\:\mathrm{point}\:\mathrm{are}\:\mathrm{drawn}\:\mathrm{on}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{number}\:\mathrm{line}\:\mathrm{A},\mathrm{B}\:\mathrm{and}\:\mathrm{C}. \\ $$$$\mathrm{Consider}\:\mathrm{a}\:\mathrm{quadractic}\:\mathrm{equation} \\ $$$${x}^{\mathrm{2}} +{ax}+{b}=\mathrm{0} \\ $$$${a}=\mathrm{Length}\:\mathrm{of}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{AB} \\ $$$${b}=\mathrm{Length}\:\mathrm{of}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{BC} \\ $$$$\mathrm{Give}\:\mathrm{construction}\:\mathrm{steps}\:\mathrm{to}\:\mathrm{identify}\:\mathrm{a}\:\mathrm{points} \\…

1-pie-3-2pi-2-e-2-11-6pi-3-e-3-25-12pi-4-e-4-137-60pi-5-e-5-log-pi-1-log-pie-1-pie-1-

Question Number 134657 by Dwaipayan Shikari last updated on 06/Mar/21 $$\frac{\mathrm{1}}{\pi{e}}+\frac{\mathrm{3}}{\mathrm{2}\pi^{\mathrm{2}} {e}^{\mathrm{2}} }+\frac{\mathrm{11}}{\mathrm{6}\pi^{\mathrm{3}} {e}^{\mathrm{3}} }+\frac{\mathrm{25}}{\mathrm{12}\pi^{\mathrm{4}} {e}^{\mathrm{4}} }+\frac{\mathrm{137}}{\mathrm{60}\pi^{\mathrm{5}} {e}^{\mathrm{5}} }+…=\frac{{log}\left(\pi\right)+\mathrm{1}−{log}\left(\pi{e}−\mathrm{1}\right)}{\pi{e}−\mathrm{1}} \\ $$ Terms of Service Privacy…