Menu Close

Category: None

Let-p-q-r-are-positive-real-numbers-0-lt-r-lt-min-p-q-Prove-that-p-r-q-r-min-pq-r-2-p-q-2r-

Question Number 71206 by naka3546 last updated on 13/Oct/19 $${Let}\:\:{p},{q},{r}\:\:{are}\:\:{positive}\:\:{real}\:\:{numbers}\:. \\ $$$$\mathrm{0}\:<\:{r}\:<\:{min}\left\{{p},{q}\right\}. \\ $$$${Prove}\:\:{that} \\ $$$$\:\:\:\:\:\sqrt{{p}−{r}}\:+\:\sqrt{{q}−{r}}\:\:\leqslant\:\:{min}\left\{\sqrt{\frac{{pq}}{{r}}}\:,\:\sqrt{\mathrm{2}\left({p}+{q}\:−\:\mathrm{2}{r}\right)}\:\right\} \\ $$ Answered by mind is power last updated…

n-0-2n-n-1-4-n-2n-1-3-pi-3-48-pi-4-ln-2-2-

Question Number 136733 by Ñï= last updated on 25/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}\frac{\mathrm{1}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\pi^{\mathrm{3}} }{\mathrm{48}}+\frac{\pi}{\mathrm{4}}{ln}^{\mathrm{2}} \mathrm{2} \\ $$ Answered by mindispower last updated on 25/Mar/21…

Question-136734

Question Number 136734 by mohammad17 last updated on 25/Mar/21 Answered by Ñï= last updated on 25/Mar/21 $${Q}\mathrm{5}::: \\ $$$${y}_{{p}} =\frac{\mathrm{1}}{{D}^{\mathrm{2}} +\mathrm{3}{D}+\mathrm{12}}\left({e}^{{x}} \mathrm{cos}\:{x}−\mathrm{cos}\:{x}\right) \\ $$$$={e}^{{x}} \frac{\mathrm{1}}{{D}^{\mathrm{2}}…

Question-71149

Question Number 71149 by naka3546 last updated on 12/Oct/19 Commented by MJS last updated on 17/Oct/19 $$\mathrm{if}\:\mathrm{the}\:\mathrm{hexagon}\:\mathrm{is}\:\mathrm{regular}\:\mathrm{it}'\mathrm{s}\:\mathrm{symmetric} \\ $$$${BE}\:\mathrm{is}\:\mathrm{the}\:\mathrm{symmetry}\:\mathrm{axis} \\ $$$${M}\in\left({BE}\right) \\ $$$${A}={C}'\:\wedge\:{F}={D}' \\ $$$$\Rightarrow\:\left({AF}\right)=\left({CD}\right)'\:\mathrm{and}\:\mathrm{similar}\:\mathrm{for}\:\mathrm{all}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{lines}…

2x-x-1-16-

Question Number 71141 by Mr. K last updated on 12/Oct/19 $$\left(\mathrm{2}{x}\right)^{{x}} =\frac{\mathrm{1}}{\mathrm{16}} \\ $$ Commented by mr W last updated on 12/Oct/19 $$\left(\mathrm{2}{x}\right)^{{x}} =\sqrt{\left(\mathrm{2}{x}\right)^{\mathrm{2}{x}} }=\sqrt{{t}^{{t}}…