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Author: Tinku Tara

Question-4584

Question Number 4584 by MrGreg last updated on 08/Feb/16 $$ \\ $$ Answered by FilupSmith last updated on 09/Feb/16 $$\:\:\:\:{S}=\mathrm{1}−\mathrm{2}+\mathrm{4}−\mathrm{8}+\mathrm{16}−\mathrm{32}+\mathrm{64}−… \\ $$$$+{S}=\mathrm{0}+\mathrm{1}−\mathrm{2}+\mathrm{4}−\mathrm{8}+\mathrm{16}−\mathrm{32}+\mathrm{64}−… \\ $$$$\mathrm{2}{S}=\mathrm{1}+\left(−\mathrm{1}+\mathrm{2}−\mathrm{4}+\mathrm{8}−\mathrm{16}+…\right) \\…

In-a-right-triangle-the-mid-point-of-the-hypotenuse-is-equidistant-from-all-the-three-vertices-of-the-triangle-

Question Number 4579 by Rasheed Soomro last updated on 08/Feb/16 $${In}\:{a}\:{right}\:{triangle},\:{the}\:{mid}-{point}\:{of}\:{the} \\ $$$${hypotenuse}\:{is}\:{equidistant}\:{from}\:{all}\:{the} \\ $$$${three}\:{vertices}\:{of}\:{the}\:{triangle}. \\ $$ Commented by Yozzii last updated on 08/Feb/16 $${Prove}\:{this}\:{theorem}?…

The-segment-between-the-mid-points-of-two-sides-of-a-triangle-is-parallel-to-the-third-side-and-half-as-long-

Question Number 4578 by Rasheed Soomro last updated on 08/Feb/16 $${The}\:{segment}\:{between}\:{the}\:{mid}-{points} \\ $$$${of}\:{two}\:{sides}\:{of}\:{a}\:{triangle}\:{is}\:{parallel} \\ $$$${to}\:{the}\:{third}\:{side}\:{and}\:{half}\:{as}\:{long}.\: \\ $$ Commented by Yozzii last updated on 08/Feb/16 $${Prove}\:{this}\:{theorem}?…

1-1-2pi-2-2-1-1-pi-2-2-1-1-1-pi-2-2-1-1-2pi-2-2-csc-2-1-pi-pi-2-

Question Number 135647 by Dwaipayan Shikari last updated on 14/Mar/21 $$…+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{2}\pi^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\pi^{\mathrm{2}} \right)^{\mathrm{2}} }+\mathrm{1}+\frac{\mathrm{1}}{\left(\mathrm{1}+\pi^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{2}\pi^{\mathrm{2}} \right)^{\mathrm{2}} }+…=\frac{{csc}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\pi}\right)}{\pi^{\mathrm{2}} } \\ $$ Terms of…

Question-70108

Question Number 70108 by naka3546 last updated on 01/Oct/19 Answered by MJS last updated on 01/Oct/19 $$\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{{x}} >\mathrm{3}^{\mathrm{36}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{18}{x}} \\ $$$$\left(−\mathrm{2ln}\:\mathrm{2}\right){x}>\mathrm{36ln}\:\mathrm{3}\:+\mathrm{18}\left(\mathrm{ln}\:\mathrm{2}\:−\mathrm{ln}\:\mathrm{3}\right){x} \\ $$$$\left(\mathrm{18ln}\:\mathrm{3}\:−\mathrm{20}\:\mathrm{ln}\:\mathrm{2}\right){x}>\mathrm{36ln}\:\mathrm{3} \\ $$$${x}>\frac{\mathrm{18ln}\:\mathrm{3}}{\mathrm{9ln}\:\mathrm{3}\:−\mathrm{10ln}\:\mathrm{2}}\approx\mathrm{6}.\mathrm{68970287}…

x-1-2-ln-1-1-x-x-dx-

Question Number 135646 by metamorfose last updated on 14/Mar/21 $$\int\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right){ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)−{x}\:{dx}=…? \\ $$ Answered by Ñï= last updated on 15/Mar/21 $$\int\left[\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right){ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)−{x}\right]{dx} \\ $$$$=\int\left[\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({ln}\left({x}+\mathrm{1}\right)−{lnx}\right)−{x}\right]{dx} \\ $$$$=\int\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right){ln}\left({x}+\mathrm{1}\right){dx}−\int\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right){lnxdx}−\int{xdx} \\…