Question Number 99645 by bramlex last updated on 22/Jun/20 $${If}\:{x},{y}\:>\:\mathrm{0}\:{then}\:\mid\sqrt{{xy}}−\frac{{x}+{y}}{\mathrm{2}}\mid\:+\:\mid\frac{{x}+{y}}{\mathrm{2}}\:+\:\sqrt{{xy}}\:\mid\:= \\ $$ Commented by bemath last updated on 22/Jun/20 $$\mathrm{since}\:\sqrt{\mathrm{xy}}\:>\:\mathrm{0}\:\&\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:>\:\mathrm{0} \\ $$$$\mathrm{so}\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:+\:\sqrt{\mathrm{xy}}\:>\:\mathrm{0} \\ $$$$\mathrm{case}\left(\mathrm{1}\right)\:\mathrm{if}\:\sqrt{\mathrm{xy}}−\frac{\mathrm{x}+\mathrm{y}}{\mathrm{2}}\:>\:\mathrm{0} \\…
Question Number 165157 by mathlove last updated on 26/Jan/22 $$\underset{{k}=\mathrm{6}} {\overset{{n}+\mathrm{6}} {\sum}}\left({k}−\mathrm{4}\right)=\frac{{an}^{\mathrm{2}} +{bn}+{c}}{\mathrm{2}} \\ $$$${a}+{b}+{c}=? \\ $$ Commented by bobhans last updated on 30/Jan/22 $$\:\underset{\mathrm{k}=\mathrm{6}}…
Question Number 99621 by student work last updated on 22/Jun/20 $$\mathrm{6}^{\mathrm{x}} =\mathrm{x}^{\mathrm{5}\:\:\:\:\:\:\:} \mathrm{x}=?\:\:\:\:\:\:\:\mathrm{help}\:\mathrm{me} \\ $$ Commented by Dwaipayan Shikari last updated on 22/Jun/20 $$\frac{{x}}{{logx}}=\frac{\mathrm{5}}{{log}\mathrm{6}} \\…
Question Number 165153 by mathlove last updated on 26/Jan/22 $${f}\left({x}\right)=\mathrm{3}{e}^{\mathrm{2}{x}} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=? \\ $$ Answered by mahdipoor last updated on 26/Jan/22 $$\Rightarrow{x}=\mathrm{3}{e}^{\mathrm{2}{y}} \Rightarrow\frac{\mathrm{1}}{\mathrm{2}}{ln}\frac{{x}}{\mathrm{3}}={y}={f}^{−\mathrm{1}} \left({x}\right)…
Question Number 34081 by Aditya 879 last updated on 30/Apr/18 $$\mathrm{2}^{{n}} −\mathrm{2}^{{n}−\mathrm{1}} =\mathrm{4}\:.{find}\:{n}^{{n}.} \\ $$ Answered by math1967 last updated on 30/Apr/18 $$\mathrm{2}^{{n}} \left(\mathrm{1}−\mathrm{2}^{−\mathrm{1}} \right)=\mathrm{4}…
Question Number 165142 by HongKing last updated on 26/Jan/22 Commented by mr W last updated on 26/Jan/22 $${how}\:{do}\:{they}\:{play}?\:{one}\:{game}\:{between} \\ $$$${each}\:{pair}?\:{you}\:{can}'{t}\:{solve}\:{if}\:{the} \\ $$$${rule}\:{of}\:{the}\:{tournament}\:{is}\:{not}\:{clear}. \\ $$ Commented…
Question Number 34056 by Learner last updated on 30/Apr/18 $${x}^{\mathrm{3}{z}} =\mathrm{1}\: \\ $$$${x}^{\mathrm{2}} ={y} \\ $$$${z}={y}^{{n}} \\ $$$${FIND}\:{THE}\:{VALUE}\:{OF}\:{n} \\ $$$${please}\:{i}\:{need}\:{your}\:{help}\:{ASAP}.\:{thanks} \\ $$ Commented by math1967…
Question Number 34044 by NECx last updated on 29/Apr/18 $${what}\:{is}\:{the}\:{remainder}\:{when}\: \\ $$$$\left(\mathrm{111}..\right)+\left(\mathrm{222}..\right)+\left(\mathrm{333}..\right)+….+\left(\mathrm{77}..\right) \\ $$$${is}\:{divided}\:{by}\:\mathrm{37} \\ $$ Commented by Rasheed.Sindhi last updated on 30/Apr/18 $$\mathrm{N}=\left(\mathrm{111}..\right)+\left(\mathrm{222}..\right)+\left(\mathrm{333}..\right)+….+\left(\mathrm{77}..\right) \\…
Question Number 99575 by MWSuSon last updated on 21/Jun/20 $$\mathrm{if}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d},\mathrm{e}\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{a}−\mathrm{4b}+\mathrm{6c}−\mathrm{4d}+\mathrm{e}. \\ $$$$\mathrm{please}\:\mathrm{help},\:\mathrm{i}\:\mathrm{do}\:\mathrm{not}\:\mathrm{know}\:\mathrm{where}\:\mathrm{to} \\ $$$$\mathrm{begin}. \\ $$ Answered by mathmax by abdo last updated…
Question Number 165098 by mnjuly1970 last updated on 26/Jan/22 Answered by Eulerian last updated on 26/Jan/22 $$\:\boldsymbol{\mathrm{Solution}}: \\ $$$$\:\mathrm{We}\:\mathrm{know}\:\mathrm{that}: \\ $$$$\:\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +……+\mathrm{n}^{\mathrm{2}} \:=\:\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6}}…