Question Number 164068 by HongKing last updated on 13/Jan/22 Answered by Rasheed.Sindhi last updated on 13/Jan/22 $$\mathrm{sin}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{8}}}\:\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{x}=\frac{\pi}{\mathrm{6}}+\mathrm{2}{k}\pi\:;\:\forall{k}\in\mathbb{Z} \\ $$ Commented by mr W…
Question Number 164071 by kdaramaths last updated on 13/Jan/22 Answered by Ar Brandon last updated on 13/Jan/22 $${Z}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{e}^{{ikx}} =\frac{\mathrm{1}−{e}^{{i}\left({n}+\mathrm{1}\right){x}} }{\mathrm{1}−{e}^{{ix}} } \\ $$$$\:\:\:=\frac{{e}^{{i}\left(\frac{\left({n}+\mathrm{1}\right){x}}{\mathrm{2}}\right)}…
Question Number 98521 by pranesh last updated on 14/Jun/20 Answered by abdomathmax last updated on 15/Jun/20 $$\mathrm{we}\:\mathrm{have}\:\mid\mathrm{x}\mid+\mathrm{x}\:\geqslant\mathrm{0}\:\:\mathrm{for}\:\mathrm{all}\:\mathrm{x} \\ $$$$\mathrm{if}\:\:\mathrm{x}\:\geqslant\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\mathrm{x}+\mathrm{x}\geqslant\mathrm{0}\:\mathrm{true} \\ $$$$\mathrm{if}\:\mathrm{x}\:\leqslant\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:−\mathrm{x}+\mathrm{x}\geqslant\mathrm{0}\:\mathrm{also}\:\mathrm{true}\:\:\mathrm{so} \\ $$$$\mathrm{x}\in\mathrm{D}_{\mathrm{f}} \:\Leftrightarrow\:\mathrm{x}>\mathrm{0}\:\mathrm{and}\:\:\mathrm{x}\neq\frac{\pi}{\mathrm{2}}\:+\mathrm{k}\pi\:\:\left(\mathrm{k}\:\in\mathrm{Z}\right) \\…
Question Number 164050 by mathlove last updated on 13/Jan/22 Answered by Ar Brandon last updated on 13/Jan/22 $${x}−\frac{\mathrm{1}}{\mathrm{8}}>\mathrm{0}\neq\mathrm{1}\Rightarrow{x}>\frac{\mathrm{1}}{\mathrm{8}}\neq\frac{\mathrm{9}}{\mathrm{8}}….\left(\mathrm{1}\right) \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{9}}>\mathrm{0}\:\Rightarrow{x}<−\frac{\mathrm{1}}{\mathrm{3}}\cup{x}>\frac{\mathrm{1}}{\mathrm{3}}\:…\left(\mathrm{2}\right) \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}−{x}\right)>\mathrm{0}\:\Rightarrow\mathrm{2}−{x}>\mathrm{0}\Rightarrow{x}<\mathrm{2}…\left(\mathrm{3}\right) \\…
Question Number 32977 by Mr eaay last updated on 08/Apr/18 $${Prove}\:{that}\:\:^{{n}} {C}_{{r}} \:\:+\:^{{n}} {C}_{{r}+\mathrm{1}} \:=\:^{{n}+\mathrm{1}} {C}_{{r}+\mathrm{1}} \\ $$$$ \\ $$ Answered by math1967 last updated…
Question Number 164039 by ArielVyny last updated on 13/Jan/22 $${consider}\:{f}\:{function}\:{Df}=\left[\mathrm{0},\mathrm{1}\right] \\ $$$${f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right)\:{c}\in\left[\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right]\:{show}\:{that}\:{f}\left({c}\right)={f}\left({c}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$ Answered by Ar Brandon last updated on 13/Jan/22 $${f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right) \\ $$$${f}\left({c}+\mathrm{0}\right)={f}\left({c}+\mathrm{1}\right)\:,\:\mathrm{since}\:{f}\:<\:\mathrm{1}-\mathrm{periodic}…
Question Number 164028 by HongKing last updated on 13/Jan/22 Answered by MJS_new last updated on 13/Jan/22 $$\mathrm{you}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$${x}\approx\mathrm{4}.\mathrm{14326024192} \\ $$ Commented by HongKing last…
Question Number 164024 by mathlove last updated on 13/Jan/22 Answered by som(math1967) last updated on 13/Jan/22 $${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =−\left({x}−{y}\right) \\ $$$$\left({x}−{y}\right)\left({x}+{y}\right)+\left({x}−{y}\right)=\mathrm{0} \\ $$$$\left({x}−{y}\right)\left({x}+{y}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}+{y}=−\mathrm{1}\:\left[{x}\neq{y}\right]…
Question Number 164027 by HongKing last updated on 13/Jan/22 Commented by HongKing last updated on 13/Jan/22 $$\mathrm{Yes}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir} \\ $$ Commented by mr W last updated…
Question Number 164019 by ajfour last updated on 13/Jan/22 $$\:\:{x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$$\:\Rightarrow\:\:{x}^{\mathrm{2}} +\frac{{d}}{{x}^{\mathrm{2}} }+{ax}+\frac{{c}}{{x}}+{b}=\mathrm{0} \\ $$$${say}\:\:\:\frac{\mathrm{1}}{{x}}={t}\:\:\Rightarrow \\ $$$$\:\:{x}^{\mathrm{2}} +{ax}+{b}+{dt}^{\mathrm{2}} +{ct}=\mathrm{0} \\ $$$${let}\:\:\:\:{x}^{\mathrm{2}}…