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Question Number 28642 by nimita last updated on 28/Jan/18 $${f}\left({x}\right)=\mathrm{4}{x}−\mathrm{1}{for}\mathrm{0}<{x}<\mathrm{4}\:{find}\:{f}\left(\mathrm{0}\right)\:,{f}\left(\mathrm{1}\right) \\ $$$${f}\left(\mathrm{1}.\mathrm{2}\right),{f}\left(\mathrm{4}\right),{f}\left(−\mathrm{1}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 94174 by i jagooll last updated on 17/May/20 $$\begin{cases}{\mathrm{x}+\mathrm{y}\:=\:\mathrm{10}}\\{\sqrt[{\mathrm{3}\:\:}]{\mathrm{x}}\:+\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{y}}\:=\:\frac{\mathrm{5}}{\mathrm{2}}\:\sqrt[{\mathrm{6}\:\:}]{\mathrm{xy}}}\end{cases}\: \\ $$$$\mathrm{find}\:\mathrm{x}\:\&\mathrm{y}?? \\ $$ Commented by behi83417@gmail.com last updated on 17/May/20 $$\mathrm{x}=\mathrm{t}^{\mathrm{6}} ,\mathrm{y}=\mathrm{s}^{\mathrm{6}} \\…
Question Number 159683 by SAMIRA last updated on 20/Nov/21 $$\mathrm{2}\:\leqslant\:\mid\boldsymbol{{x}}−\mathrm{2}\mid\:\leqslant\:\mathrm{6} \\ $$ Commented by tounghoungko last updated on 20/Nov/21 $$\:\mid{x}−\mathrm{2}\mid\geqslant\mathrm{2}\:\wedge\:\mid{x}−\mathrm{2}\mid\:\leqslant\:\mathrm{6} \\ $$$$\Rightarrow\left({x}−\mathrm{2}−\mathrm{2}\right)\left({x}−\mathrm{2}+\mathrm{2}\right)\geqslant\:\mathrm{0}\:\wedge\:\left({x}−\mathrm{2}−\mathrm{6}\right)\left({x}−\mathrm{2}+\mathrm{6}\right)\leqslant\:\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{4}\right){x}\:\geqslant\:\mathrm{0}\:\wedge\:\left({x}−\mathrm{8}\right)\left({x}+\mathrm{4}\right)\leqslant\:\mathrm{0} \\…
Question Number 94124 by M±th+et+s last updated on 17/May/20 $$\mathrm{20}+{a}={a}\:{cosh}\left(\frac{\mathrm{75}}{{a}}\right) \\ $$$${a}=? \\ $$ Commented by MJS last updated on 17/May/20 $$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$$\mathrm{I}\:\mathrm{get}\:{a}\approx\mathrm{143}.\mathrm{84} \\…
Question Number 159654 by HongKing last updated on 19/Nov/21 $$\mathrm{Find}:\:\:\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{x}\:\centerdot\:\mathrm{arctan}^{\mathrm{2}} \left(\mathrm{x}\right)}{\left(\mathrm{x}\:+\:\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)}\:\mathrm{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 94095 by Jidda28 last updated on 16/May/20 $$\mathrm{How}\:\mathrm{many}\:\mathrm{subgroups}\:\mathrm{do}\:\mathrm{Z}_{\mathrm{3}} \oplus\mathrm{Z}_{\mathrm{16}\:} \:\mathrm{has}?\:\mathrm{Justify}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 28554 by naka3546 last updated on 27/Jan/18 $$\frac{\left({a}\:−\:{b}\right)}{\left({c}\:−\:{d}\right)}\:\:=\:\:\mathrm{3} \\ $$$$\frac{\left({a}\:−\:{c}\right)}{\left({b}\:−\:{d}\right)}\:\:=\:\:\mathrm{4} \\ $$$$\frac{\left({a}\:−\:{d}\right)}{\left({b}\:−\:{c}\right)}\:\:=\:\:? \\ $$$$ \\ $$ Answered by ajfour last updated on 27/Jan/18…
Question Number 28546 by abdo imad last updated on 26/Jan/18 $${let}\:{give}\:{w}={e}^{{i}\frac{\mathrm{2}\pi}{{n}}} \:\:\:{and}\:\:{S}=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:{w}^{{k}^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\:\:{S}=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:{w}^{\left({q}+{k}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:{find}\:\mid{S}\mid. \\ $$ Terms…
Question Number 28544 by abdo imad last updated on 26/Jan/18 $${if}\:\:{a}_{\mathrm{1}} \:,{a}_{\mathrm{2}} ,…{a}_{\mathrm{14}\:} {are}\:{roots}\:{of}\:{the}\:{polynomial} \\ $$$${p}\left({x}\right)={x}^{\mathrm{14}} +{x}^{\mathrm{8}} \:\mathrm{2}{x}+\mathrm{1}\:\:\:{calculate}\:\:\sum_{{i}=\mathrm{1}} ^{\mathrm{14}} \:\:\frac{\mathrm{1}}{\left({a}_{{i}} −\mathrm{1}\right)^{\mathrm{2}} }\:\:. \\ $$ Commented…