Question Number 130362 by greg_ed last updated on 24/Jan/21 $$\mathrm{f}\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mid\mathrm{2}{x}−\mathrm{3}\mid}} \\ $$$$\mathrm{Domain}\:\mathrm{D}_{\mathrm{f}} \:=\:? \\ $$ Answered by ajfour last updated on 24/Jan/21 $${x}^{\mathrm{2}} >\mid\mathrm{2}{x}−\mathrm{3}\mid…
Question Number 130354 by greg_ed last updated on 24/Jan/21 $$\mathrm{f}\left({x}\right)=\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{1}\:\leqslant\:\mathrm{f}\left({x}\right)\:\leqslant\:\mathrm{2} \\ $$ Answered by MJS_new last updated on 24/Jan/21 $${f}\left({x}\right)=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}}…
Question Number 130334 by pete last updated on 24/Jan/21 $$\mathrm{Complete}\:\mathrm{the}\:\mathrm{square}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\mathrm{y}^{\mathrm{2}} \:+\mathrm{8y}+\mathrm{9k}\:\mathrm{and}\:\mathrm{hence}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{k}\:\mathrm{that}\:\mathrm{makes}\:\mathrm{it}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}. \\ $$ Answered by TheSupreme last updated on 24/Jan/21 $$\left({y}+{a}\right)={y}^{\mathrm{2}}…
Question Number 130331 by ayoubbacmath0 last updated on 24/Jan/21 $$\mathrm{f}\left(\mathrm{x}\right)={a}\mathrm{x}^{\mathrm{3}} +\mathrm{bx}^{\mathrm{2}} +\mathrm{c} \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)=\mathrm{3}{a}\mathrm{x}^{\mathrm{2}} +\mathrm{2bx} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=−\mathrm{2} \\ $$$$\Rightarrow{a}\left(\mathrm{0}\right)^{\mathrm{3}} +\mathrm{b}\left(\mathrm{0}\right)^{\mathrm{2}} +\mathrm{c}=−\mathrm{2} \\ $$$$\Rightarrow\mathrm{c}=−\mathrm{2} \\ $$$$\mathrm{f}\left(−\mathrm{2}\right)=\mathrm{2}…
Question Number 64758 by Tawa1 last updated on 21/Jul/19 $$\mathrm{Solve}:\:\:\:\:\:\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{5x}^{\mathrm{3}} \:−\:\mathrm{4x}^{\mathrm{2}} \:+\:\mathrm{7x}\:−\:\mathrm{1}\:\:=\:\:\mathrm{0} \\ $$ Answered by ajfour last updated on 21/Jul/19 $${x}={t}−\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${say}\:\:{we}\:{then}\:{get}…
Question Number 130290 by bramlexs22 last updated on 24/Jan/21 $$\:\underset{\mathrm{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{\mathrm{j}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{i}+\mathrm{j}+\mathrm{k}} }\:?\:\mathrm{where}\:\mathrm{i}\neq\mathrm{j}\neq\mathrm{k} \\ $$ Answered by EDWIN88 last updated on…
Question Number 130257 by pete last updated on 23/Jan/21 $${n}\:\mathrm{is}\:\mathrm{an}\:\mathrm{integer} \\ $$$$\mathrm{prove}\:\mathrm{algebraically}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{n}\left({n}+\mathrm{1}\right)\:\mathrm{and}\:\frac{\mathrm{1}}{\mathrm{2}}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\:\mathrm{is}\:\mathrm{always} \\ $$$$\mathrm{a}\:\mathrm{square}\:\mathrm{number}. \\ $$$$\mathrm{note}:\:{write}\:{your}\:{expression}\:{in}\:{a}\:{form} \\ $$$${that}\:{clearly}\:{shows}\:{a}\:{square}\:{number}. \\ $$ Commented by Dwaipayan…
Question Number 130242 by Raxreedoroid last updated on 23/Jan/21 $$\mathrm{Find}\:\:\:{m} \\ $$$${n}=\left({p}_{\mathrm{1}} {p}_{\mathrm{2}} \right)^{{m}} \\ $$$$\mathrm{2}{n}=\left({p}_{\mathrm{1}} ^{\mathrm{1}+{log}_{{p}_{\mathrm{1}} {p}_{\mathrm{2}} } {n}} \right)\left({p}_{\mathrm{2}} ^{\mathrm{1}+{log}_{{p}_{\mathrm{1}} {p}_{\mathrm{2}} } {n}}…
Question Number 64631 by Mikael last updated on 19/Jul/19 $$\sqrt{\frac{\mathrm{1}+\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\mathrm{2}}}+\mathrm{2}{x}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{To}\:\mathrm{solve}\:\mathrm{in}\:\mathbb{R} \\ $$ Commented by behi83417@gmail.com last updated on 20/Jul/19 $$−\mathrm{1}<\mathrm{x}\leqslant\mathrm{1},\mathrm{let}:\:\mathrm{x}=\mathrm{cost} \\…
Question Number 64604 by mathmax by abdo last updated on 19/Jul/19 $${find}\:\:{all}\:{integr}\:{naturals}\:{n}\:{and}\:{k}\:{wich}\:{verify} \\ $$$${k}!=\left(\mathrm{2}^{{n}} −\mathrm{1}\right)\left(\mathrm{2}^{{n}} −\mathrm{2}\right)\left(\mathrm{2}^{{n}} −\mathrm{4}\right)…\left(\mathrm{2}^{{n}} −\mathrm{2}^{{n}−\mathrm{1}} \right) \\ $$ Terms of Service Privacy…