Question Number 152828 by mathdanisur last updated on 01/Sep/21 $$\underset{\:\mathrm{1}} {\overset{\:\infty} {\int}}\:\frac{\sqrt{\mathrm{x}}\:\mathrm{ln}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:\mathrm{dx}\:=\:? \\ $$ Answered by Ar Brandon last updated on 01/Sep/21 $${f}\left(\kappa\right)=\int_{\mathrm{1}} ^{\infty}…
Question Number 152827 by mathdanisur last updated on 01/Sep/21 Answered by MJS_new last updated on 02/Sep/21 $${p}=\frac{{xy}}{{x}+{y}}\:\Leftrightarrow\:{y}=\frac{{px}}{{x}−{p}}\:\Leftrightarrow\:{x}=\frac{{py}}{{y}−{p}} \\ $$$$\Rightarrow\:{x}>{p}\wedge{y}>{p}\:\wedge\:\left({x}−{p}\right)\mid\left({px}\right)\:\wedge\:\left({y}−{p}\right)\mid\left({py}\right) \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:{y}={x}\:\Rightarrow\:{y}={x}=\mathrm{2}{p} \\ $$$$\left(\mathrm{2}\right)\:{y}={p}+\mathrm{1}\:\Rightarrow\:{x}={p}\left({p}+\mathrm{1}\right)…
Question Number 21733 by Joel577 last updated on 02/Oct/17 $$\mathrm{If}\:{p}\:\mathrm{is}\:\mathrm{one}\:\mathrm{of}\:\:\mathrm{roots}\:\mathrm{from}\:{x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\:+\:\mathrm{6}\:=\:\mathrm{0} \\ $$$$\mathrm{then}\:{p}^{\mathrm{4}} \:+\:\mathrm{16}{p}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:… \\ $$ Answered by mrW1 last updated on 02/Oct/17 $$\mathrm{p}^{\mathrm{2}} −\mathrm{2p}+\mathrm{6}=\mathrm{0}…
Question Number 152791 by mathdanisur last updated on 01/Sep/21 Answered by MJS_new last updated on 02/Sep/21 $$\left({xy}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{16}\left({x}−\mathrm{1}\right)\left({y}−\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{18}{xy}+\mathrm{16}{x}+\mathrm{16}{y}−\mathrm{15}\geqslant\mathrm{0} \\ $$$${x}=\mathrm{3}+{p}\wedge{y}=\mathrm{3}+{q}\wedge{p}\geqslant\mathrm{0}\wedge{q}\geqslant\mathrm{0} \\…
Question Number 87253 by peter frank last updated on 03/Apr/20 $${Three}\:{pair}\:{of}\:{socks}\:{are} \\ $$$${placed}\:{in}\:{a}\:{box}.{If}\:{two} \\ $$$${socks}\:{are}\:{drawn}\:{at} \\ $$$${random}\:{from}\:{the}\:{box} \\ $$$${What}\:{is}\:{the}\:{probability} \\ $$$$\left({a}\right){of}\:{drawing}\:\:{a}\:{match} \\ $$$${pair} \\ $$$$\left({b}\right){of}\:{drawing}\:{a}\:{socks}…
Question Number 152790 by mathdanisur last updated on 01/Sep/21 Commented by MJS_new last updated on 01/Sep/21 $$\mathrm{you}\:\mathrm{can}\:\mathrm{do}\:\mathrm{it}\:\mathrm{yourself}. \\ $$$$\mathrm{simply}\:\mathrm{let}\:{x}=\frac{{t}}{{a}}\:\mathrm{and}\:\mathrm{solve}\:\mathrm{it}! \\ $$ Terms of Service Privacy…
Question Number 152779 by Lekhraj last updated on 01/Sep/21 Commented by mr W last updated on 01/Sep/21 $${what}\:{is}\:{the}\:{question}? \\ $$ Commented by Lekhraj last updated…
Question Number 152764 by mnjuly1970 last updated on 01/Sep/21 Commented by prakash jain last updated on 01/Sep/21 $$\left.\mathrm{1}\right)\:{x}={n}+{f} \\ $$$${n}\geqslant\mathrm{0} \\ $$$${n}=\left({n}+{f}\right)\left({n}+{f}\right)=\left({n}+{f}\right)^{\mathrm{2}} \\ $$$${f}^{\mathrm{2}} +\mathrm{2}{nf}+{n}\left({n}−\mathrm{1}\right)=\mathrm{0}…
Question Number 152757 by EDWIN88 last updated on 01/Sep/21 $${Find}\:{all}\:{complex}\:{number}\:{z}\:{such} \\ $$$${that}\:\left(\mathrm{3}{z}+\mathrm{1}\right)\left(\mathrm{4}{z}+\mathrm{1}\right)\left(\mathrm{6}{z}+\mathrm{1}\right)\left(\mathrm{12}{z}+\mathrm{1}\right)=\mathrm{2} \\ $$ Answered by john_santu last updated on 01/Sep/21 $${note}\:{that}\:\mathrm{8}\left(\mathrm{3}{z}+\mathrm{1}\right)\mathrm{6}\left(\mathrm{4}{z}+\mathrm{1}\right)\mathrm{4}\left(\mathrm{6}{z}+\mathrm{1}\right)\mathrm{2}\left(\mathrm{12}{z}+\mathrm{1}\right)=\mathrm{768} \\ $$$$\left(\mathrm{24}{z}+\mathrm{8}\right)\left(\mathrm{24}{z}+\mathrm{6}\right)\left(\mathrm{24}{z}+\mathrm{4}\right)\left(\mathrm{24}{z}+\mathrm{2}\right)=\mathrm{768} \\…
Question Number 87224 by jagoll last updated on 03/Apr/20 $$\mathrm{how}\:\mathrm{to}\:\mathrm{simply}\:\mathrm{the}\: \\ $$$$\mathrm{boolean}\:\mathrm{algebra}\:\left(\mathrm{X}+\mathrm{Y}+\mathrm{Z}\right)\left(\mathrm{X}'\:+\mathrm{Y}+\mathrm{Z}\right) \\ $$$$\left(\mathrm{X}+\mathrm{Y}'+\mathrm{Z}\right)\: \\ $$ Commented by jagoll last updated on 03/Apr/20 $$\mathrm{how}\:\mathrm{sir}? \\…