Question Number 212686 by mr W last updated on 21/Oct/24 $${in}\:{how}\:{many}\:{ways}\:{can}\:{a}\:{teacher} \\ $$$${divide}\:{his}\:\mathrm{10}\:{studens}\:{into}\:\mathrm{4}\:{groups} \\ $$$${such}\:{that}\:{each}\:{group}\:{has}\:{at}\:{least}\:\mathrm{2}\: \\ $$$${students}? \\ $$ Commented by Spillover last updated on…
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Question Number 212711 by RojaTaniya last updated on 21/Oct/24 $$\:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\:{y}={f}\left({x}\right),\:{z}={f}\left({y}\right),\:{f}\left({z}\right)=? \\ $$ Answered by MATHEMATICSAM last updated on 21/Oct/24 $${y}\:=\:{f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}\:−\:{x}} \\ $$$${f}\left({y}\right)\:=\:{f}\left(\frac{\mathrm{1}}{\mathrm{1}\:−\:{x}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{1}\:−\:{x}}}\:=\:\frac{\mathrm{1}\:−\:{x}}{−\:{x}}\:=\:\frac{{x}\:−\:\mathrm{1}}{{x}} \\…
Question Number 212673 by golsendro last updated on 21/Oct/24 $$\:\:\:\: \\ $$ Answered by Frix last updated on 21/Oct/24 $$\sqrt{\mathrm{2}{x}+\mathrm{3}}={x}^{\mathrm{2}} −{x}−\mathrm{3} \\ $$$$\mathrm{Squaring}\:\&\:\mathrm{transforming} \\ $$$$\:\:\:\:\:\left[\mathrm{introduces}\:\mathrm{false}\:\mathrm{solutions}!\right]…
Question Number 212690 by Spillover last updated on 21/Oct/24 Answered by mr W last updated on 21/Oct/24 $$\mathrm{10}×\left(\mathrm{2}{n}\right)=\left(\mathrm{2}\sqrt{\mathrm{15}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{n}=\mathrm{3} \\ $$$$\Rightarrow{R}={n}+\frac{\mathrm{10}}{\mathrm{2}}=\mathrm{8} \\ $$$${painted}\:{area}\:=\pi\left({R}^{\mathrm{2}} −{n}^{\mathrm{2}}…
Question Number 212668 by Spillover last updated on 20/Oct/24 Answered by efronzo1 last updated on 21/Oct/24 $$\:\:\mathrm{BC}^{\mathrm{2}} =\:\mathrm{64}+\mathrm{25}−\mathrm{40}\:=\:\mathrm{49} \\ $$$$\:\:\mathrm{BC}^{\mathrm{2}} =\:\mathrm{2r}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \Rightarrow\mathrm{r}=\sqrt{\frac{\mathrm{BC}^{\mathrm{2}} }{\mathrm{3}}}=\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{3}} \\…
Question Number 212669 by Spillover last updated on 20/Oct/24 Commented by Spillover last updated on 21/Oct/24 $${ans}\:{X}=\:\mathrm{35}\sqrt{\mathrm{2}}\: \\ $$ Answered by mr W last updated…
Question Number 212654 by Spillover last updated on 20/Oct/24 Answered by mr W last updated on 20/Oct/24 Commented by mr W last updated on 20/Oct/24…
Question Number 212648 by issac last updated on 20/Oct/24 $$\mathrm{prove}\:\mathrm{the}\:\mathrm{Following}\:\mathrm{Equation}. \\ $$$$\:{J}_{\nu} \left({z}\right)\:\mathrm{and}\:{Y}_{\nu} \left({z}\right)\:\mathrm{are}\:\:\mathrm{Bessel}\:\mathrm{function} \\ $$$${J}_{−\nu−\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right)=\left(−\mathrm{1}\right)^{\nu+\mathrm{1}} {Y}_{\nu+\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right) \\ $$$${Y}_{−\nu−\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right)=\left(−\mathrm{1}\right)^{\nu} {J}_{\nu+\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right) \\…
Question Number 212651 by hardmath last updated on 20/Oct/24 Answered by Ghisom last updated on 20/Oct/24 $${x}+\mathrm{3}{y}+\mathrm{5}{z}\leqslant\mathrm{15} \\ $$$${x}+{y}+{z}\leqslant\mathrm{7} \\ $$$$\mathrm{2}{x}+{y}+\mathrm{4}{z}\leqslant\mathrm{12} \\ $$$$======== \\ $$$${x}+\mathrm{3}{y}+\mathrm{5}{z}\leqslant\mathrm{15}…