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Question-212736

Question Number 212736 by RojaTaniya last updated on 22/Oct/24 Answered by efronzo1 last updated on 22/Oct/24 $$\:\:\:\:\frac{\mathrm{x}}{\mathrm{4}}\:=\:\frac{\mathrm{9}}{\mathrm{15}}\:\Rightarrow\mathrm{x}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\:\:\:\:\frac{\mathrm{y}}{\mathrm{5}}=\:\frac{\mathrm{9}}{\mathrm{15}}\Rightarrow\mathrm{y}=\:\mathrm{3} \\ $$$$\:\:\:\:\:\frac{\mathrm{z}}{\mathrm{6}}=\:\frac{\mathrm{9}}{\mathrm{15}}\:\Rightarrow\mathrm{z}=\frac{\mathrm{18}}{\mathrm{5}} \\ $$ Commented by…

in-how-many-ways-can-a-teacher-divide-his-10-studens-into-4-groups-such-that-each-group-has-at-least-2-students-

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…

f-x-1-1-x-y-f-x-z-f-y-f-z-

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-212673

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-212690

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-212668

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}} \\…