Question Number 209510 by mr W last updated on 12/Jul/24 $${three}\:{points}\:{are}\:{randomly}\:{selected} \\ $$$${on}\:{a}\:{circle}\:{to}\:{form}\:{a}\:{triangle}.\: \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{probability}\:{that}\:{the}\:{center} \\ $$$${of}\:{the}\:{circle}\:{lies}\:{inside}\:{the}\:{triangle}. \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{probability}\:{that}\:{the}\: \\ $$$${triangle}\:{is}\:{an}\:{acute}\:{triangle}. \\ $$ Terms of…
Question Number 209495 by Tawa11 last updated on 11/Jul/24 Commented by Tawa11 last updated on 11/Jul/24 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{missing}\:\mathrm{angle}\:\left(\mathrm{the}\:\mathrm{red}\right) \\ $$ Answered by mr W last updated…
Question Number 209506 by Ismoiljon_008 last updated on 11/Jul/24 Commented by Ismoiljon_008 last updated on 11/Jul/24 $${Help}\:{please} \\ $$ Commented by mr W last updated…
Question Number 209436 by hardmath last updated on 10/Jul/24 $$\begin{cases}{\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{1}}\\{\mathrm{42x}\:+\:\mathrm{44y}\:+\:\mathrm{30z}\:=\:\mathrm{42}}\end{cases} \\ $$$$\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{1},\mathrm{0},\mathrm{0}\right)\:\mathrm{yes},\:\mathrm{but}\:\mathrm{solution}… \\ $$ Commented by mr W last updated on 10/Jul/24 $${you}\:{mean}\:{integer}\:{solutions}. \\ $$…
Question Number 209432 by Tawa11 last updated on 10/Jul/24 A body is projected vertically upwards with a speed of 20m/s. Find the time in seconds…
Question Number 209465 by RoseAli last updated on 10/Jul/24 Answered by mr W last updated on 10/Jul/24 $${x}−\mathrm{2}=\mathrm{0}\:\Rightarrow{x}=\mathrm{2} \\ $$$${x}−\mathrm{2}=\pm\mathrm{1}\:\Rightarrow{x}=\mathrm{1}\:{or}\:\mathrm{3} \\ $$$$\mathrm{10}{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{3}{x}\:\Rightarrow\left(\mathrm{2}{x}−\mathrm{1}\right)\left(\mathrm{5}{x}+\mathrm{1}\right)=\mathrm{0}\:\Rightarrow{x}=−\frac{\mathrm{1}}{\mathrm{5}}\:{or}\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{5}\:{solutions}…
Question Number 209433 by essaad last updated on 10/Jul/24 Answered by A5T last updated on 10/Jul/24 $${S}_{{p}} =\frac{\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}×…×\left(\mathrm{2}{p}\right)}{\mathrm{2}×\mathrm{4}×\mathrm{6}×…×\mathrm{2}{p}}=\frac{\left(\mathrm{2}{p}\right)!}{\mathrm{2}^{{p}} \left(\mathrm{1}×\mathrm{2}×…×{p}\right)} \\ $$$$=\frac{\left(\mathrm{2}{p}\right)!}{\mathrm{2}^{{p}} {p}!}\Rightarrow{C} \\ $$ Terms…
Question Number 209434 by justenspi last updated on 10/Jul/24 $${Help} \\ $$ Commented by justenspi last updated on 10/Jul/24 Commented by Berbere last updated on…
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Question Number 209456 by peter frank last updated on 10/Jul/24 Answered by Ghisom last updated on 10/Jul/24 $${c}=\mathrm{cos}\:\alpha\:\:\:\:\:{s}=\mathrm{sin}\:\alpha\:\:\:\:\:{t}=\mathrm{tan}\:\alpha\:=\frac{{s}}{{c}} \\ $$$${c}=\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }=\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${s}=\sqrt{\mathrm{1}−{c}^{\mathrm{2}} }=\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}}…