Question Number 226115 by fantastic2 last updated on 20/Nov/25 $${can}\:{we}\:{find}\:{the}\:{black}\:{area} \\ $$$$\left({the}\:{shadow}\:{of}\:{earth}\right){on}\:{the}\:{moon}\:{during} \\ $$$${moon}\:{eclipse}\:{after}\:{time}\:{t}\:{of} \\ $$$${its}\:{starting}? \\ $$$${Given}: \\ $$$${Moon}\:{radius}={R}_{{m}} \\ $$$${Earth}\:{rafius}={R}_{{e}} \\ $$$${Sun}\:{radius}={R}_{{s}} \\…
Question Number 226111 by Linton last updated on 19/Nov/25 $${Show}\:{that}\:{the}\:{equation} \\ $$$$\frac{\mathrm{1}}{{sin}\theta+{cos}\theta}\:+\:\frac{\mathrm{1}}{{sin}\theta−{cos}\theta}\:=\:\mathrm{1} \\ $$$${may}\:{be}\:{express}\:{in}\:{the}\:{form} \\ $$$${a}\left({sin}\theta\right)^{\mathrm{2}} +{bsin}\theta+{c}=\mathrm{0}\:{where}\:{a}\:{b}\: \\ $$$${and}\:{c}\:{are}\:{constants}\:{to}\:{be}\:{found}. \\ $$ Answered by Frix last…
Question Number 226112 by Spillover last updated on 19/Nov/25 Answered by Spillover last updated on 20/Nov/25 Answered by Spillover last updated on 20/Nov/25 Terms of…
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Question Number 226015 by mr W last updated on 18/Nov/25 Answered by fantastic2 last updated on 18/Nov/25 $${f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)={x}…\left({i}\right) \\ $$$${put}\:{x}=\mathrm{2}\:{in}\:{i} \\ $$$${f}\left(\mathrm{2}\right)+{f}\left(−\mathrm{1}\right)=\mathrm{2}….\mathrm{1} \\ $$$${put}\:−\mathrm{1}\:{in}\:{i} \\…
Question Number 226053 by Linton last updated on 18/Nov/25 $${The}\:{coefficient}\:{of}\:{x}^{\mathrm{2}} \:{in}\:{the}\:{expansion} \\ $$$${of}\:\left(\mathrm{1}+\:\left(\mathrm{2}/{p}\right){x}\right)^{\mathrm{5}} \:+\:\left(\mathrm{1}+{px}\right)^{\mathrm{6}} \:{is}\:\mathrm{70}. \\ $$$${Find}\:{the}\:{possible}\:{values}\:{of}\:{the} \\ $$$${constant}\:{p}. \\ $$ Answered by mr W…
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Question Number 225934 by gregori last updated on 16/Nov/25 $$\:\: \begin{cases}{\lceil\:\frac{\mathrm{8}−\mathrm{2}{x}}{\mathrm{3}}\:\rceil\:;\:{x}\geqslant\:\mathrm{0}}\\{\lfloor\:\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{4}}\:\rfloor\:;\:{x}<\mathrm{0}}\end{cases}. \\ $$$$\left.\:\: − −\mathrm{1}\right)+\: \\ $$$$ \\ $$ Answered by efronzo1 last updated on…
Question Number 225932 by Rojarani last updated on 16/Nov/25 $$\:{If},\:\frac{{by}+{cz}}{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }=\frac{{cz}+{ax}}{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{{ax}+{by}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\:{then}\:{prove}\:{that},\:\frac{{x}}{{a}}=\frac{{y}}{{b}}=\frac{{z}}{{c}} \\ $$ Answered by som(math1967) last updated…
Question Number 225938 by Spillover last updated on 16/Nov/25 Answered by Spillover last updated on 17/Nov/25 Answered by Spillover last updated on 17/Nov/25 Answered by…