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 209456 by peter frank last updated on 10/Jul/24 Answered by Ghisom last updated on 10/Jul/24 $${c}=\mathrm{cos}\:\alpha\:\:\:\:\:{s}=\mathrm{sin}\:\alpha\:\:\:\:\:{t}=\mathrm{tan}\:\alpha\:=\frac{{s}}{{c}} \\ $$$${c}=\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }=\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${s}=\sqrt{\mathrm{1}−{c}^{\mathrm{2}} }=\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}}…
Question Number 209455 by peter frank last updated on 10/Jul/24 Answered by Spillover last updated on 10/Jul/24 Commented by peter frank last updated on 11/Jul/24…
Question Number 209404 by Tawa11 last updated on 09/Jul/24 Answered by mr W last updated on 09/Jul/24 $${s}=\mathrm{4}{t}+{at}^{\mathrm{2}} +{bt}^{\mathrm{3}} \\ $$$${v}=\frac{{ds}}{{dt}}=\mathrm{4}+\mathrm{2}{at}+\mathrm{3}{bt}^{\mathrm{2}} \\ $$$${a}=\frac{{dv}}{{dt}}=\mathrm{2}{a}+\mathrm{6}{bt} \\ $$$${at}\:{v}_{{max}}…
Question Number 209415 by RoseAli last updated on 09/Jul/24 Commented by Frix last updated on 09/Jul/24 $$\mathrm{This}\:\mathrm{is}\:\mathrm{just}\:\mathrm{a}\:\mathrm{term}. \\ $$$$\mathrm{No}\:\mathrm{question}\:\Rightarrow\:\mathrm{No}\:\mathrm{answer} \\ $$$$ \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{like}\:\mathrm{these}: \\ $$$$\bullet\:\mathrm{5}{ab}^{\mathrm{2}}…
Question Number 209394 by Tawa11 last updated on 08/Jul/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 209357 by mnjuly1970 last updated on 08/Jul/24 $$ \\ $$$$\:\:\:\:\:\:{Evaluate}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{B}_{{n}} =\:\underset{{k}=\mathrm{3}} {\overset{{n}} {\prod}}\:\frac{\:{k}^{\:\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} \:+\:{k}\:−\mathrm{6}}=\:? \\ $$ Answered by…
Question Number 209385 by efronzo1 last updated on 08/Jul/24 Answered by Rasheed.Sindhi last updated on 08/Jul/24 $${p}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+\mathrm{1}\Rightarrow{p}\left(\mathrm{1}\right)={a}+{b}+\mathrm{1} \\ $$$${q}\left({x}\right)={bx}^{\mathrm{2}} +{ax}+\mathrm{1}\Rightarrow{q}\left(\mathrm{1}\right)={b}+{a}+\mathrm{1} \\ $$$${p}\left(\:{q}\left(\mathrm{1}\right)\:\right)={q}\left(\:{p}\left(\mathrm{1}\right)\:\right) \\ $$$${p}\left({a}+{b}+\mathrm{1}\right)={q}\left({a}+{b}+\mathrm{1}\right)…
Question Number 209309 by hardmath last updated on 06/Jul/24 $$\mathrm{m}\:,\:\mathrm{n}\:\in\:\mathbb{N} \\ $$$$\mathrm{m}\:\geqslant\:\mathrm{2}\:\:\:\mathrm{and}\:\:\:\mathrm{n}\:\geqslant\:\mathrm{2} \\ $$$$\mathrm{p}\:>\:\mathrm{0}\:\:\:\mathrm{and}\:\:\:\mathrm{q}\:>\:\mathrm{0} \\ $$$$\mathrm{p}\:+\:\mathrm{q}\:=\:\mathrm{1} \\ $$$$\mathrm{Prove}\:\mathrm{that}:\:\:\:\left(\mathrm{1}−\mathrm{q}^{\boldsymbol{\mathrm{n}}} \right)^{\boldsymbol{\mathrm{m}}} \:+\:\left(\mathrm{1}−\mathrm{p}^{\boldsymbol{\mathrm{m}}} \right)^{\boldsymbol{\mathrm{n}}} \:\geqslant\:\mathrm{1} \\ $$ Terms…
Question Number 209290 by klipto last updated on 06/Jul/24 $$\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}−^{\mathrm{1000}} \sqrt{\left(\mathrm{1}+\mathrm{8000}\boldsymbol{\mathrm{a}}\right)}=\mathrm{1000} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{a}} \\ $$ Commented by mr W last updated on 06/Jul/24 $${you}\:{can}\:{only}\:{approximate}!…