Question Number 210679 by peter frank last updated on 16/Aug/24 Answered by mr W last updated on 16/Aug/24 $${RH}_{\mathrm{16}°{C}} =\frac{\mathrm{273}.\mathrm{15}+\mathrm{16}}{\mathrm{273}.\mathrm{15}+\mathrm{20}}×\frac{\mathrm{17}.\mathrm{5}}{\mathrm{13}.\mathrm{6}}×\mathrm{36\%}\approx\mathrm{46\%} \\ $$ Commented by peter…
Question Number 210689 by Tawa11 last updated on 16/Aug/24 Answered by mm1342 last updated on 16/Aug/24 $${AM}={MD}={a}\Rightarrow{BC}=\frac{\mathrm{2}}{\mathrm{3}}{a} \\ $$$$\frac{{s}}{{S}}=\frac{\frac{{BC}×{h}}{\mathrm{2}}}{\frac{\left({BC}+{AD}\right){h}}{\mathrm{2}}}=\frac{{BC}}{{BC}+{AD}}=\frac{\frac{\mathrm{2}}{\mathrm{3}}{a}}{\mathrm{2}{a}+\frac{\mathrm{2}}{\mathrm{3}}{a}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\checkmark \\ $$$$ \\ $$…
Question Number 210666 by universe last updated on 15/Aug/24 $$\:\:\mathrm{prove}\:\mathrm{that}\:\mathrm{p}\left(\mathrm{n}\right)\:\mathrm{is}\:\mathrm{integer}\:\forall\:\mathrm{n}\in\mathbb{Z} \\ $$$$\:\:\:\mathrm{p}\left(\mathrm{n}\right)\:=\:\frac{\mathrm{3n}^{\mathrm{7}} +\mathrm{7n}^{\mathrm{3}} +\mathrm{11n}}{\mathrm{21}} \\ $$ Answered by A5T last updated on 15/Aug/24 $${Let}\:{f}\left({n}\right)=\mathrm{3}{n}^{\mathrm{7}} +\mathrm{7}{n}^{\mathrm{3}}…
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Question Number 210643 by ChantalYah last updated on 14/Aug/24 Answered by mahdipoor last updated on 14/Aug/24 $$\left.\mathrm{1}\right){e}^{{x}} ={t} \\ $$$${t}^{\mathrm{3}} −\mathrm{3}{t}−\frac{\mathrm{4}}{{t}}=\mathrm{0}\Rightarrow{t}^{\mathrm{4}} −\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}=\mathrm{0}\Rightarrow \\ $$$${t}^{\mathrm{2}}…
Question Number 210639 by ChantalYah last updated on 14/Aug/24 $${given}\:{that}\:{the}\:{roots} \\ $$$$\:{of}\:{the}\:{equation} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\left(\mathrm{4}+\mathrm{2}{k}\right){x}+\mathrm{2}{k}=\mathrm{0} \\ $$$${are}\:\alpha\:{and}\:\beta \\ $$$${find}\:{the}\:{value}\:{of}\:{k} \\ $$$${for}\:{which}\:\beta=\mathrm{3}\alpha \\ $$ Answered by…
Question Number 210629 by peter frank last updated on 14/Aug/24 Answered by Rasheed.Sindhi last updated on 14/Aug/24 $$\begin{cases}{{x}={a}+\left({a}+{d}\right)+\left({a}+\mathrm{2}{d}\right)+,…+\left({a}+\left({m}−\mathrm{1}\right){d}\right)}\\{{y}=\left({a}+{md}\right)+\left({a}+\left({m}+\mathrm{1}\right){d}\right)+…+\left({a}+\left(\mathrm{2}{m}−\mathrm{1}\right){d}\right)}\\{{z}=\left({a}+\mathrm{2}{md}\right)+\left({a}+\left(\mathrm{2}{m}+\mathrm{1}\right){d}\right)+…+\left({a}+\left(\mathrm{3}{m}−\mathrm{1}\right){d}\right)}\end{cases}\: \\ $$$$\begin{cases}{{x}−{ma}=\left({d}\right)+\left(\mathrm{2}{d}\right)+,…+\left(\left({m}−\mathrm{1}\right){d}\right)}\\{{y}−{ma}=\left({md}\right)+\left(\left({m}+\mathrm{1}\right){d}\right)+…+\left(\left(\mathrm{2}{m}−\mathrm{1}\right){d}\right)}\\{{z}−{ma}=\left(\mathrm{2}{md}\right)+\left(\left(\mathrm{2}{m}+\mathrm{1}\right){d}\right)+…+\left(\left(\mathrm{3}{m}−\mathrm{1}\right){d}\right)}\end{cases}\:\: \\ $$$$\begin{cases}{\frac{{x}−{ma}}{{d}}=\left(\mathrm{1}\right)+\left(\mathrm{2}\right)+,…+\left(\left({m}−\mathrm{1}\right)\right)}\\{\frac{{y}−{ma}}{{d}}=\left({m}\right)+\left(\left({m}+\mathrm{1}\right)\right)+…+\left(\left(\mathrm{2}{m}−\mathrm{1}\right)\right)\:}\\{\frac{{z}−{ma}}{{d}}=\left(\mathrm{2}{m}\right)+\left(\left(\mathrm{2}{m}+\mathrm{1}\right)\right)+…+\left(\left(\mathrm{3}{m}−\mathrm{1}\right)\right)}\end{cases}\: \\ $$$$…\:\: \\…