Question Number 168091 by leicianocosta last updated on 02/Apr/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 168090 by leicianocosta last updated on 02/Apr/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 168084 by mathocean1 last updated on 02/Apr/22 $${solve}\:{in}\:\mathbb{R} \\ $$$${arcsin}\left({x}\right)+{arcsin}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)=\frac{\pi}{\mathrm{2}} \\ $$ Commented by MJS_new last updated on 02/Apr/22 $${x}\geqslant\mathrm{0} \\ $$…
Question Number 168086 by mokys last updated on 02/Apr/22 Answered by som(math1967) last updated on 03/Apr/22 $$\mathrm{4}.\:{Angle}\:{between}\:\mathrm{2}{x}+{y}−\mathrm{2}{z}=\mathrm{5} \\ $$$${and}\:\mathrm{3}{x}−\mathrm{6}{y}−\mathrm{2}{z}=\mathrm{7} \\ $$$$\:\theta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}×\mathrm{3}+\mathrm{1}×\left(−\mathrm{6}\right)+\left(−\mathrm{2}\right)×\left(−\mathrm{2}\right)}{\:\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\left(−\mathrm{2}\right)^{\mathrm{2}} }\sqrt{\mathrm{3}^{\mathrm{2}}…
Question Number 168081 by aurpeyz last updated on 02/Apr/22 $$ \\ $$$$\mathrm{A}\:\mathrm{cook}\:\mathrm{puts}\:\mathrm{9}.\mathrm{00}\:\mathrm{g}\:\mathrm{of}\:\mathrm{water}\:\mathrm{in}\:\mathrm{a}\:\mathrm{2}.\mathrm{00L}\:\mathrm{pressure}\:\mathrm{cooker}\: \\ $$$$\mathrm{th}{a}\mathrm{t}\:\mathrm{is}\:\mathrm{then}\:\mathrm{warmed}\:\mathrm{to}\:\mathrm{500}\:\mathrm{degree}\:\mathrm{C}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{pressur}{e}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{container}\:\mathrm{A}.\:\mathrm{1}.\mathrm{89}\:\mathrm{times}\:\mathrm{106Pa}\:\mathrm{B}.\: \\ $$$$\mathrm{1}.\mathrm{89times}\:\mathrm{105Pa}\:\mathrm{C}.\:\mathrm{1}.\mathrm{99}\:\mathrm{times}\:\mathrm{104Pa}\:\mathrm{D}.\:\mathrm{1}.\mathrm{89}\:\mathrm{times}\:\mathrm{103P} \\ $$$$\mathrm{a}\:\mathrm{E}.\:\mathrm{1}.\mathrm{99}\:\mathrm{times}\:\mathrm{109Pa} \\ $$ Terms of Service…
Question Number 168071 by aurpeyz last updated on 02/Apr/22 $$ \\ $$$$\mathrm{A}\:\mathrm{clean}\:\mathrm{tube}\:\mathrm{of}\:\mathrm{diameter}\:\mathrm{2}.\mathrm{5}\:\mathrm{mm}\:\mathrm{is}\:\mathrm{imersed}\:\mathrm{in}\:\mathrm{a}\:\mathrm{liquidi} \\ $$$$\mathrm{wth}\:\mathrm{a}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{surface}\:\mathrm{tension}\:\:\mathrm{0}.\mathrm{4}\:\mathrm{nm}.\:\mathrm{The}\:\mathrm{anglef} \\ $$$$\mathrm{o}\:\mathrm{contact}\:\mathrm{of}\:\mathrm{the}\:\mathrm{liquid}\:\mathrm{with}\:\mathrm{the}\:\mathrm{glass}\:\mathrm{can}\:\mathrm{be}\:\mathrm{assumed}\:\mathrm{as1} \\ $$$$\mathrm{35}^{\mathrm{0}} .\:\mathrm{The}\:\mathrm{density}\:\mathrm{of}\:\mathrm{liquid}\:\mathrm{13600}\:\mathrm{kgm3}.\:\mathrm{What}\:\mathrm{would}\: \\ $$$$\mathrm{bethe}\:\mathrm{level}\:\mathrm{of}\:\mathrm{the}\:\mathrm{liquid}\:\mathrm{in}\:\mathrm{the}\:\mathrm{tube}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{the}\:\mathrm{freeu} \\ $$$$\mathrm{srface}\:\mathrm{of}\:\mathrm{the}\:\mathrm{liuid}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{tube}\:\mathrm{Answer}\:\mathrm{should}\:\mathrm{be}\:\mathrm{inm} \\ $$$$\mathrm{m}…
Question Number 168064 by MathsFan last updated on 02/Apr/22 $$\:{q}=\frac{{m}}{\:\sqrt{{p}}}+\frac{{p}^{\mathrm{2}} }{{m}} \\ $$$$\:{make}\:\:{p}\:\:{the}\:{subject} \\ $$ Commented by MJS_new last updated on 02/Apr/22 $$\mathrm{you}\:\mathrm{can}'\mathrm{t}.\:\mathrm{this}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\left(\sqrt{{p}}\right)^{\mathrm{5}}…
Question Number 168053 by lapache last updated on 01/Apr/22 $${Solve}\:{the}\:{system}\: \\ $$$$\begin{cases}{{x}+\mathrm{2}{y}+\mathrm{3}{z}+\mathrm{2}{t}=\mathrm{2}}\\{\mathrm{2}{x}+\mathrm{5}{y}−\mathrm{8}{z}+\mathrm{6}{t}=\mathrm{5}}\\{\mathrm{3}{x}+\mathrm{4}{y}−\mathrm{5}{z}+\mathrm{2}{t}=\mathrm{4}}\end{cases} \\ $$ Answered by MJS_new last updated on 01/Apr/22 $$\mathrm{3}\:\mathrm{equations};\:\mathrm{4}\:\mathrm{variables}\:\Rightarrow\:\mathrm{parametric}\:\mathrm{solution} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\\{{t}}\end{pmatrix}\:\:=\begin{pmatrix}{\mathrm{2}{p}}\\{\mathrm{1}−\mathrm{2}{p}}\\{\mathrm{0}}\\{{p}}\end{pmatrix}\:\:\mathrm{with}\:{p}\in\mathbb{R} \\…
Question Number 168013 by leicianocosta last updated on 31/Mar/22 Answered by nurtani last updated on 31/Mar/22 $$\:\:\:\:\:^{\mathrm{9}} \sqrt{{x}}+\frac{\:^{\mathrm{9}} \sqrt{{x}^{\mathrm{8}} }}{{x}}=\frac{\mathrm{17}}{\mathrm{4}} \\ $$$$\Leftrightarrow\:{x}^{\frac{\mathrm{1}}{\mathrm{9}}} +\frac{{x}^{\frac{\mathrm{8}}{\mathrm{9}}} }{{x}}=\frac{\mathrm{17}}{\mathrm{4}} \\…
Question Number 168000 by aurpeyz last updated on 31/Mar/22 $${Divide}\:{a}^{\frac{\mathrm{5}}{\mathrm{2}}} −\mathrm{5}{a}^{\mathrm{2}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{10}{a}^{\frac{\mathrm{3}}{\mathrm{2}}} {b}^{\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{10}{ab}+\mathrm{5}{a}^{\frac{\mathrm{1}}{\mathrm{2}}} {b}^{\frac{\mathrm{4}}{\mathrm{3}}} \:{by}\:{a}^{\frac{\mathrm{1}}{\mathrm{2}}} −{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$ Terms of Service Privacy Policy…