Question Number 33860 by anicetoluisjoao@gmail.com last updated on 26/Apr/18 $$\sqrt{\left[{x}+\mathrm{3}{x}\right]×\mathrm{4}}\:\:=\left(\begin{vmatrix}{\mathrm{2}\:\:\:\:\mathrm{4}\:\:\:\mathrm{5}}\\{\mathrm{3}\:\:\:\:\mathrm{5}\:\:\:\:\mathrm{7}}\end{vmatrix}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 33822 by mondodotto@gmail.com last updated on 25/Apr/18 $$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{following}}\:\boldsymbol{\mathrm{table}}\:\boldsymbol{\mathrm{shows}}\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{distributuons}}\:\boldsymbol{\mathrm{of}}\:\mathrm{100}\:\boldsymbol{\mathrm{families}}\:\boldsymbol{\mathrm{according}} \\ $$$$\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{their}}\:\boldsymbol{\mathrm{expenditure}}\:\boldsymbol{\mathrm{per}}\:\boldsymbol{\mathrm{week}}. \\ $$$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{mode}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{be}}\:\mathrm{24}. \\ $$$$\mid\frac{\boldsymbol{\mathrm{expenditure}}}{\boldsymbol{\mathrm{Number}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{families}}}\mid\frac{\mathrm{10}−\mathrm{20}}{\mathrm{x}}\mid\frac{\mathrm{20}−\mathrm{30}}{\mathrm{27}}\mid\frac{\mathrm{30}−\mathrm{40}}{\boldsymbol{\mathrm{y}}}\mid\frac{\mathrm{40}−\mathrm{50}}{\mathrm{15}}\mid \\ $$$$\left(\boldsymbol{\mathrm{a}}\right)\:\boldsymbol{\mathrm{calculate}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{missing}}\:\boldsymbol{\mathrm{frequency}} \\ $$$$\left(\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{calculate}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{mean}} \\ $$$$\left(\boldsymbol{\mathrm{c}}\right)\boldsymbol{\mathrm{calculate}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{median}} \\…
Question Number 33819 by mondodotto@gmail.com last updated on 25/Apr/18 $$\boldsymbol{\mathrm{given}}\:\:\:\boldsymbol{\mathrm{a}}=\mathrm{3}\boldsymbol{\mathrm{i}}−\mathrm{4}\boldsymbol{\mathrm{j}}+\mathrm{2}\boldsymbol{\mathrm{k}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{b}}=−\mathrm{2}\boldsymbol{\mathrm{i}}+\boldsymbol{\mathrm{j}}−\mathrm{3}\boldsymbol{\mathrm{k}} \\ $$$$\boldsymbol{\mathrm{find}}\:\:\:\left(\boldsymbol{\mathrm{i}}\right)\:\boldsymbol{\mathrm{v}}=\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{ii}}\right)\boldsymbol{\mathrm{v}}\:\boldsymbol{\mathrm{into}}\:\boldsymbol{\mathrm{position}}\:\boldsymbol{\mathrm{vector}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{iii}}\right)\boldsymbol{\mathrm{unit}}\:\boldsymbol{\mathrm{vector}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{v}} \\ $$ Terms of Service Privacy Policy…
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Question Number 99321 by Ar Brandon last updated on 20/Jun/20 $$\begin{bmatrix}{\boldsymbol{\mathrm{Angle}}\left(\boldsymbol{\theta}\right)}&{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\theta}\right)}&{\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\theta}\right)\left[\right.}\\{\mathrm{0}^{°} }&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{15}°}&{\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}}}&{\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}}\\{\mathrm{18}°}&{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}}&{\frac{\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}}{\mathrm{2}\sqrt{\mathrm{2}}}}\\{\mathrm{30}°}&{\frac{\mathrm{1}}{\mathrm{2}}}&{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\\{\mathrm{36}°}&{\frac{\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}{\mathrm{2}\sqrt{\mathrm{2}}}}&{\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}}\\{\mathrm{45}°}&{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}&{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\\{\mathrm{54}°}&{\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}}&{\frac{\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}{\mathrm{2}\sqrt{\mathrm{2}}}}\\{\mathrm{60}°}&{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}&{\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{72}°}&{\frac{\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}}{\mathrm{2}\sqrt{\mathrm{2}}}}&{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}}\\{\mathrm{75}°}&{\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}}&{\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}}}\\{\mathrm{90}°}&{\mathrm{1}}&{\mathrm{0}}\end{bmatrix} \\ $$$$\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\theta}\right)=\boldsymbol{\mathrm{cos}}\left(\mathrm{90}°−\boldsymbol{\theta}\right) \\ $$ Commented by bobhans last updated on 20/Jun/20 $$\mathrm{waw}…\mathrm{cooll} \\…
Question Number 99306 by Ar Brandon last updated on 20/Jun/20 $$\mathrm{Hello}\:\mathrm{Mr}\:\mathrm{administrator}.\:\mathrm{Sorry}\:\mathrm{to}\:\mathrm{trouble}\:\mathrm{with}\:\mathrm{my}\:\mathrm{question}.\:\mathrm{The} \\ $$$$\mathrm{app}\:\mathrm{is}\:\mathrm{perfectly}\:\mathrm{working}\:\mathrm{for}\:\mathrm{me}.\:\mathrm{But}\:\mathrm{just}\:\mathrm{wanted}\:\mathrm{to}\:\mathrm{know}\:\mathrm{if}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{possible}\:\mathrm{to}\:\mathrm{copy}\:\mathrm{some}\:\mathrm{previously}\:\mathrm{saved}\:\mathrm{equations}\:\mathrm{then}\:\mathrm{paste} \\ $$$$\:\mathrm{them}\:\mathrm{in}\:\mathrm{the}\:\mathrm{text}\:\mathrm{box}\:\mathrm{in}\:\mathrm{the}\:\mathrm{forum},\:\mathrm{appart}\:\mathrm{from}\:\mathrm{exporting}\:\mathrm{them}\:\mathrm{as}\:\mathrm{images}? \\ $$$$\mathrm{By}\:\mathrm{the}\:\mathrm{way}\:\mathrm{how}\:\mathrm{can}\:\mathrm{we}\:\mathrm{do}\:\mathrm{to}\:\mathrm{save}\:\mathrm{and}\:\mathrm{recover}\:\mathrm{drafts}?\:\mathrm{It}\:\mathrm{does}\:\mathrm{so}\:\mathrm{usually}\:\mathrm{out}\:\mathrm{of} \\ $$$$\mathrm{my}\:\mathrm{will}.\:\mathrm{How}\:\mathrm{can}\:\mathrm{we}\:\mathrm{manage}\:\mathrm{this}\:\mathrm{situation}? \\ $$$$\mathcal{T}\mathrm{hanks}\:\mathrm{in}\:\mathrm{advance}\:! \\ $$…
Question Number 33767 by artibunja last updated on 23/Apr/18 Answered by MJS last updated on 23/Apr/18 $$\left({x}−\mathrm{3}\right)^{\mathrm{2}} >\mathrm{0}\:\mathrm{with}\:{x}\in\mathbb{R}\backslash\left\{\mathrm{3}\right\}\:\Rightarrow\:\mathrm{multiplication} \\ $$$$\mathrm{with}\:\mathrm{denominator}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{change}\:\mathrm{sign} \\ $$$${x}^{\mathrm{2}} −\mathrm{7}\mid{x}\mid+\mathrm{10}<\mathrm{0} \\ $$$$\mathrm{case}\:\mathrm{1}.:…
Question Number 33764 by .none. last updated on 23/Apr/18 $$\mathrm{2}{x}−{y}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}−\mathrm{2}} \\ $$$${x}+\mathrm{2}{y}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}+\mathrm{2}} \\ $$$$\left[\mathrm{3}{x}^{\mathrm{2}} −\mathrm{8}{xy}−\mathrm{3}{y}^{\mathrm{2}} \right]? \\ $$ Answered by Rasheed.Sindhi last updated on 23/Apr/18…