Question Number 101766 by 175mohamed last updated on 04/Jul/20 Commented by mr W last updated on 04/Jul/20 $$\mathrm{1}\:{m}^{\mathrm{3}} =\mathrm{100}^{\mathrm{3}} \:{cm}^{\mathrm{3}} \\ $$$$\frac{\mathrm{100}^{\mathrm{3}} \:{cm}^{\mathrm{3}} }{\mathrm{5}×\mathrm{4}×\mathrm{1}\:{cm}^{\mathrm{3}} }=\mathrm{50}\:\mathrm{000}…
Question Number 35789 by ajfour last updated on 23/May/18 Commented by ajfour last updated on 23/May/18 $${How}\:{many}\:{skew}\:{pair}\:{of}\:{lines} \\ $$$${are}\:{there}\:\left({when}\:{a}\:{line}\:{is}\:{drawn}\right. \\ $$$${by}\:{joining}\:{any}\:{two}\:{vertices} \\ $$$$\left.{of}\:{the}\:{cube}\right)\:? \\ $$…
Question Number 35430 by ajfour last updated on 18/May/18 Commented by ajfour last updated on 19/May/18 $${Solution}\:{to}\:{Q}.\mathrm{35422}\:\left({Find}\:{P}\:{Q}\right) \\ $$ Answered by ajfour last updated on…
Question Number 35422 by ajfour last updated on 18/May/18 Commented by ajfour last updated on 19/May/18 $${Sphere}\:{of}\:{radius}\:{R}\:{touches} \\ $$$${plane}\:{at}\:{B}.\:{Line}\:{APQ}\:{intersects} \\ $$$${sphere}\:{at}\:{P}\:{and}\:{Q}.\: \\ $$$${minimum}\:{angle}\:{that}\:{line}\:{makes} \\ $$$${with}\:{plane}\:{is}\:\theta.\:{From}\:{this}\:…
Question Number 100850 by Dwaipayan Shikari last updated on 28/Jun/20 $$\int_{\mathrm{0}} ^{\mathrm{102}} \left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)…..\left({x}−\mathrm{100}\right)×\left(\frac{\mathrm{1}}{{x}−\mathrm{1}}+\frac{\mathrm{1}}{{x}−\mathrm{2}}+…+\frac{\mathrm{1}}{{x}−\mathrm{100}}\right){dx} \\ $$ Commented by Dwaipayan Shikari last updated on 28/Jun/20 $$\int_{\mathrm{0}} ^{\mathrm{102}}…
Question Number 166134 by amin96 last updated on 13/Feb/22 Answered by MJS_new last updated on 14/Feb/22 $$\mathrm{you}\:\mathrm{need}\:\mathrm{the}\:\mathrm{trigonometric}\:\mathrm{method}\:\mathrm{to}\:\mathrm{get} \\ $$$${x}_{\mathrm{1}} =−\mathrm{cos}\:\frac{\pi}{\mathrm{9}} \\ $$$${x}_{\mathrm{2}} =\mathrm{sin}\:\frac{\pi}{\mathrm{18}} \\ $$$${x}_{\mathrm{3}}…
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Question Number 34231 by byaw last updated on 03/May/18 Answered by MJS last updated on 03/May/18 $$\mathrm{136}°−\mathrm{46}°=\mathrm{90}°\:\Rightarrow\:\mathrm{cos}\:\mathrm{136}°=−\mathrm{sin}\:\mathrm{46}°;\:\mathrm{sin}\:\mathrm{136}°=\mathrm{cos}\:\mathrm{46}° \\ $$$${v}_{\mathrm{1}} =\left(\mathrm{20}\:\angle\mathrm{136}°\right)=\begin{pmatrix}{\mathrm{20cos}\:\mathrm{136}°}\\{\mathrm{20sin}\:\mathrm{136}°}\end{pmatrix}=\begin{pmatrix}{−\mathrm{20sin}\:\mathrm{46}°}\\{\mathrm{20cos}\:\mathrm{46}°}\end{pmatrix} \\ $$$${v}_{\mathrm{2}} =\left(\mathrm{5}\:\angle\mathrm{46}°\right)=\begin{pmatrix}{\mathrm{5cos}\:\mathrm{46}°}\\{\mathrm{5sin}\:\mathrm{46}°}\end{pmatrix} \\ $$$${v}={v}_{\mathrm{1}}…
Question Number 34230 by byaw last updated on 03/May/18 Answered by MJS last updated on 03/May/18 $$\mathrm{aerial}: \\ $$$$\mathrm{bottom}\:{C}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{top}\:{B}=\begin{pmatrix}{\mathrm{0}}\\{{a}}\end{pmatrix} \\ $$$$\mathrm{surveyor}: \\ $$$$\mathrm{position}\:\mathrm{1}\:{A}_{\mathrm{1}}…
Question Number 164639 by cortano1 last updated on 20/Jan/22 Answered by mr W last updated on 20/Jan/22 $${x}={a}\:\mathrm{cos}\:\mathrm{2}\pi{t}\:\Rightarrow\frac{{dx}}{{dt}}=−\mathrm{2}\pi{a}\:\mathrm{sin}\:\mathrm{2}\pi{t} \\ $$$${y}={a}\:\mathrm{sin}\:\mathrm{2}\pi{t}\:\Rightarrow\frac{{dy}}{{dt}}=\mathrm{2}\pi{a}\:\mathrm{cos}\:\mathrm{2}\pi{t} \\ $$$${z}={bt}\:\Rightarrow\frac{{dz}}{{dt}}={b} \\ $$$${at}\:{t}=\mathrm{1}: \\…