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Question Number 82729 by TawaTawa last updated on 23/Feb/20 Commented by Kunal12588 last updated on 23/Feb/20 $${In}\:{the}\:\bigtriangleup{ABC},\:{let}\:{G}\:{be}\:{the}\:{centroid},\:{and}\:{let}\:{I} \\ $$$${be}\:{the}\:{center}\:{of}\:{the}\:{inscribed}\:{circle}.\:{Let}\:\alpha \\ $$$${and}\:\beta\:{be}\:{the}\:{angles}\:{at}\:{the}\:{vertices}\:{A}\:{and}\:{B} \\ $$$${respectively}.\:{Suppose}\:{that}\:{the}\:{segment}\:{IG}\:\parallel\:{AB} \\ $$$${and}\:{that}\:\beta\:=\:\mathrm{2}\:{tan}^{−\mathrm{1}}…
Question Number 82721 by M±th+et£s last updated on 23/Feb/20 $${show}\:{that}\: \\ $$$$\int{xe}^{−{x}^{\mathrm{6}} } \:{sin}\left({x}^{\mathrm{3}} \right)\:{dx}=\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}\right)}{\mathrm{3}}\:\mathrm{1}{F}\mathrm{1}\left[\frac{\mathrm{5}}{\mathrm{6}};\frac{\mathrm{3}}{\mathrm{2}};\frac{−\mathrm{1}}{\mathrm{4}}\right] \\ $$ Commented by mind is power last updated on…
Question Number 17179 by tawa tawa last updated on 01/Jul/17 Answered by Tinkutara last updated on 02/Jul/17 $$\boldsymbol{\mathrm{Q}}.\mathrm{1}.\:\mathrm{Moles}\:\mathrm{of}\:\mathrm{H}_{\mathrm{2}} \:=\:\frac{\mathrm{10}}{\mathrm{2}}\:=\:\mathrm{5} \\ $$$$\mathrm{N}_{\mathrm{2}} \:=\:\frac{\mathrm{50}}{\mathrm{28}}\:\approx\:\mathrm{1}.\mathrm{786} \\ $$$$\mathrm{CO}_{\mathrm{2}} \:=\:\frac{\mathrm{40}}{\mathrm{44}}\:\approx\:\mathrm{0}.\mathrm{9}…
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Question Number 148222 by BHOOPENDRA last updated on 26/Jul/21 Answered by iloveisrael last updated on 26/Jul/21 $$\mathrm{v}_{\mathrm{3}} =\lambda\mathrm{v}_{\mathrm{1}} +\alpha\mathrm{v}_{\mathrm{2}} \\ $$$$\begin{pmatrix}{−\mathrm{3}}\\{\:\:\:\:\mathrm{4}}\\{\:\:\:\:\mathrm{7}}\end{pmatrix}\:=\:\begin{pmatrix}{\:\:\lambda}\\{\:\:\mathrm{0}}\\{−\lambda}\end{pmatrix}\:+\begin{pmatrix}{\:\:\mathrm{0}}\\{\mathrm{2}\alpha}\\{\mathrm{2}\alpha}\end{pmatrix} \\ $$$$\:\begin{cases}{\lambda=−\mathrm{3}}\\{\alpha=\mathrm{2}}\end{cases}\Rightarrow\mathrm{v}_{\mathrm{3}} =−\mathrm{3v}_{\mathrm{1}} +\mathrm{2v}_{\mathrm{2}}…
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Question Number 17129 by virus last updated on 01/Jul/17 Answered by ajfour last updated on 01/Jul/17 $$\mathrm{v}_{\mathrm{avg}} =\frac{\mid\bigtriangleup\boldsymbol{\mathrm{r}}\mid}{\bigtriangleup\mathrm{t}}=\frac{\mathrm{2rsin}\:\left(\theta/\mathrm{2}\right)}{\mathrm{r}\theta/\mathrm{v}} \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{v}\left[\frac{\mathrm{sin}\:\left(\theta/\mathrm{2}\right)}{\left(\theta/\mathrm{2}\right)}\right]\:, \\ $$$$\mathrm{a}_{\mathrm{avg}} =\frac{\mid\bigtriangleup\boldsymbol{\mathrm{v}}\mid}{\bigtriangleup\mathrm{t}}=\frac{\mathrm{2vsin}\:\left(\theta/\mathrm{2}\right)}{\mathrm{r}\theta/\mathrm{v}} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{v}^{\mathrm{2}}…
Question Number 17117 by tawa tawa last updated on 01/Jul/17 $$\mathrm{A}\:\mathrm{clock}\:\mathrm{has}\:\mathrm{a}\:\mathrm{pendulum}\:\mathrm{made}\:\mathrm{of}\:\mathrm{iron}\:\mathrm{rod}\:\mathrm{of}\:\mathrm{length}\:\mathrm{2}.\mathrm{5m}, \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{clock}\:\mathrm{keeps}\:\mathrm{accurate}\:\mathrm{time}\:\mathrm{at}\:\mathrm{0}°\mathrm{C}.\:\mathrm{By}\:\mathrm{how}\:\mathrm{much}\:\mathrm{time}\:\mathrm{will}\:\mathrm{it}\:\mathrm{be}\:\mathrm{late} \\ $$$$\mathrm{running}\:\mathrm{at}\:\mathrm{a}\:\mathrm{temperature}\:\mathrm{30}°\mathrm{C}\:\mathrm{for}\:\mathrm{1}\:\mathrm{day}.\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{linear}\:\mathrm{expansivity}\:\mathrm{of} \\ $$$$\mathrm{iron}\:\mathrm{is}\:\:\mathrm{1}.\mathrm{2}\:×\:\mathrm{10}^{−\mathrm{5}} \mathrm{per}\:\mathrm{k}. \\ $$ Answered by ajfour last updated…