Question Number 71572 by TawaTawa last updated on 17/Oct/19 Commented by TawaTawa last updated on 17/Oct/19 $$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{How}\:? \\ $$ Answered by Kunal12588 last updated on…
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Question Number 71538 by mr W last updated on 16/Oct/19 Commented by mr W last updated on 16/Oct/19 $${find}\:{the}\:{position}\:{of}\:{center}\:{of}\:{mass} \\ $$$${of}\:{the}\:{parabolic}\:{arc} \\ $$ Answered by…
Question Number 5998 by Kasih last updated on 09/Jun/16 $${determine}\:{equation}\:{of}\:{circle}\:{that}\:{offensive}\:{the} \\ $$$${both}\:{of}\:{coordinate}\:{and}\:{through}\:\left(\mathrm{2},−\mathrm{1}\right) \\ $$ Commented by Rasheed Soomro last updated on 09/Jun/16 $${What}\:{do}\:{you}\:{mean}\:{by}\:“{offensive}''? \\ $$…
Question Number 5982 by Kasih last updated on 08/Jun/16 $${the}\:{center}\:{of}\:{circle}\:{in}\:\mathrm{2}{x}+{y}−\mathrm{11}=\mathrm{0} \\ $$$${determine}\:{the}\:{equation}\:{of}\:{circle}\:{that} \\ $$$${passing}\:{through}\:\left(−\mathrm{1},\mathrm{3}\right),\left(\mathrm{7},−\mathrm{1}\right) \\ $$ Commented by Rasheed Soomro last updated on 08/Jun/16 $${the}\:{center}\:{of}\:{circle}\:{in}\:\mathrm{2}{x}+{y}−\mathrm{11}=\mathrm{0}…
Question Number 71517 by TawaTawa last updated on 16/Oct/19 Answered by mind is power last updated on 16/Oct/19 $$\mathrm{AD}=\mathrm{x} \\ $$$$\frac{\mathrm{x}}{\mathrm{sin}\left(\theta\right)}=\frac{\mathrm{DC}}{\mathrm{sin}\left(\theta\right)}\Rightarrow\mathrm{1}=\frac{\mathrm{x}}{\mathrm{DC}} \\ $$$$\frac{\mathrm{x}}{\mathrm{sin}\left(\mathrm{3}\theta\right)}=\frac{\mathrm{DC}}{\mathrm{sin}\left(\mathrm{10}\theta\right)}\Rightarrow\frac{\mathrm{x}}{\mathrm{DC}}=\frac{\mathrm{sin}\left(\mathrm{3}\theta\right)}{\mathrm{sin}\left(\mathrm{10}\theta\right)} \\ $$$$…
Question Number 136990 by EDWIN88 last updated on 28/Mar/21 $$ \\ $$Given a cube ABCD.EFGH where point Z is the midpoint of AE, point R…
Question Number 71428 by TawaTawa last updated on 15/Oct/19 Answered by ajfour last updated on 15/Oct/19 Commented by ajfour last updated on 15/Oct/19 $${P}\:\left[{r}\mathrm{cos}\:\left(\theta−\phi\right),\:{r}\mathrm{sin}\:\left(\theta−\phi\right)\right] \\…
Question Number 71423 by ajfour last updated on 15/Oct/19 Commented by ajfour last updated on 15/Oct/19 $${MjS}\:{Sir}… \\ $$ Commented by mr W last updated…
Question Number 71410 by TawaTawa last updated on 15/Oct/19 Answered by MJS last updated on 15/Oct/19 $${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{{p}}\\{\mathrm{0}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:\:{D}=\begin{pmatrix}{\mathrm{0}}\\{{q}}\end{pmatrix} \\ $$$${DC}:\:{y}={q} \\ $$$${AC}:\:{y}=\frac{{q}}{{p}}{x} \\ $$$${BE}:\:{y}=−\frac{{p}}{{q}}{x}+\frac{{p}^{\mathrm{2}} }{{q}} \\…