Question Number 74301 by ~blr237~ last updated on 21/Nov/19 $${Prove}\:{that}\:\:{S}=\left\{\left({x},{y},{z}\right)\in\mathbb{R}^{\mathrm{3}} \backslash\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={z}^{\mathrm{2}} \:\right\}\:{is}\:{a}\:{surface}\: \\ $$$${and}\:{find}\:{out}\:{if}\:{possible}\:{the}\:{tangent}\:{plan}\:{in}\:{O}\left(\mathrm{0},\mathrm{0},\mathrm{0}\right). \\ $$ Answered by mind is power last updated…
Question Number 8756 by trapti rathaur@ gmail.com last updated on 25/Oct/16 $${show}\:{that}\:{every}\:{sphere}\:{through}\:{the}\:{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{ax}+{r}^{\mathrm{2}} =\mathrm{0},{z}=\mathrm{0} \\ $$$$,{z}=\mathrm{0}\:\:\:\:\:\:\:{cuts}\:{orthogonally}\:{every}\:{sphere}\:{through}\:{the}\:{circle}\: \\ $$$${x}^{\mathrm{2}} +{z}^{\mathrm{2}} ={r}^{\mathrm{2}} ,\:{y}={o}\:. \\ $$ Terms…
Question Number 8755 by trapti rathaur@ gmail.com last updated on 25/Oct/16 $${find}\:{the}\:{equation}\:{of}\:{the}\:{sphere}\:{which}\:{touches}\:{the}\:{plane}\: \\ $$$$\mathrm{3}{x}+\mathrm{2}{y}−{z}+\mathrm{2}=\mathrm{0}\:{at}\:{the}\:{point}\:\left(\mathrm{1},−\mathrm{2},\mathrm{1}\right)\:{and}\:{cuts}\:{orthogonally}\:{the} \\ $$$${the}\:{sphere}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{6}{y}+\mathrm{4}=\mathrm{0} \\ $$ Terms of Service Privacy…
Question Number 74280 by ~blr237~ last updated on 21/Nov/19 $${Let}\:\:{consider}\:\alpha\::\:{I}\rightarrow\mathbb{R}^{\mathrm{2}} \:\:{a}\:{parametric}\:{curve}\:{defined}\:{as} \\ $$$$\forall\:{t}\in{I}\:\:\:\alpha\left({t}\right)=\left(\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{3}} −\mathrm{1}}\:,\frac{\mathrm{2}{t}}{{t}^{\mathrm{3}} −\mathrm{1}}\right)\: \\ $$$${Prove}\:{that}\:{for}\:{a},{b},{c}\in{I}\:\:\: \\ $$$$\:\:\alpha\left({a}\right),\alpha\left({b}\right),\alpha\left({c}\right)\:{are}\:{on}\:{the}\:{same}\:{lign}\:{iff}\:\:{abc}={a}+{b}+{c}+\mathrm{1} \\ $$ Commented by MJS…
Question Number 8608 by tawakalitu last updated on 17/Oct/16 $$\mathrm{if}\:\mathrm{any}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{is} \\ $$$$\left(−\:\mathrm{1},\:\mathrm{2}\right)\:\mathrm{and}\:\mathrm{any}\:\mathrm{one}\:\mathrm{side}\:\mathrm{is}\:\:\mathrm{x}\:−\:\sqrt{\mathrm{3y}}\:+\:\mathrm{5}\:=\:\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{are}\:?? \\ $$ Answered by sandy_suhendra last updated on 18/Oct/16 Answered by…
Question Number 73919 by necxxx last updated on 16/Nov/19 Commented by necxxx last updated on 16/Nov/19 $${Good}\:{day}\:{sirs}.\:{This}\:{question}\:{was}\:{formed} \\ $$$${and}\:{solved}\:{by}\:{some}\:{of}\:{us}\:{here}.\:{I}\:{really}\: \\ $$$${do}\:{not}\:{remember}\:{the}\:{question}\:{or} \\ $$$${approaches}\:{applied}.\:{Please}\:{help}. \\ $$$${Thanks}\:{in}\:{advance}.…
Question Number 139422 by 676597498 last updated on 26/Apr/21 $${C}_{{n}} ^{\:\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{8}}{P}_{{n}\:} ^{\:\mathrm{8}} \\ $$$${find}\:{n} \\ $$ Commented by mohammad17 last updated on 27/Apr/21 $${can}\:{not}\:{find}\:{value}\:{of}\:{n}\:{in}\:{this}\:{form}…
Find-the-equation-of-the-perpendicular-bisector-of-the-line-joining-the-points-5-4-to-the-point-9-3-
Question Number 8243 by lepan last updated on 04/Oct/16 $${Find}\:{the}\:{equation}\:{of}\:{the}\:{perpendicular}\:{bisector}\:{of}\:{the}\:{line}\:{joining}\:{the}\:{points}\:\left(−\mathrm{5},\mathrm{4}\right)\:{to}\:{the}\:{point}\:\left(\mathrm{9},−\mathrm{3}\right) \\ $$$$ \\ $$ Answered by sandy_suhendra last updated on 04/Oct/16 $$\mathrm{let}\:\mathrm{A}\left(−\mathrm{5},\mathrm{4}\right)\:\mathrm{and}\:\mathrm{B}\left(\mathrm{9},−\mathrm{3}\right) \\ $$$$\mathrm{P}\:\mathrm{is}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{AB}\:\mathrm{so}\:\mathrm{P}\left(\frac{−\mathrm{5}+\mathrm{9}}{\mathrm{2}}\:,\:\frac{\mathrm{4}−\mathrm{3}}{\mathrm{2}}\:\right)=\mathrm{P}\left(\mathrm{2},\frac{\mathrm{1}}{\mathrm{2}}\right) \\…
Question Number 8150 by trapti rathaur@ gmail.com last updated on 02/Oct/16 $${show}\:{that}\:{the}\:{plane}\:\:\:{x}+\mathrm{2}{y}−\mathrm{3}{z}+{d}=\mathrm{0}\:\:{is}\:{perpendiculr}\:{to}\:{each}\:{of} \\ $$$${the}\:{plane}\:{is}\:\:\:\mathrm{2}{x}+\mathrm{5}{y}+\mathrm{4}{z}+\mathrm{1}=\mathrm{0}\:\:{and}\:\:\:\mathrm{4}{x}+\mathrm{7}{y}+\mathrm{3}{z}+\mathrm{2}=\mathrm{0}\:.\: \\ $$ Commented by 123456 last updated on 03/Oct/16 $$\alpha:{x}+\mathrm{2}{y}−\mathrm{3}{z}+{d}=\mathrm{0} \\…
Question Number 73671 by Tanmay chaudhury last updated on 14/Nov/19 Terms of Service Privacy Policy Contact: info@tinkutara.com