Question Number 213045 by MrGaster last updated on 29/Oct/24 $$\overset{\rightharpoonup} {{a}},\overset{\rightharpoonup} {{b}ls}\:{the}\:{unit}\:{vector},\mid\overset{\rightharpoonup} {{a}}+\overset{\rightharpoonup} {{b}}\mid=\mathrm{1} \\ $$$${ask}\:\mid\overset{\rightharpoonup} {{a}}−\overset{\rightharpoonup} {{b}}\mid? \\ $$ Answered by mr W last…
Question Number 212616 by im13 last updated on 19/Oct/24 $$ \\ $$$$ \\ $$$${if}\:\:\mathrm{f}\left(\mathrm{x}\right)=\:^{{a}} {x}={x}^{{x}^{{x}^{\iddots^{{x}} } } } \left({there}\:{are}\:{a}\:{x}'{s};{a}\in\mathbb{N}\:{and}\:{a}\neq\mathrm{0}\right) \\ $$$$\int{f}\left({x}\right){dx}=? \\ $$ Commented by…
Question Number 210554 by Batmath last updated on 12/Aug/24 Answered by MrGaster last updated on 02/Nov/24 $$\int_{\mathrm{0}} ^{\infty} \left[\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({px}\right)\right]{x}^{\mathrm{2}{q}−\mathrm{1}} {dx}=\int_{\mathrm{0}} ^{\infty} \left[\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left({tpx}\right)}{{t}}{dt}\right]{x}^{\mathrm{2}{q}−\mathrm{1}} {dx}…
Question Number 209888 by estherwealth last updated on 24/Jul/24 $${n}_{\mathrm{0}} =\frac{{Z}^{\mathrm{2}} .{p}.\left(\mathrm{1}−{p}\right)}{{C}^{\mathrm{2}{m}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 207712 by efronzo1 last updated on 24/May/24 Answered by Frix last updated on 24/May/24 $$\mathrm{For}\:{x}={q}\:\mathrm{and}\:{A}=\begin{pmatrix}{{p}}\\{\mathrm{0}}\end{pmatrix}\:\mathrm{the}\:\mathrm{area}\:\mathrm{is}\:\left({q}−{p}\right)\sqrt{{p}} \\ $$$$\frac{{d}\left[\left({q}−{p}\right)\sqrt{{p}}\right]}{{dp}}=\mathrm{0} \\ $$$$\frac{{q}−\mathrm{3}{p}}{\mathrm{2}\sqrt{{p}}}=\mathrm{0}\:\Rightarrow\:{p}=\frac{{q}}{\mathrm{3}} \\ $$$$\mathrm{Max}\:\mathrm{area}\:=\frac{\mathrm{2}\sqrt{\mathrm{3}}{q}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{9}} \\…
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Question Number 204273 by mustafazaheen last updated on 10/Feb/24 $$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{ln}\left(\frac{\mathrm{2}}{\mathrm{4}}\right)} } \\ $$$$\mathrm{Domain}\:\mathrm{f}\left(\mathrm{x}\right)\:=? \\ $$ Answered by Mathspace last updated on 11/Feb/24 $${f}\left({x}\right)=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} }=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{−{ln}\mathrm{2}} }…
Question Number 203900 by necx122 last updated on 01/Feb/24 $${Let}\:{A}\:\in\:{R}^{{N}×{N}} \:{be}\:{a}\:{symmetric}\:{positive} \\ $$$${definite}\:{matrix}\:{and}\:{b}\:\in\:{R}^{{N}} \:{a}\:{vector}. \\ $$$${If}\:{x}\:\in\:{R}^{{N}} ,\:{evaluate}\:{the}\:{integral} \\ $$$${Z}\left({A},{b}\right)\:=\:\int{e}^{−\frac{\mathrm{1}}{\mathrm{2}}{x}^{{T}} {Ax}\:+\:{b}^{{T}} {x}} {dx}\:{as}\:{a}\:{function} \\ $$$${of}\:{A}\:{and}\:{b}. \\…
Question Number 201947 by Mingma last updated on 16/Dec/23 Answered by Frix last updated on 18/Dec/23 $$\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\:\mathrm{cos}\:\left(\pi−\alpha−\beta\right)\:+ \\ $$$$\:\:\:\:\:+\mathrm{sin}\:\beta\:\mathrm{sin}\:\left(\pi−\alpha−\beta\right)\:\mathrm{cos}\:\alpha\:+ \\ $$$$\:\:\:\:\:+\mathrm{sin}\:\left(\pi−\alpha−\beta\right)\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta\:= \\ $$$$=\frac{\mathrm{3}−\left(\mathrm{cos}\:\mathrm{2}\alpha\:+\mathrm{cos}\:\mathrm{2}\beta\:+\mathrm{cos}\:\mathrm{2}\left(\alpha+\beta\right)\right.}{\mathrm{4}}= \\ $$$$\:\:\:\:\:\left[\mathrm{Let}\:\alpha={u}−{v}\wedge\beta={u}+{v}\right]…
Question Number 201144 by sts313 last updated on 30/Nov/23 $$\left(\frac{\mathrm{14}}{\mathrm{15}}\right)^{\mathrm{6}} ×\left(\frac{\mathrm{45}}{\mathrm{28}}\right)^{\mathrm{6}} = \\ $$ Answered by mathlove last updated on 01/Dec/23 $$\left(\frac{\mathrm{14}}{\mathrm{15}}\right)^{\mathrm{6}} ×\left(\frac{\mathrm{3}×\mathrm{15}}{\mathrm{2}×\mathrm{14}}\right)^{\mathrm{6}} =\frac{\cancel{\mathrm{14}^{\mathrm{6}} }}{\cancel{\mathrm{15}^{\mathrm{6}}…