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Category: Vector Calculus

11-If-5-2-236-and-10-3-162-then-the-value-of-15-10-20-40-5-80-is-a-5-398-b-4-398-c-3-398-d-6-398-12-If-x-3-1-3-then-x

Question Number 222943 by Rajakumarselvi last updated on 11/Jul/25 $$\mathrm{11}.\:\mathrm{If}\:\sqrt{\mathrm{5}}\:=\:\mathrm{2}.\mathrm{236}\:\mathrm{and}\:\sqrt{\mathrm{10}}\:=\:\mathrm{3}.\mathrm{162}\:\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{15}}{\:\sqrt{\mathrm{10}}+\sqrt{\mathrm{20}}+\sqrt{\mathrm{40}}−\sqrt{\mathrm{5}}−\sqrt{\mathrm{80}}}\:\mathrm{is} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{5}.\mathrm{398}\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{4}.\mathrm{398}\:\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{3}.\mathrm{398}\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{6}.\mathrm{398} \\ $$$$\mathrm{12}.\:\mathrm{If}\:\mathrm{x}=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{3}}\:\:\mathrm{then}\:\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} \:}=? \\ $$$$\left(\mathrm{a}\right)\:\frac{\mathrm{28}\sqrt{\mathrm{3}}\:+\mathrm{15}}{\mathrm{8}}\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\frac{\mathrm{28}\sqrt{\mathrm{3}}−\mathrm{15}}{\mathrm{8}}\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\frac{\mathrm{27}\sqrt{\mathrm{3}}−\mathrm{35}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\frac{\mathrm{27}\sqrt{\mathrm{3}}+\mathrm{35}}{\mathrm{4}} \\ $$$$\mathrm{13}.\:\mathrm{Simplify}\:\:\sqrt[{\mathrm{5}}]{\mathrm{x}^{\mathrm{4}} \sqrt[{\mathrm{4}}]{\mathrm{x}^{\mathrm{3}_{\:} } \sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}}}}} \\…

lim-x-3-x-1-3-x-0-

Question Number 216698 by sniper237 last updated on 16/Feb/25 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:^{\mathrm{3}} \sqrt{{x}+\mathrm{1}}\:−^{\mathrm{3}} \sqrt{{x}}\:\overset{?} {=}\:\mathrm{0} \\ $$ Answered by MrGaster last updated on 16/Feb/25 $$\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}−\:^{\mathrm{3}} \sqrt{{x}}=\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}}…

Question-210554

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-207712

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}} \\…

f-x-1-x-1-ln-2-4-Domain-f-x-

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}} }…

Let-A-R-N-N-be-a-symmetric-positive-definite-matrix-and-b-R-N-a-vector-If-x-R-N-evaluate-the-integral-Z-A-b-e-1-2-x-T-Ax-b-T-x-dx-as-a-function-of-A-and-b-

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}. \\…