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Question Number 11035 Answers: 1 Comments: 0
$${A}=\begin{bmatrix}{\mathrm{1}}&{−\mathrm{1}}\\{−\mathrm{4}}&{−\mathrm{2}}\end{bmatrix} \\ $$$${A}^{\mathrm{2016}} =.....? \\ $$
Question Number 11034 Answers: 0 Comments: 1
$${u}=\begin{bmatrix}{\mathrm{1}}&{−\mathrm{1}}&{\mathrm{2}}\end{bmatrix} \\ $$$${A}=\begin{bmatrix}{\mathrm{3}}&{\mathrm{2}}\\{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$${v}=\begin{bmatrix}{\mathrm{2}}&{−\mathrm{1}}\end{bmatrix} \\ $$$${uAv}^{{t}} =....? \\ $$
Question Number 11214 Answers: 1 Comments: 0
Question Number 11212 Answers: 1 Comments: 0
Question Number 11217 Answers: 0 Comments: 0
$$\exists{e}_{{i}} :{i}\in\mathbb{N}\:\:\:\:\:\:\:\:{e}_{{i}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{basis}\:\mathrm{vector} \\ $$$$\boldsymbol{{A}}\in\mathbb{C}^{{n}} \\ $$$$\boldsymbol{{A}}=\underset{{i}\in\mathbb{N}} {\sum}{e}_{{i}} {A}_{{i}} =\begin{pmatrix}{{A}_{\mathrm{1}} }\\{{A}_{\mathrm{2}} }\\{\vdots}\\{{A}_{{n}} }\end{pmatrix}\:\:\:=\:\mid{A}\rangle \\ $$$$\langle{B}\mid=\left(\mid{B}^{\ast} \rangle\right)^{\mathrm{T}} \\ $$$$\mid{A}\rangle\langle{B}\mid=??? \\ $$
Question Number 11215 Answers: 0 Comments: 0
Question Number 11024 Answers: 1 Comments: 0
$${P}\left({x}+\mathrm{1}\right)×{P}\left({x}−\mathrm{1}\right)=\mathrm{4}{x}^{\mathrm{2}} +\mathrm{8}{x}+{a}−\mathrm{5} \\ $$$${a}=? \\ $$
Question Number 11023 Answers: 1 Comments: 0
$$\left({x}^{\mathrm{3}} +\mathrm{6}\right)×{P}\left({x}\right)+\mathrm{6}{x}={ax}^{\mathrm{3}} +\mathrm{2}{ax}+{b}+\mathrm{3} \\ $$$$\Rightarrow{b}=? \\ $$
Question Number 11019 Answers: 2 Comments: 0
Question Number 11020 Answers: 2 Comments: 0
Question Number 11013 Answers: 0 Comments: 0
$$\mathrm{Euler}\:\mathrm{vs}.\:\mathrm{Newton} \\ $$
Question Number 11011 Answers: 2 Comments: 0
$${if}\:{tan}\left({xy}\right)={x}\:{then}\:\frac{{dy}}{{dx}}= \\ $$
Question Number 11009 Answers: 1 Comments: 0
$$\mathrm{If}\:\:{x}^{\mathrm{3}} \:=\:{y}^{\mathrm{3}} \\ $$$$\mathrm{Is}\:\mathrm{it}\:\mathrm{always}\:\mathrm{true}\:\mathrm{that}\:{x}\:=\:{y}\:? \\ $$
Question Number 11005 Answers: 1 Comments: 0
$${prove}\:{that}\:\frac{\mathrm{9}}{\mathrm{4}}\:<\:\left(\mathrm{log}\:_{\mathrm{2}} \mathrm{3}\right)^{\mathrm{2}} \:<\:\frac{\mathrm{25}}{\mathrm{9}}\:\:. \\ $$
Question Number 10995 Answers: 1 Comments: 1
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{3}>\left(\mathrm{log}_{\mathrm{2}} \mathrm{3}\right)^{\mathrm{2}} >\mathrm{2}. \\ $$
Question Number 10994 Answers: 0 Comments: 3
$$\Gamma\left(\mathrm{2407}\right)\:=\:\left(\mathrm{2406}\right)!\:=\:\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {x}^{\mathrm{2406}} \:\mathrm{dx} \\ $$$$\mathrm{How}\:\mathrm{evaluate}\:\int{e}^{−{x}} {x}^{\mathrm{2406}} \:\mathrm{dx}\:??? \\ $$
Question Number 10989 Answers: 1 Comments: 0
$${find}\:{the}\:{image}\:{of}\:{the}\:{point}\left(\mathrm{5},\mathrm{2}\right)\: \\ $$$${under}\:{a}\:{rotation}\:{of}\:\mathrm{90}°\:{clockwise} \\ $$
Question Number 10987 Answers: 1 Comments: 0
$$\mathrm{If}\: \\ $$$$\bullet\:{f}\left(\mathrm{2}{x}\:+\:\mathrm{1}\right)\:+\:{g}\left(\mathrm{3}\:−\:{x}\right)\:=\:{x} \\ $$$$\bullet\:{f}\left(\frac{\mathrm{3}{x}\:+\:\mathrm{5}}{\:{x}\:+\:\mathrm{1}}\right)\:+\:\mathrm{2}{g}\left(\frac{\mathrm{2}{x}\:+\:\mathrm{1}}{{x}\:+\:\mathrm{1}}\right)\:=\:\frac{{x}}{{x}\:+\:\mathrm{1}} \\ $$$$\mathrm{for}\:\mathrm{every}\:{x}\:\in\:\mathbb{R},\:\:{x}\:\neq\:−\mathrm{1} \\ $$$$\mathrm{Find}\:{f}\left({x}\right)\:!! \\ $$
Question Number 10981 Answers: 1 Comments: 1
Question Number 10970 Answers: 1 Comments: 0
$${x}^{{x}^{{x}^{\iddots^{\mathrm{2}} } } } \:=\:\mathrm{2} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}\:? \\ $$
Question Number 10973 Answers: 3 Comments: 1
$$\mid\mid{x}\mid+\mathrm{2x}\mid\leqslant\mathrm{3},\:\mathrm{interval}\:\mathrm{x}=...? \\ $$$$\mathrm{A}.−\mathrm{3}\leqslant{x}\leqslant\mathrm{3} \\ $$$$\mathrm{B}.\:{x}\geqslant\mathrm{0} \\ $$$$\mathrm{C}.\:{x}\leqslant\mathrm{0} \\ $$$$\mathrm{D}.\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\ $$$$\mathrm{E}.\:{x}\leqslant\mathrm{2} \\ $$
Question Number 10956 Answers: 0 Comments: 4
$${y}\:=\:\frac{{x}\:−\:\mathrm{2}}{\mathrm{2}\left({x}\:−\:\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} } \\ $$$$\mathrm{Let}\:\:{p}\:=\:{x}\:−\:\mathrm{1} \\ $$$$\Rightarrow\:{y}\:=\:\frac{{p}\:−\:\mathrm{1}}{\mathrm{2}{p}^{\mathrm{3}/\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{Is}\:\mathrm{it}\:\mathrm{true}\:\mathrm{that}\:\:\:\frac{{dy}}{{dx}}\:\:\:=\:\:\frac{{dy}}{{dp}}\:\:? \\ $$
Question Number 10955 Answers: 2 Comments: 0
$$\mathrm{If}\:\:\frac{\mathrm{cos}\:\theta}{\mathrm{1}\:−\:\mathrm{sin}\:\theta}\:=\:{a}\:\:\:\:\:\:\:\:\:\:\:{a}\:\neq\:\frac{\pi}{\mathrm{2}}\:+\:\mathrm{2}{k}\pi \\ $$$$\mathrm{So},\:\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\:=\:... \\ $$$$\left(\mathrm{A}\right)\:\:\frac{{a}}{{a}\:+\:\mathrm{1}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\:\frac{{a}\:+\:\mathrm{1}}{{a}\:−\:\mathrm{1}} \\ $$$$\left(\mathrm{B}\right)\:\:\frac{\mathrm{1}}{{a}\:+\:\mathrm{1}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{E}\right)\:\:\frac{{a}}{{a}\:−\:\mathrm{1}} \\ $$$$\left(\mathrm{C}\right)\:\:\frac{{a}\:−\:\mathrm{1}}{{a}\:+\:\mathrm{1}} \\ $$
Question Number 10948 Answers: 1 Comments: 0
$$\mathrm{If}\:\mathrm{cos}^{−\mathrm{1}} \frac{{x}}{{a}}+\mathrm{cos}^{−\mathrm{1}} \frac{{y}}{{b}}=\alpha \\ $$$${prove}\: \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\mathrm{2}{xy}}{{ab}}\mathrm{cos}\:\alpha+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{sin}^{\mathrm{2}} \alpha \\ $$
Question Number 10947 Answers: 0 Comments: 0
$$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{following} \\ $$$$\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\:=\:\frac{\mathrm{cot}\:\frac{{A}}{\mathrm{2}}+\mathrm{cot}\:\frac{{B}}{\mathrm{2}}+\mathrm{cot}\:\frac{{C}}{\mathrm{2}}}{\mathrm{cot}\:{A}+\mathrm{cot}\:{B}+\mathrm{cot}\:{C}} \\ $$
Question Number 10944 Answers: 1 Comments: 2
$$\mathrm{Find}\:\mathrm{all}\:\mathrm{ordered}\:\mathrm{pairs}\:\left(\mathrm{a},\mathrm{b}\right)\:\mathrm{so}\:\mathrm{that}\:\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}}\:\mathrm{is}\:\mathrm{an}\:\mathrm{integer}. \\ $$$$\left(\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{are}\:\mathrm{integers}\right). \\ $$
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