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Question Number 170028 Answers: 0 Comments: 1

verify that ⟨p_x ∣x^ ∣Ψ⟩=ih(d/dp_x )<p_x ∣Ψ>

$${verify}\:{that}\:\:\langle{p}_{{x}} \mid\hat {{x}}\mid\Psi\rangle={ih}\frac{{d}}{{dp}_{{x}} }<{p}_{{x}} \mid\Psi> \\ $$

Question Number 170025 Answers: 1 Comments: 0

Question Number 170022 Answers: 0 Comments: 0

2. [((1 2)),((2 −1)) ] Soln: [((1 2)),((2 −1)) ]= [((1 0)),((0 1)) ].A ⇒ [((1 2)),((0 −5)) ]= [(( 1 0)),((−2 1)) ].A [R_2 →R_2 −2R_1 ] ⇒ [((1 2)),((0 1)) ]= [(( 1 0)),(((2/5) −(1/5))) ].A [R_2 →(−(1/5))R_2 ] ⇒ [((1 0)),((0 1)) ]= [(((1/5) (2/5))),(((2/5) −(1/5))) ].A [R_1 →R_1 −2R_2 ] ∴A^(−1) =(1/5) [((1 2)),((2 −1)) ]

Question Number 170020 Answers: 0 Comments: 0

log _((x+(1/4))) (2) < log _x (4)

$$\:\:\:\:\mathrm{log}\:_{\left({x}+\frac{\mathrm{1}}{\mathrm{4}}\right)} \left(\mathrm{2}\right)\:<\:\mathrm{log}\:_{{x}} \left(\mathrm{4}\right) \\ $$

Question Number 170018 Answers: 1 Comments: 0

lim_(x→(π/2)) (1+sin (π−2x))^(5/(sin (x−(π/2)))) .ln (4.((cos (π−2x)−1)/(((π/2)−x)^2 )))=?

$$\:\:\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{sin}\:\left(\pi−\mathrm{2}{x}\right)\right)^{\frac{\mathrm{5}}{\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{2}}\right)}} .\mathrm{ln}\:\left(\mathrm{4}.\frac{\mathrm{cos}\:\left(\pi−\mathrm{2}{x}\right)−\mathrm{1}}{\left(\frac{\pi}{\mathrm{2}}−{x}\right)^{\mathrm{2}} }\right)=? \\ $$

Question Number 170015 Answers: 1 Comments: 0

verify that the force F=xyi+xyj+yzk is conservative

$${verify}\:{that}\:{the}\:{force}\: \\ $$$$\:{F}={xyi}+{xyj}+{yzk}\:{is} \\ $$$$\:{conservative} \\ $$

Question Number 170014 Answers: 1 Comments: 0

Question Number 170003 Answers: 1 Comments: 1

prove that Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 )=(π^2 /(12))

$${prove}\:{that} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$

Question Number 169989 Answers: 0 Comments: 0

Prove or disprove: For any extension E of a field F, F(u)=F[u] ∀ u∈E. Where F(u) is a smallest subfield of E containing F and u and F[u]={f(u)∣f(x)∈F[x]}, F[x] is a polynomial ring over F.

$$ \\ $$$$\mathrm{Prove}\:\mathrm{or}\:\mathrm{disprove}: \\ $$$$\mathrm{For}\:\mathrm{any}\:\mathrm{extension}\:{E}\:\mathrm{of}\:\mathrm{a}\:\mathrm{field}\:{F}, \\ $$$${F}\left({u}\right)={F}\left[{u}\right]\:\:\:\:\forall\:{u}\in{E}. \\ $$$$\mathrm{Where}\:{F}\left({u}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{smallest}\:\mathrm{subfield} \\ $$$$\mathrm{of}\:{E}\:\mathrm{containing}\:{F}\:\mathrm{and}\:{u}\:\mathrm{and} \\ $$$${F}\left[{u}\right]=\left\{{f}\left({u}\right)\mid{f}\left({x}\right)\in{F}\left[{x}\right]\right\},\:{F}\left[{x}\right]\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{polynomial}\:\mathrm{ring}\:\mathrm{over}\:{F}. \\ $$

Question Number 169987 Answers: 1 Comments: 0