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Question Number 220456 Answers: 0 Comments: 0

a, b, c, d ≥ 1 ; a + b + c = d show that; ab + bc + ca + (1/a) + (1/b) + (1/c) ≥ 2d − 3 + (9/d)

$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:{a},\:{b},\:{c},\:{d}\:\geqslant\:\mathrm{1}\:\:\:\:\:\:;\:\:\:{a}\:+\:{b}\:+\:{c}\:=\:{d} \\ $$$$\:\:\:\:\:\:\:\:{show}\:{that};\: \\ $$$$\:\:\:{ab}\:+\:{bc}\:+\:{ca}\:+\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:\geqslant\:\mathrm{2}{d}\:−\:\mathrm{3}\:+\:\frac{\mathrm{9}}{{d}}\:\:\:\: \\ $$

Question Number 220454 Answers: 0 Comments: 0

In the CO_2 molecule, each oxygen atom forms a double bond with central carbon atom. Given that the Bohr radius (a_o ) is approximately 0.529 A^o and the experimental C=O bond length is about 1.16 A^(o ) , calculate : a) The approximate region (in the terms of Bohr radii) where the shared electrons are most likely to be found between C and O. b) Compare the calculated bond region to the sum of the covalent radii of carbon and oxygen, and comment on the effect of π bonding on the contraction of bond length.

$$\:\mathrm{In}\:\mathrm{the}\:{CO}_{\mathrm{2}} \:\mathrm{molecule},\:\mathrm{each}\:\mathrm{oxygen}\:\mathrm{atom}\:\mathrm{forms}\:\mathrm{a}\:\mathrm{double}\:\mathrm{bond}\:\mathrm{with}\:\mathrm{central}\:\mathrm{carbon}\:\mathrm{atom}. \\ $$$$\:\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{Bohr}\:\mathrm{radius}\:\left({a}_{{o}} \right)\:\mathrm{is}\:\mathrm{approximately}\:\mathrm{0}.\mathrm{529}\:\overset{\mathrm{o}} {\mathrm{A}}\:\mathrm{and}\:\mathrm{the}\:\mathrm{experimental}\:{C}={O}\: \\ $$$$\:\:\mathrm{bond}\:\mathrm{length}\:\mathrm{is}\:\mathrm{about}\:\mathrm{1}.\mathrm{16}\:\overset{\mathrm{o}\:} {\mathrm{A}},\:\mathrm{calculate}\:: \\ $$$$\: \\ $$$$\left.\:\mathrm{a}\right)\:\mathrm{The}\:\mathrm{approximate}\:\mathrm{region}\:\left(\mathrm{in}\:\mathrm{the}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{Bohr}\:\mathrm{radii}\right)\:\mathrm{where}\:\mathrm{the}\:\mathrm{shared}\:\mathrm{electrons}\:\mathrm{are} \\ $$$$\:\mathrm{most}\:\mathrm{likely}\:\mathrm{to}\:\mathrm{be}\:\mathrm{found}\:\mathrm{between}\:{C}\:\mathrm{and}\:{O}. \\ $$$$\left.\:\mathrm{b}\right)\:\mathrm{Compare}\:\mathrm{the}\:\mathrm{calculated}\:\mathrm{bond}\:\mathrm{region}\:\mathrm{to}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{covalent}\:\mathrm{radii}\:\mathrm{of}\:\mathrm{carbon}\:\mathrm{and} \\ $$$$\:\mathrm{oxygen},\:\mathrm{and}\:\mathrm{comment}\:\mathrm{on}\:\mathrm{the}\:\mathrm{effect}\:\mathrm{of}\:\pi\:\mathrm{bonding}\:\mathrm{on}\:\mathrm{the}\:\mathrm{contraction}\:\mathrm{of}\:\mathrm{bond}\:\mathrm{length}. \\ $$

Question Number 220447 Answers: 0 Comments: 1

Question Number 220499 Answers: 3 Comments: 1

Question Number 220405 Answers: 2 Comments: 0

9^x^2 −3^(x+1) =0

$$\mathrm{9}^{{x}^{\mathrm{2}} } −\mathrm{3}^{{x}+\mathrm{1}} =\mathrm{0} \\ $$

Question Number 220403 Answers: 1 Comments: 0

Question Number 220493 Answers: 0 Comments: 1

Let ABC be a triangle such CosA+cosB+cosC=(√2) SinA+sinB+sinC=1+(√2) Find A,B,C

$${Let}\:{ABC}\:{be}\:{a}\:{triangle}\:{such} \\ $$$${CosA}+{cosB}+{cosC}=\sqrt{\mathrm{2}} \\ $$$${SinA}+{sinB}+{sinC}=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$${Find}\:{A},{B},{C} \\ $$

Question Number 220502 Answers: 1 Comments: 0

each J_ν (z),Y_ν (z) are linear independent....?? W_(Ronskian) {J_ν ^ (z),Y_ν (z)}= determinant (((J_ν (z)),( Y_ν (z))),((J_ν ′(z)),(Y_ν ′(z)))) =J_ν ^((1)) (z)Y_ν (z)−J_ν (z)Y_ν ^((1)) (z) J_ν ^((1)) (z)Y_ν (z)=J_(ν−1) (z)Y_ν (z)−(ν/z)J_ν (z)Y_ν (z) J_ν (z)Y_ν ^((1)) (z)=Y_(ν−1) (z)J_ν (z)−(ν/z)J_ν (z)Y_ν (z) J_(ν−1) (z)Y_ν (z)−J_ν (z)Y_(ν−1) (z).... .....damn..... Result is (2/(πz)) ......

$$\mathrm{each}\:{J}_{\nu} \left({z}\right),{Y}_{\nu} \left({z}\right)\:\mathrm{are}\:\mathrm{linear}\:\mathrm{independent}....?? \\ $$$${W}_{\mathrm{Ronskian}} \left\{{J}_{\nu} ^{\:} \left({z}\right),{Y}_{\nu} \left({z}\right)\right\}=\begin{vmatrix}{{J}_{\nu} \left({z}\right)}&{\:{Y}_{\nu} \left({z}\right)}\\{{J}_{\nu} '\left({z}\right)}&{{Y}_{\nu} '\left({z}\right)}\end{vmatrix} \\ $$$$={J}_{\nu} ^{\left(\mathrm{1}\right)} \left({z}\right){Y}_{\nu} \left({z}\right)−{J}_{\nu} \left({z}\right){Y}_{\nu} ^{\left(\mathrm{1}\right)} \left({z}\right) \\ $$$${J}_{\nu} ^{\left(\mathrm{1}\right)} \left({z}\right){Y}_{\nu} \left({z}\right)={J}_{\nu−\mathrm{1}} \left({z}\right){Y}_{\nu} \left({z}\right)−\frac{\nu}{{z}}{J}_{\nu} \left({z}\right){Y}_{\nu} \left({z}\right) \\ $$$${J}_{\nu} \left({z}\right){Y}_{\nu} ^{\left(\mathrm{1}\right)} \left({z}\right)={Y}_{\nu−\mathrm{1}} \left({z}\right){J}_{\nu} \left({z}\right)−\frac{\nu}{{z}}{J}_{\nu} \left({z}\right){Y}_{\nu} \left({z}\right) \\ $$$${J}_{\nu−\mathrm{1}} \left({z}\right){Y}_{\nu} \left({z}\right)−{J}_{\nu} \left({z}\right){Y}_{\nu−\mathrm{1}} \left({z}\right).... \\ $$$$.....\mathrm{damn}..... \\ $$$$\mathrm{Result}\:\mathrm{is}\:\frac{\mathrm{2}}{\pi{z}}\:...... \\ $$

Question Number 220495 Answers: 0 Comments: 6

Question Number 220396 Answers: 0 Comments: 2

Question Number 220395 Answers: 2 Comments: 0

Find: Ω =Σ_(x=1) ^∞ Σ_(y=1) ^∞ (1/(x^2 y^3 (x^2 + 1)(y + 2))) = ?

$$\mathrm{Find}:\:\:\:\Omega\:=\underset{\boldsymbol{\mathrm{x}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\underset{\boldsymbol{\mathrm{y}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:\mathrm{y}^{\mathrm{3}} \:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left(\mathrm{y}\:+\:\mathrm{2}\right)}\:=\:? \\ $$

Question Number 220393 Answers: 1 Comments: 1

Question Number 220391 Answers: 3 Comments: 0

Question Number 220390 Answers: 1 Comments: 0

sinθ + sin(π + θ) + sin(2π + θ) + ... + sin(nπ + θ) = ? when n is an odd integer.

$$\mathrm{sin}\theta\:+\:\mathrm{sin}\left(\pi\:+\:\theta\right)\:+\:\mathrm{sin}\left(\mathrm{2}\pi\:+\:\theta\right)\:+\:...\: \\ $$$$+\:\mathrm{sin}\left({n}\pi\:+\:\theta\right)\:=\:?\:\mathrm{when}\:{n}\:\mathrm{is}\:\mathrm{an}\:\mathrm{odd} \\ $$$$\mathrm{integer}. \\ $$

Question Number 220388 Answers: 0 Comments: 1

Question Number 220380 Answers: 3 Comments: 0

lim_(n→∞) tan[(π/4)+(1/n)]^n =?

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}tan}\left[\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right]^{{n}} =? \\ $$

Question Number 220378 Answers: 0 Comments: 0

Question Number 220375 Answers: 1 Comments: 2

Question Number 220366 Answers: 2 Comments: 0

solve the system of equation using gaussian elimination method x+2y+3z=10 2x−3y+z=1 3x+y−2z=9

$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{system}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{equation}} \\ $$$$\boldsymbol{\mathrm{using}}\:\boldsymbol{\mathrm{gaussian}}\:\boldsymbol{\mathrm{elimination}}\:\boldsymbol{\mathrm{method}} \\ $$$$\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{y}}+\mathrm{3}\boldsymbol{\mathrm{z}}=\mathrm{10} \\ $$$$\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{3}\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}=\mathrm{1} \\ $$$$\mathrm{3}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}−\mathrm{2}\boldsymbol{\mathrm{z}}=\mathrm{9} \\ $$

Question Number 220365 Answers: 3 Comments: 0

Question Number 220362 Answers: 0 Comments: 0

Question Number 220377 Answers: 1 Comments: 0

Prove equation ∫_0 ^( ∞) f(u)g(u)e^(−uρ) du=(1/(2πi)) ∫_(−∞i+𝛄) ^( +∞i+𝛄) F(u)G(u−ρ)du F(u)=∫_0 ^( ∞) f(t)e^(−ut) dt G(u)=∫_0 ^( ∞) g(t)e^(−ut) dt

$$\mathrm{Prove}\:\mathrm{equation} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({u}\right)\mathrm{g}\left({u}\right){e}^{−{u}\rho} \mathrm{d}{u}=\frac{\mathrm{1}}{\mathrm{2}\pi\boldsymbol{{i}}}\:\int_{−\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} ^{\:+\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} \:\:{F}\left({u}\right){G}\left({u}−\rho\right)\mathrm{d}{u} \\ $$$${F}\left({u}\right)=\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({t}\right){e}^{−{ut}} \mathrm{d}{t} \\ $$$${G}\left({u}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{g}\left({t}\right){e}^{−{ut}} \mathrm{d}{t} \\ $$

Question Number 220340 Answers: 2 Comments: 0

Question Number 220320 Answers: 1 Comments: 3

Question Number 220353 Answers: 1 Comments: 3

Question Number 220307 Answers: 1 Comments: 0

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