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Question Number 220583 Answers: 0 Comments: 0

Question Number 220581 Answers: 1 Comments: 0

An object is thrown vertically upward from the top of a building 20m high. If the object passes the point it was thrown 4 seconds on it way down, find the. (a) Velocity at which the object was thrown. (b) time taken when the object is 10m above the level it was thrown. (c) time taken when the object is 10m below the level it was thrown. (d) Velocity with which the object hit the ground. [Take g = 10m/s²]

Question Number 220580 Answers: 0 Comments: 0

∫_(−∞i) ^(+∞i) ((atan(w))/w)e^(3iw) dw=??

$$\int_{−\infty\boldsymbol{{i}}} ^{+\infty\boldsymbol{{i}}} \:\frac{\mathrm{atan}\left({w}\right)}{{w}}{e}^{\mathrm{3}\boldsymbol{{i}}{w}} \mathrm{d}{w}=?? \\ $$

Question Number 220577 Answers: 1 Comments: 0

Calculate the perimeter of a rectangle whose area is represented by the polynomial 25x^2 −35x+12(Given that the length and breath are not constant)

$${Calculate}\:{the}\:{perimeter}\:{of}\:{a}\:{rectangle} \\ $$$${whose}\:{area}\:{is}\:{represented}\:{by}\:{the}\:{polynomial} \\ $$$$\mathrm{25}{x}^{\mathrm{2}} −\mathrm{35}{x}+\mathrm{12}\left({Given}\:{that}\:{the}\:{length}\:{and}\:{breath}\:{are}\:{not}\:{constant}\right) \\ $$

Question Number 220579 Answers: 1 Comments: 0

A=7×19×31×43×.....upto 29 terms find the last four digits of A.

$$\:{A}=\mathrm{7}×\mathrm{19}×\mathrm{31}×\mathrm{43}×.....{upto}\:\mathrm{29}\:{terms} \\ $$$$\:{find}\:{the}\:{last}\:{four}\:{digits}\:{of}\:{A}. \\ $$

Question Number 220563 Answers: 1 Comments: 0

Calculate the exact value of : I=∫_0 ^4 e^(−x^2 ) dx

$${Calculate}\:{the}\:{exact}\:{value}\:{of}\:: \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{4}} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$

Question Number 220562 Answers: 1 Comments: 0

∫_0 ^( ∞) ((ln(z^2 +1))/(z^2 +1)) dz=I I(t)=∫_0 ^( ∞) ((ln(z^2 +1))/(z^2 +1))e^(−zt) dz I′(t)=−∫_0 ^( ∞) ((z∙ln(z^2 +1))/(z^2 +1))e^(−zt) dz I′(t)=−∫_0 ^( ∞) ((z^2 ln(z^2 +1)+ln(z^2 +1)−ln(z^2 +1))/(z(z^2 +1)))e^(−zt) dz I′(t)=−∫_0 ^( ∞) ((ln(z^2 +1))/z)e^(−zt ) dz+∫_0 ^( ∞) ((ln(z^2 +1))/(z(z^2 +1)))e^(−zt) dz I′′(t)=∫_0 ^∞ ln(z^2 +1)e^(−zt) dz−∫_0 ^( ∞) ((ln(z^2 +1))/(z^2 +1))e^(−zt) dz I^((2)) (t)+I(t)=∫_0 ^( ∞) ln(z^2 +1)e^(−zt) dz........(A) ∫_0 ^( ∞) ln(z^2 +1)e^(−zt) dz cos^2 (α)+sin^2 (α)=1 → 1+tan^2 (α)=sec^2 (α) z=tan(u) → u=tan^(−1) (z) (dz/du)=sec^2 (u) → dz=sec^2 (u)du ∫_0 ^( ∞) ln(z^2 +1)e^(−st) dz=∫_0 ^( (π/2)) sec^2 (u)∙ln(sec^2 (u))e^(−s∙tan(u)) du i can′t Calculate anymore.... can′t solve ODE....(A) that Equation(A) Seems to Weird cus Feynman trick ...... How can i solve ∫_0 ^( ∞) ((ln(z^2 +1))/(z^2 +1)) dz or....Should i do Complex integral.. ∮_( C) f(z) dz.....??

Question Number 220560 Answers: 3 Comments: 0

sinα=0.8 ⇒ ((BE)/(EF))=?

$${sin}\alpha=\mathrm{0}.\mathrm{8}\:\Rightarrow\:\frac{{BE}}{{EF}}=? \\ $$

Question Number 220544 Answers: 1 Comments: 0

Prove that inequality; ∫_( 0) ^( 1) ((ln(1 + x^2 ))/(1 + x^2 )) dx < ∫_( 0) ^( 1) ((x ln(1 + x^2 ))/(1 + x^2 )) dx + (1/3)

$$ \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{inequality}}; \\ $$$$\:\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:{dx}\:<\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{x}\:\mathrm{ln}\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)}{\mathrm{1}\:+\:{x}^{\mathrm{2}} \:}\:{dx}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 220546 Answers: 0 Comments: 3

Question Number 220540 Answers: 2 Comments: 0

Find: 𝛀 = Σ_(n=1) ^∞ (((−1)^(n+1) )/(n^3 ∙(n + 1)^3 ∙(2n + 1)^2 )) = ?

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

Question Number 220538 Answers: 0 Comments: 0

Question Number 220526 Answers: 2 Comments: 0

Question Number 220519 Answers: 0 Comments: 1

Solve in R ((15)/(x^2 - 3x + 4)) + (7/(x^2 + 7x)) + ((10)/(x^2 + 4x - 21)) + 1 = 0

$$\mathrm{Solve}\:\mathrm{in}\:\:\:\mathbb{R} \\ $$$$\frac{\mathrm{15}}{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{3x}\:+\:\mathrm{4}}\:\:+\:\:\frac{\mathrm{7}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{7x}}\:\:+\:\:\frac{\mathrm{10}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4x}\:-\:\mathrm{21}}\:\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$

Question Number 220511 Answers: 3 Comments: 0

AB=2CE & DE=2(√3)+4 CE ⊥AB & AD⊥BC & AB=AC & EF ⊥BC BF=?

$${AB}=\mathrm{2}{CE}\:\:\&\:\:{DE}=\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4}\:\:\: \\ $$$${CE}\:\bot{AB}\:\:\:\&\:\:\:{AD}\bot{BC}\:\:\&\:\:{AB}={AC}\:\:\&\:{EF}\:\bot{BC}\: \\ $$$${BF}=? \\ $$$$ \\ $$

Question Number 220486 Answers: 1 Comments: 2

z ∈ C and λ > 0 Then prove that: ∣z + 2λ∣ + ∣z + λ∣ ≥ ∣z + ((3λ − λ(√3)i)/2)∣

$$\mathrm{z}\:\in\:\mathbb{C}\:\:\:\mathrm{and}\:\:\:\lambda\:>\:\mathrm{0} \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mid\mathrm{z}\:+\:\mathrm{2}\lambda\mid\:+\:\mid\mathrm{z}\:+\:\lambda\mid\:\geqslant\:\mid\mathrm{z}\:+\:\frac{\mathrm{3}\lambda\:−\:\lambda\sqrt{\mathrm{3}}\mathrm{i}}{\mathrm{2}}\mid \\ $$

Question Number 220480 Answers: 1 Comments: 0

Can you guys teach me about Weber function E_ν (z) and Anger function J_ν (z)?? Let′s Consider n-dimensional Euclidean Space and function f , f;R^n →R Helmholtz Equation defined as (▽^2 +k^2 )f=0 and in 2-dimensional Solution is f(r,θ)=Σ_(ℓ=0) ^∞ (a_ℓ ^ cos(ℓθ)+b_ℓ sin(ℓθ))(c_ℓ ^ J_ℓ (kr)+d_ℓ Y_ℓ (kr)) When I solved this equation I knew from the separation of variables that each of the Bessel functions J_ν (z) and Y_ν (z) comes out as a basis for solution But, When Bessel Equation not equal to 0 in other word x^2 y^((2)) (x)+xy^((1)) (x)+(x^2 −ν^2 )y(x)=(((x−ν)sin(νπ))/π) (A) and x^2 y^((2)) (x)+xy^((1)) (x)+(x^2 −ν^2 )y(x)=−((x+ν+(x−ν)sin(πν))/π) (B) and Each Solution as Follows Solution (A) {Weber}=C_1 J_ν (x)+C_2 Y_ν (x)+E_ν (x) Solution (B) {Anger}=C_1 J_ν (x)+C_2 Y_ν (x)+J_ν (x) I know how the Bessel function works aka J_ν (z) and Y_ν (z) but I don′t know How these two functions (each Weber function and Anger function) work... I′d like to know what its for or is it just a nonlinear differential equation thats been create by these weirdo mathematicians for their intellectual play???

$$\mathrm{Can}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{teach}\:\mathrm{me}\:\mathrm{about} \\ $$$$\mathrm{Weber}\:\mathrm{function}\:\boldsymbol{\mathrm{E}}_{\nu} \left({z}\right)\:\mathrm{and}\:\mathrm{Anger}\:\mathrm{function}\:\boldsymbol{\mathrm{J}}_{\nu} \left({z}\right)?? \\ $$$$\: \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{Consider}\:{n}-\mathrm{dimensional}\:\mathrm{Euclidean}\:\mathrm{Space} \\ $$$$\mathrm{and}\:\mathrm{function}\:{f}\:,\:{f};\mathbb{R}^{{n}} \rightarrow\mathbb{R} \\ $$$$\mathrm{Helmholt}{z}\:\mathrm{Equation}\:\mathrm{defined}\:\mathrm{as} \\ $$$$\left(\bigtriangledown^{\mathrm{2}} +{k}^{\mathrm{2}} \right){f}=\mathrm{0}\:\mathrm{and}\:\mathrm{in}\:\:\mathrm{2}-\mathrm{dimensional}\:\mathrm{Solution}\:\mathrm{is} \\ $$$${f}\left({r},\theta\right)=\underset{\ell=\mathrm{0}} {\overset{\infty} {\sum}}\left({a}_{\ell} ^{\:} \mathrm{cos}\left(\ell\theta\right)+{b}_{\ell} \mathrm{sin}\left(\ell\theta\right)\right)\left({c}_{\ell} ^{\:} {J}_{\ell} \left({kr}\right)+{d}_{\ell} {Y}_{\ell} \left({kr}\right)\right) \\ $$$$\: \\ $$$$\mathrm{When}\:\mathrm{I}\:\mathrm{solved}\:\mathrm{this}\:\mathrm{equation}\:\mathrm{I}\:\mathrm{knew}\:\mathrm{from}\:\mathrm{the}\: \\ $$$$\mathrm{separation}\:\mathrm{of}\:\mathrm{variables}\:\mathrm{that}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{Bessel}\:\mathrm{functions} \\ $$$${J}_{\nu} \left({z}\right)\:\mathrm{and}\:{Y}_{\nu} \left({z}\right)\:\mathrm{comes}\:\mathrm{out}\:\mathrm{as}\:\mathrm{a}\:\mathrm{basis}\:\mathrm{for}\:\mathrm{solution} \\ $$$$\mathrm{But}, \\ $$$$\mathrm{When}\:\mathrm{Bessel}\:\mathrm{Equation}\:\mathrm{not}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{0} \\ $$$$\mathrm{in}\:\mathrm{other}\:\mathrm{word} \\ $$$${x}^{\mathrm{2}} {y}^{\left(\mathrm{2}\right)} \left({x}\right)+{xy}^{\left(\mathrm{1}\right)} \left({x}\right)+\left({x}^{\mathrm{2}} −\nu^{\mathrm{2}} \right){y}\left({x}\right)=\frac{\left({x}−\nu\right)\mathrm{sin}\left(\nu\pi\right)}{\pi} \\ $$$$\left(\mathrm{A}\right) \\ $$$$\mathrm{and} \\ $$$${x}^{\mathrm{2}} {y}^{\left(\mathrm{2}\right)} \left({x}\right)+{xy}^{\left(\mathrm{1}\right)} \left({x}\right)+\left({x}^{\mathrm{2}} −\nu^{\mathrm{2}} \right){y}\left({x}\right)=−\frac{{x}+\nu+\left({x}−\nu\right)\mathrm{sin}\left(\pi\nu\right)}{\pi} \\ $$$$\left(\mathrm{B}\right)\: \\ $$$$\mathrm{and}\:\mathrm{Each}\:\mathrm{Solution}\:\mathrm{as}\:\mathrm{Follows} \\ $$$$\mathrm{Solution}\:\left(\mathrm{A}\right)\:\left\{\mathrm{Weber}\right\}={C}_{\mathrm{1}} {J}_{\nu} \left({x}\right)+{C}_{\mathrm{2}} {Y}_{\nu} \left({x}\right)+\boldsymbol{\mathrm{E}}_{\nu} \left({x}\right) \\ $$$$\mathrm{Solution}\:\left(\mathrm{B}\right)\:\left\{\mathrm{Anger}\right\}={C}_{\mathrm{1}} {J}_{\nu} \left({x}\right)+{C}_{\mathrm{2}} {Y}_{\nu} \left({x}\right)+\boldsymbol{\mathrm{J}}_{\nu} \left({x}\right) \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{know}\:\mathrm{how}\:\mathrm{the}\:\mathrm{Bessel}\:\mathrm{function}\:\mathrm{works}\:\mathrm{aka}\:{J}_{\nu} \left({z}\right)\:\mathrm{and}\:{Y}_{\nu} \left({z}\right) \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{How}\:\mathrm{these}\:\mathrm{two}\:\mathrm{functions} \\ $$$$\left(\mathrm{each}\:\mathrm{Weber}\:\mathrm{function}\:\mathrm{and}\:\mathrm{Anger}\:\mathrm{function}\right)\:\mathrm{work}... \\ $$$$\:\mathrm{I}'\mathrm{d}\:\mathrm{like}\:\mathrm{to}\:\mathrm{know}\:\mathrm{what}\:\mathrm{its}\:\mathrm{for}\:\mathrm{or}\:\mathrm{is}\:\mathrm{it} \\ $$$$\mathrm{just}\:\mathrm{a}\:\mathrm{nonlinear}\:\mathrm{differential}\:\mathrm{equation}\:\mathrm{thats}\:\mathrm{been} \\ $$$$\mathrm{create}\:\mathrm{by}\:\mathrm{these}\:\mathrm{weirdo}\:\mathrm{mathematicians}\:\mathrm{for}\:\mathrm{their} \\ $$$$\mathrm{intellectual}\:\mathrm{play}??? \\ $$

Question Number 220474 Answers: 0 Comments: 1

Question Number 220473 Answers: 0 Comments: 0

lim_(n→∞) (tan((π/4)+(1/n)))^n →^(t=(1/n)) =lim_(t→0) [tan((π/4)+t)]^(1/t) ⇒lim_(t→0) [tan((π/4)+t)−1]×(1/t) =lim_(t→0) (((1+tant)/(1−tant))−1)×(1/t) =lim_(t→0) (((2tant)/(1−tant)))×(1/t)=2 ⇒Ans=e^2 ✓

$${lim}_{{n}\rightarrow\infty} \left({tan}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right)\right)^{{n}} \:\:\:\:\:\overset{{t}=\frac{\mathrm{1}}{{n}}} {\rightarrow} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left[{tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)\right]^{\frac{\mathrm{1}}{{t}}} \\ $$$$\Rightarrow{lim}_{{t}\rightarrow\mathrm{0}} \left[{tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)−\mathrm{1}\right]×\frac{\mathrm{1}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{\mathrm{1}+{tant}}{\mathrm{1}−{tant}}−\mathrm{1}\right)×\frac{\mathrm{1}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{\mathrm{2}{tant}}{\mathrm{1}−{tant}}\right)×\frac{\mathrm{1}}{{t}}=\mathrm{2} \\ $$$$\Rightarrow{Ans}={e}^{\mathrm{2}} \:\:\checkmark \\ $$$$ \\ $$

Question Number 220469 Answers: 0 Comments: 0

∫_0 ^( 1) ((6x(1 − x))/((x + 1)(x^2 + 1) ln (x + 1))) dx

$$ \\ $$$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{\mathrm{6}{x}\left(\mathrm{1}\:−\:{x}\right)}{\left({x}\:+\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:\mathrm{ln}\:\left({x}\:+\:\mathrm{1}\right)}\:{dx} \\ $$$$ \\ $$

Question Number 220468 Answers: 3 Comments: 0

a+b+c=1, a^2 +b^2 +c^2 =1 (a,b,c ∈R) a^(10) +b^(10) +c^(10) =1, a^4 +b^4 +c^4 =?

$$\:{a}+{b}+{c}=\mathrm{1},\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1}\:\:\left({a},{b},{c}\:\in{R}\right) \\ $$$$\:{a}^{\mathrm{10}} +{b}^{\mathrm{10}} +{c}^{\mathrm{10}} =\mathrm{1},\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =? \\ $$

Question Number 220459 Answers: 3 Comments: 0

find Σ_(n=1) ^∞ (1/(n^2 +a^2 ))=? (a∈R)

$${find}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{a}^{\mathrm{2}} }=?\:\:\:\:\:\:\:\:\:\:\left({a}\in{R}\right) \\ $$

Question Number 220457 Answers: 0 Comments: 0

Find: 𝛀 = ∫_0 ^( ∞) ∫_0 ^( (𝛑/2)) (((x+1)sin^2 (x)ln(y^3 +1))/(xy(y^2 +1))) dxdy = ?

$$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\:\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\left(\mathrm{x}+\mathrm{1}\right)\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{y}^{\mathrm{3}} +\mathrm{1}\right)}{\mathrm{xy}\left(\mathrm{y}^{\mathrm{2}} +\mathrm{1}\right)}\:\mathrm{dxdy}\:=\:? \\ $$

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.

Question Number 220447 Answers: 0 Comments: 1

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