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Question Number 36167 Answers: 0 Comments: 2

let give I = ∫_0 ^∞ (dx/((x^2 +i)^2 )) 1) extract Re(I) and Im(I) 2) find the value of I 3) calculate Re(I) and Im(I) .

$${let}\:{give}\:\:{I}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{i}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{extract}\:{Re}\left({I}\right)\:{and}\:{Im}\left({I}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:{I} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{Re}\left({I}\right)\:{and}\:{Im}\left({I}\right)\:. \\ $$

Question Number 36166 Answers: 0 Comments: 1

Find the middle term in the expansion of (x^ + (3/x))^9

$$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{term}\:\mathrm{in}\: \\ $$$$\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\:\left(\mathrm{x}^{} \:+\:\frac{\mathrm{3}}{\mathrm{x}}\right)^{\mathrm{9}} \\ $$

Question Number 36163 Answers: 0 Comments: 3

Question Number 36154 Answers: 0 Comments: 0

Q. If x≠y≠z and determinant ((x,x^3 ,(x^4 −1)),(y,y^3 ,(y^4 −1)),((z ),z^3 ,(z^4 −1)))=0 Prove that xyz(xy+yz+zx)=(x+y+z) please help.

$${Q}.\:\:{If}\:{x}\neq{y}\neq{z}\:\:{and}\:\:\begin{vmatrix}{{x}}&{{x}^{\mathrm{3}} }&{{x}^{\mathrm{4}} −\mathrm{1}}\\{{y}}&{{y}^{\mathrm{3}} }&{{y}^{\mathrm{4}} −\mathrm{1}}\\{{z}\:}&{{z}^{\mathrm{3}} }&{{z}^{\mathrm{4}} −\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$$$ \\ $$$${Prove}\:{that}\:\:{xyz}\left({xy}+{yz}+{zx}\right)=\left({x}+{y}+{z}\right) \\ $$$$ \\ $$$${please}\:{help}. \\ $$

Question Number 36153 Answers: 0 Comments: 1

(((x+yi−2)^2 )/(x−yi+1))

$$\frac{\left({x}+{yi}−\mathrm{2}\right)^{\mathrm{2}} }{{x}−{yi}+\mathrm{1}} \\ $$

Question Number 36140 Answers: 1 Comments: 1

Question Number 36148 Answers: 0 Comments: 0

[2^x −^(+ 3) 1 4^(2y+x) x6]=[3^(0−7) 2x]

$$ \\ $$$$\:\:\left[\overset{\mathrm{x}} {\mathrm{2}}\overset{+\:\:\mathrm{3}} {−}\mathrm{1}\:\:\:\:\overset{\mathrm{2y}+\mathrm{x}} {\mathrm{4}x6}\right]=\left[\overset{\mathrm{0}−\mathrm{7}} {\mathrm{3}}\:\mathrm{2x}\right] \\ $$

Question Number 36132 Answers: 0 Comments: 7

a+b=10.........(i) ab+c=0..........(ii) ac+d=6..........(iii) ad=−1...........(iv) (a,b,c,d)=? Note: This problem is related to solve the equation (t^4 +10t+6t−1=0) of Q#35844

$${a}+{b}=\mathrm{10}.........\left(\mathrm{i}\right) \\ $$$${ab}+{c}=\mathrm{0}..........\left(\mathrm{ii}\right) \\ $$$${ac}+{d}=\mathrm{6}..........\left(\mathrm{iii}\right) \\ $$$${ad}=−\mathrm{1}...........\left(\mathrm{iv}\right) \\ $$$$\left({a},{b},{c},{d}\right)=? \\ $$$$\mathcal{N}{ote}:\:{This}\:{problem}\:{is}\:{related}\:{to}\:{solve} \\ $$$${the}\:{equation}\:\left({t}^{\mathrm{4}} +\mathrm{10}{t}+\mathrm{6}{t}−\mathrm{1}=\mathrm{0}\right)\:{of} \\ $$$${Q}#\mathrm{35844} \\ $$

Question Number 36128 Answers: 3 Comments: 3

∫sin^8 xdx ∫sin^6 xdx

$$\int\boldsymbol{\mathrm{sin}}^{\mathrm{8}} \boldsymbol{{xdx}} \\ $$$$\int\boldsymbol{\mathrm{sin}}^{\mathrm{6}} \boldsymbol{{xdx}} \\ $$

Question Number 36126 Answers: 0 Comments: 4

x^4 +10x^3 +6x−1 =^(?) (x^2 +(((√5)−1)/2))(x^2 +10x−(((√5)+1)/2))

$$\mathrm{x}^{\mathrm{4}} +\mathrm{10x}^{\mathrm{3}} +\mathrm{6x}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{?} {=}\left({x}^{\mathrm{2}} +\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)\left({x}^{\mathrm{2}} +\mathrm{10}{x}−\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right) \\ $$

Question Number 36120 Answers: 1 Comments: 0

3(√(200×1080))

$$\mathrm{3}\sqrt{\mathrm{200}×\mathrm{1080}} \\ $$

Question Number 36119 Answers: 0 Comments: 3

3(√(433^ ))

$$\mathrm{3}\sqrt{\mathrm{43}\hat {\mathrm{3}}} \\ $$

Question Number 36115 Answers: 0 Comments: 1

Question Number 36110 Answers: 0 Comments: 0

{Δ1 3 6 / ×<⌈+2/ 47

$$\left\{\Delta\mathrm{1}\:\mathrm{3}\:\mathrm{6}\:/\:×<\lceil+\mathrm{2}/\right. \\ $$$$\mathrm{47} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 36104 Answers: 0 Comments: 1

If f:R→R is a function such that ∣ f(x) − f(y)∣ ≤ ∣ sin x − sin y ∣∀x,y∈R, Then f(x) is (1) Bijective (2) many−one (3) periodic (4) non−periodic

$$\mathrm{If}\:\boldsymbol{\mathrm{f}}:\boldsymbol{\mathrm{R}}\rightarrow\boldsymbol{\mathrm{R}}\:\mathrm{is}\:\mathrm{a}\:\mathrm{function}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mid\:\mathrm{f}\left({x}\right)\:−\:\mathrm{f}\left(\mathrm{y}\right)\mid\:\leqslant\:\mid\:\mathrm{sin}\:{x}\:−\:\mathrm{sin}\:\mathrm{y}\:\mid\forall{x},\mathrm{y}\in\mathbb{R}, \\ $$$$\mathrm{Then}\:\mathrm{f}\left({x}\right)\:\mathrm{is}\: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Bijective} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{many}−\mathrm{one} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{periodic} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{non}−\mathrm{periodic} \\ $$

Question Number 36103 Answers: 3 Comments: 0

convert 0.26999999...into fraction (a/b) where a≠0

$$\boldsymbol{\mathrm{convert}}\:\mathrm{0}.\mathrm{26999999}...\boldsymbol{\mathrm{into}}\:\boldsymbol{\mathrm{fraction}}\: \\ $$$$\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}}\:\boldsymbol{\mathrm{where}}\:\boldsymbol{\mathrm{a}}\neq\mathrm{0} \\ $$

Question Number 36101 Answers: 2 Comments: 2

Question Number 36099 Answers: 0 Comments: 4

Question Number 36096 Answers: 1 Comments: 4

Question Number 36092 Answers: 1 Comments: 1

solve for 0°≤ θ ≤ 360° the equation cos(θ + (π/3))= (1/2)

$$\:\mathrm{solve}\:\mathrm{for}\:\mathrm{0}°\leqslant\:\theta\:\leqslant\:\mathrm{360}°\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\mathrm{cos}\left(\theta\:+\:\frac{\pi}{\mathrm{3}}\right)=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Question Number 36091 Answers: 0 Comments: 1

Given the position vectors v_1 = 2i − 2j and v_2 = 2j, show that the unit vector in the direction of the vector v_1 − v_(2 ) is (1/(√5))(i−2j)

$$\mathrm{Given}\:\mathrm{the}\:\mathrm{position}\:\mathrm{vectors} \\ $$$${v}_{\mathrm{1}} =\:\mathrm{2}{i}\:−\:\mathrm{2}{j}\:{and}\:{v}_{\mathrm{2}} =\:\mathrm{2}{j}, \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{unit}\:\mathrm{vector}\:\mathrm{in}\: \\ $$$$\mathrm{the}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vector}\: \\ $$$${v}_{\mathrm{1}} −\:{v}_{\mathrm{2}\:\:\:} \mathrm{is}\:\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\left(\mathrm{i}−\mathrm{2j}\right) \\ $$

Question Number 36080 Answers: 2 Comments: 1

i want to know how α^2 + β^2 = (α+β)^2 − 2αβ why not α^2 +β^2 = (α+β)^2 + 2αβ?

$$\:\mathrm{i}\:\mathrm{want}\:\mathrm{to}\:\mathrm{know}\:\mathrm{how}\: \\ $$$$\alpha^{\mathrm{2}} +\:\beta^{\mathrm{2}} =\:\left(\alpha+\beta\right)^{\mathrm{2}} −\:\mathrm{2}\alpha\beta\:\mathrm{why}\:\mathrm{not}\: \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\:\left(\alpha+\beta\right)^{\mathrm{2}} +\:\mathrm{2}\alpha\beta? \\ $$

Question Number 36068 Answers: 0 Comments: 1

Why is it not advisable to use small incident angle when performing experiment on refraction using a triangular prism?

$${Why}\:{is}\:{it}\:{not}\:{advisable}\:{to}\:{use} \\ $$$${small}\:{incident}\:{angle}\:{when}\:{performing} \\ $$$${experiment}\:{on}\:{refraction}\:{using}\:{a} \\ $$$${triangular}\:{prism}? \\ $$

Question Number 36061 Answers: 1 Comments: 2

Question Number 36059 Answers: 1 Comments: 0

Question Number 36057 Answers: 2 Comments: 1

find the value of ∫_0 ^(π/4) ((cosx)/(sinx +tanx))dx

$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\frac{{cosx}}{{sinx}\:+{tanx}}{dx}\: \\ $$

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