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AllQuestion and Answers: Page 1325

Question Number 80982 Answers: 0 Comments: 3

Question Number 80977 Answers: 1 Comments: 6

Question Number 80974 Answers: 0 Comments: 2

Show that gcd (a , a + x) ∣ x hence show that any two consecutive integers are coprime

$$\:\mathrm{Show}\:\mathrm{that}\:\mathrm{gcd}\:\left({a}\:,\:{a}\:+\:{x}\right)\:\mid\:{x} \\ $$$${hence}\:{show}\:{that}\:{any}\:{two}\:{consecutive} \\ $$$${integers}\:{are}\:{coprime} \\ $$

Question Number 80973 Answers: 0 Comments: 0

Given that f(x) = { ((2x−7, 0 < x < 6)),((2^x , 7 < x < 8)) :} and f is periodic of period 4. find f(200)

$$\mathrm{Given}\:\mathrm{that}\:{f}\left({x}\right)\:=\:\begin{cases}{\mathrm{2}{x}−\mathrm{7},\:\:\mathrm{0}\:<\:{x}\:<\:\mathrm{6}}\\{\mathrm{2}^{{x}} ,\:\:\:\mathrm{7}\:<\:{x}\:<\:\mathrm{8}}\end{cases} \\ $$$$\mathrm{and}\:\mathrm{f}\:\mathrm{is}\:\mathrm{periodic}\:\mathrm{of}\:\mathrm{period}\:\mathrm{4}. \\ $$$$\mathrm{find}\:{f}\left(\mathrm{200}\right) \\ $$

Question Number 80943 Answers: 1 Comments: 0

Find equation of a plane passing through the points (x_1 , y_1 , z_1 ), (x_2 , y_2 , z_2 ) and perpendicular to the plane ax+by+cz=d

$$\mathrm{Find}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{plane}\:\mathrm{passing}\:\mathrm{through}\:\mathrm{the} \\ $$$$\mathrm{points}\:\left({x}_{\mathrm{1}} ,\:{y}_{\mathrm{1}} ,\:{z}_{\mathrm{1}} \right),\:\left({x}_{\mathrm{2}} ,\:{y}_{\mathrm{2}} ,\:{z}_{\mathrm{2}} \right)\:\mathrm{and}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{plane}\:{ax}+{by}+{cz}={d} \\ $$

Question Number 80941 Answers: 0 Comments: 2

cos x−2cos y=−(√3) sin (x−y)=((2(√2))/3) what is sin x−2sin y ?

$$\mathrm{cos}\:{x}−\mathrm{2cos}\:{y}=−\sqrt{\mathrm{3}} \\ $$$$\mathrm{sin}\:\left({x}−{y}\right)=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$${what}\:{is}\:\mathrm{sin}\:{x}−\mathrm{2sin}\:{y}\:? \\ $$

Question Number 80951 Answers: 1 Comments: 3

Question Number 80930 Answers: 0 Comments: 3

Question Number 80929 Answers: 1 Comments: 5

show that cos(π/7)cos((2π)/7)cos((4π)/7)=−(1/8)

$$\mathrm{show}\:\mathrm{that} \\ $$$${cos}\frac{\pi}{\mathrm{7}}\mathrm{cos}\frac{\mathrm{2}\pi}{\mathrm{7}}\mathrm{cos}\frac{\mathrm{4}\pi}{\mathrm{7}}=−\frac{\mathrm{1}}{\mathrm{8}} \\ $$

Question Number 80927 Answers: 1 Comments: 2

Question Number 80926 Answers: 0 Comments: 2

i^i

$${i}^{{i}} \\ $$

Question Number 80925 Answers: 0 Comments: 1

∫_(−∞) ^∞ ((cos(x))/(1+x^2 )) dx =(π/e)

$$\int_{−\infty} ^{\infty} \frac{{cos}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:=\frac{\pi}{{e}} \\ $$

Question Number 80924 Answers: 1 Comments: 3

show that ∫_0 ^∞ (x^((π/5)−1) /(1+x^(2π) )) dx =φ

$${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\frac{\pi}{\mathrm{5}}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{2}\pi} }\:{dx}\:=\phi\: \\ $$

Question Number 80921 Answers: 0 Comments: 2

∫_0 ^(π/2) ((xdx)/(sin x+cos x)) = ?

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{xdx}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:=\:? \\ $$

Question Number 80917 Answers: 1 Comments: 0

((√(18−7x−x^2 ))/(2x+9)) ≥ ((√(18−7x−x^2 ))/(x+8))

$$\frac{\sqrt{\mathrm{18}−\mathrm{7}{x}−{x}^{\mathrm{2}} }}{\mathrm{2}{x}+\mathrm{9}}\:\geqslant\:\frac{\sqrt{\mathrm{18}−\mathrm{7}{x}−{x}^{\mathrm{2}} }}{{x}+\mathrm{8}} \\ $$

Question Number 80914 Answers: 0 Comments: 0

(1) Integrate F(x, y) = x^2 over the region bounded by y = x^2 , x = 2 and x = 1 (2) Integrate G(x, y) = x^2 + y^2 over the region bounded by the triangle x = y, y = 1 and y = 0

$$\left(\mathrm{1}\right) \\ $$$$\mathrm{Integrate}\:\:\mathrm{F}\left(\mathrm{x},\:\mathrm{y}\right)\:\:=\:\:\mathrm{x}^{\mathrm{2}} \:\:\:\mathrm{over}\:\mathrm{the}\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{by}\:\:\:\mathrm{y}\:\:=\:\:\mathrm{x}^{\mathrm{2}} , \\ $$$$\mathrm{x}\:\:=\:\:\mathrm{2}\:\:\mathrm{and}\:\mathrm{x}\:\:=\:\:\mathrm{1} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{Integrate}\:\:\:\:\mathrm{G}\left(\mathrm{x},\:\mathrm{y}\right)\:\:=\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:\:\:\:\mathrm{over}\:\mathrm{the}\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\: \\ $$$$\mathrm{triangle}\:\:\:\:\mathrm{x}\:\:=\:\:\mathrm{y},\:\:\mathrm{y}\:\:=\:\:\mathrm{1}\:\:\mathrm{and}\:\:\mathrm{y}\:\:=\:\:\mathrm{0} \\ $$

Question Number 80912 Answers: 0 Comments: 2

Question Number 80907 Answers: 1 Comments: 0

a)∫e^x tan xdx b)∫xtan xdx

$$\left.{a}\right)\int{e}^{{x}} \mathrm{tan}\:{xdx} \\ $$$$\left.{b}\right)\int{x}\mathrm{tan}\:{xdx} \\ $$

Question Number 80900 Answers: 1 Comments: 1

Question Number 80896 Answers: 1 Comments: 0

α is a real ∈ ]0;(π/2)[. we give this (E_α ):2x^2 −2x(√2)(cosα)+cos2α=0 1. show that Δ=8sin^2 x i showed it. 2.Solve E_α in R. please help me for this question.

$$\left.\alpha\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\in\:\right]\mathrm{0};\frac{\pi}{\mathrm{2}}\left[.\:\mathrm{we}\:\mathrm{give}\:\mathrm{this}\:\right. \\ $$$$\left(\mathrm{E}_{\alpha} \right):\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}\sqrt{\mathrm{2}}\left({cos}\alpha\right)+\mathrm{cos2}\alpha=\mathrm{0} \\ $$$$\mathrm{1}.\:\mathrm{show}\:\mathrm{that}\:\Delta=\mathrm{8sin}^{\mathrm{2}} {x} \\ $$$${i}\:{showed}\:{it}. \\ $$$$\mathrm{2}.{S}\mathrm{olve}\:\mathrm{E}_{\alpha} \:\mathrm{in}\:\mathbb{R}. \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{for}\:\mathrm{this}\:\mathrm{question}. \\ $$

Question Number 80890 Answers: 1 Comments: 0

(((1+i)/(2−i))+((2+i)/(1−i)))^3 =500x+500yi find x,y

$$\left(\frac{\mathrm{1}+\mathrm{i}}{\mathrm{2}−\mathrm{i}}+\frac{\mathrm{2}+\mathrm{i}}{\mathrm{1}−\mathrm{i}}\right)^{\mathrm{3}} =\mathrm{500}{x}+\mathrm{500}{y}\mathrm{i} \\ $$$${find}\:{x},{y} \\ $$

Question Number 80889 Answers: 0 Comments: 3

find Z if arg(z−3)=π and arg(z+i)=(π/4)

$${find}\:{Z} \\ $$$${if}\:{arg}\left(\mathrm{z}−\mathrm{3}\right)=\pi \\ $$$${and}\:{arg}\left(\mathrm{z}+\mathrm{i}\right)=\frac{\pi}{\mathrm{4}} \\ $$

Question Number 80885 Answers: 0 Comments: 3

Question Number 80884 Answers: 0 Comments: 0

lim_(u→i∞) Σ_(k=0) ^∞ (((−1)^k 6^k u^(k+1) )/((6k+1)!!!!!!e^(−u^6 ) )) =?

$$\underset{{u}\rightarrow{i}\infty} {{lim}}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} \:\mathrm{6}^{{k}} \:{u}^{{k}+\mathrm{1}} }{\left(\mathrm{6}{k}+\mathrm{1}\right)!!!!!!{e}^{−{u}^{\mathrm{6}} } }\:=? \\ $$

Question Number 80882 Answers: 1 Comments: 4

Question Number 80881 Answers: 1 Comments: 0

v=−(((2b+3cp)p)/(3+bp^2 )) ((3v^2 +2bpv+3cp+b)/(1+bp^2 +cp^3 )) > 0 b,c ∈ R , b<0 Any non-zero real value of p in terms of b,c obeying above condition?

$${v}=−\frac{\left(\mathrm{2}{b}+\mathrm{3}{cp}\right){p}}{\mathrm{3}+{bp}^{\mathrm{2}} } \\ $$$$\:\frac{\mathrm{3}{v}^{\mathrm{2}} +\mathrm{2}{bpv}+\mathrm{3}{cp}+{b}}{\mathrm{1}+{bp}^{\mathrm{2}} +{cp}^{\mathrm{3}} }\:>\:\mathrm{0}\:\: \\ $$$${b},{c}\:\in\:\mathbb{R}\:,\:{b}<\mathrm{0} \\ $$$${Any}\:{non}-{zero}\:{real}\:{value}\:{of}\:{p} \\ $$$${in}\:{terms}\:{of}\:{b},{c}\:\:{obeying}\:{above} \\ $$$${condition}? \\ $$

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