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Question Number 187025 Answers: 1 Comments: 0

Question Number 187044 Answers: 0 Comments: 0

f(x,y)={(x+y),0<x<1, 0<y<1 0, otherwise find: (1) σxy (2) p(0<x<((1/4))∣((2/5))>y>((1/3))) note: f(x,y) is the joint PDF

$${f}\left({x},{y}\right)=\left\{\left({x}+{y}\right),\mathrm{0}<{x}<\mathrm{1},\:\mathrm{0}<{y}<\mathrm{1}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0},\:{otherwise} \\ $$$${find}: \\ $$$$\left(\mathrm{1}\right)\:\sigma{xy} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\:{p}\left(\mathrm{0}<{x}<\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\mid\left(\frac{\mathrm{2}}{\mathrm{5}}\right)>{y}>\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right) \\ $$$$ \\ $$$${note}:\:{f}\left({x},{y}\right)\:{is}\:{the}\:{joint}\:{PDF} \\ $$

Question Number 187020 Answers: 0 Comments: 1

(a/x)=(b/y)=(c/z)=(1/3) ,a−2b+c=2 and 2y−3z=1 x=? how is solution

$$\frac{{a}}{{x}}=\frac{{b}}{{y}}=\frac{{c}}{{z}}=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:,{a}−\mathrm{2}{b}+{c}=\mathrm{2}\:\:{and}\:\:\mathrm{2}{y}−\mathrm{3}{z}=\mathrm{1}\:\:\:\:{x}=? \\ $$$${how}\:{is}\:{solution} \\ $$

Question Number 187011 Answers: 0 Comments: 0

a∈C determinant (((a 1 … 1)),((1 ⋱ ⋱ ⋮)),((⋮ ⋱ ⋱ 1)),((1 … 1 a)))=?

$$\mathrm{a}\in\mathbb{C} \\ $$$$\begin{vmatrix}{\mathrm{a}\:\:\:\mathrm{1}\:\:\:\:\ldots\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\ddots\:\:\ddots\:\:\vdots}\\{\vdots\:\:\ddots\:\:\ddots\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\ldots\:\:\:\:\mathrm{1}\:\:\:\mathrm{a}}\end{vmatrix}=? \\ $$

Question Number 187010 Answers: 0 Comments: 0

a,b∈C determinant (((a+b a 0 … 0)),(( b a+b ⋱ ⋱ ⋮)),(( 0 ⋱ ⋱ ⋱ 0)),(( ⋮ ⋱ ⋱ ⋱ a)),(( 0 … 0 b a+b)))=?

$$\mathrm{a},\mathrm{b}\in\mathbb{C} \\ $$$$\:\begin{vmatrix}{\mathrm{a}+\mathrm{b}\:\:\:\:\:\:\:\:\:\mathrm{a}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\ldots\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\:\:\:\mathrm{b}\:\:\:\:\:\:\:\:\:\:\mathrm{a}+\mathrm{b}\:\:\:\:\:\:\ddots\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\:\:\:\:\vdots}\\{\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\:\:\vdots\:\:\:\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\:\ddots\:\:\:\:\:\:\ddots\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}}\\{\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\ldots\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}+\mathrm{b}}\end{vmatrix}=? \\ $$

Question Number 187008 Answers: 2 Comments: 0

(a/3)=(b/4)=(c/5) 3a+c=42 b=? how is solution

$$\frac{{a}}{\mathrm{3}}=\frac{{b}}{\mathrm{4}}=\frac{{c}}{\mathrm{5}}\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{a}+{c}=\mathrm{42}\:\:\:\:\:\:\:\:{b}=? \\ $$$${how}\:{is}\:{solution} \\ $$

Question Number 187007 Answers: 0 Comments: 2

prove that 7^(k+1) ∣ 13^(7k) + 1 , ∀k∈N

$${prove}\:\:{that} \\ $$$$\mathrm{7}^{{k}+\mathrm{1}} \mid\:\mathrm{13}^{\mathrm{7}{k}} \:+\:\mathrm{1}\:,\:\forall{k}\in\mathbb{N} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 186999 Answers: 0 Comments: 0

Question Number 186998 Answers: 1 Comments: 1

Question Number 186996 Answers: 1 Comments: 0

Question Number 186989 Answers: 0 Comments: 0

Question Number 186983 Answers: 3 Comments: 0

Question Number 186960 Answers: 1 Comments: 0

Question Number 186958 Answers: 0 Comments: 0

Question Number 186952 Answers: 2 Comments: 0

Question Number 186951 Answers: 2 Comments: 0

Question Number 186950 Answers: 0 Comments: 0

Question Number 186941 Answers: 0 Comments: 1

((sinx)/x^b )=((Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!))x^(2n+1) )/x^b ) prove that

$$\frac{{sinx}}{{x}^{{b}} }=\frac{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{x}^{\mathrm{2}{n}+\mathrm{1}} }{{x}^{{b}} } \\ $$$${prove}\:{that} \\ $$

Question Number 186929 Answers: 3 Comments: 0

If { ((B+R+P=−1)),((B^2 +R^2 +P^2 =17)),((B^3 +R^3 +P^3 =11)) :} then B^5 +R^5 +P^5 =?

$$\:\:{If}\:\begin{cases}{{B}+{R}+{P}=−\mathrm{1}}\\{{B}^{\mathrm{2}} +{R}^{\mathrm{2}} +{P}^{\mathrm{2}} =\mathrm{17}}\\{{B}^{\mathrm{3}} +{R}^{\mathrm{3}} +{P}^{\mathrm{3}} =\mathrm{11}}\end{cases} \\ $$$$\:{then}\:{B}^{\mathrm{5}} +{R}^{\mathrm{5}} +{P}^{\mathrm{5}} \:=? \\ $$

Question Number 186926 Answers: 1 Comments: 1

Question Number 186924 Answers: 3 Comments: 0

Question Number 186923 Answers: 0 Comments: 0

Question Number 186925 Answers: 0 Comments: 1

Question Number 186921 Answers: 0 Comments: 0

Question Number 186912 Answers: 0 Comments: 2

Question Number 186911 Answers: 1 Comments: 0

∫_0 ^∞ (1/(1+a^x +a^(x/2) ))dx = (1/(ln a))[ln 3−(π/(3(√3)))]

$$ \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{a}^{{x}} +{a}^{\frac{{x}}{\mathrm{2}}} }{dx}\:=\:\frac{\mathrm{1}}{\mathrm{ln}\:{a}}\left[\mathrm{ln}\:\mathrm{3}−\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\right] \\ $$

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