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Question Number 196166    Answers: 1   Comments: 1

for any a,b∈[−1,1] calculate: arccos(a)+arcsin(b)=? arcsin(a)+arcsin(b)=? arccos(a)+arccos(b)=?

$$\boldsymbol{{for}}\:\boldsymbol{{any}}\:\boldsymbol{{a}},\boldsymbol{{b}}\in\left[−\mathrm{1},\mathrm{1}\right]\:\:\boldsymbol{{calculate}}:\: \\ $$$$\boldsymbol{{arccos}}\left(\boldsymbol{{a}}\right)+\boldsymbol{{arcsin}}\left(\boldsymbol{{b}}\right)=? \\ $$$$\boldsymbol{{arcsin}}\left(\boldsymbol{{a}}\right)+\boldsymbol{{arcsin}}\left(\boldsymbol{{b}}\right)=? \\ $$$$\boldsymbol{{arccos}}\left(\boldsymbol{{a}}\right)+\boldsymbol{{arc}}{cos}\left(\boldsymbol{{b}}\right)=? \\ $$

Question Number 196164    Answers: 2   Comments: 0

calcul: lim_(n→+∞ ) Un=((((2n)!)/(n!n^n )))^(1/n)

$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathrm{calcul}: \\ $$$$ \\ $$$$\:\:\:{lim}_{{n}\rightarrow+\infty\:\:\:\:} {Un}=\sqrt[{{n}}]{\frac{\left(\mathrm{2}{n}\right)!}{{n}!{n}^{{n}} }} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 196165    Answers: 3   Comments: 0

{: ((x^(x+y) =y^(24) )),((y^(x+y) =x^6 )) }⇒(x,y)=(?,?)

$$\left.\begin{matrix}{{x}^{{x}+{y}} ={y}^{\mathrm{24}} }\\{{y}^{{x}+{y}} ={x}^{\mathrm{6}} }\end{matrix}\right\}\Rightarrow\left({x},{y}\right)=\left(?,?\right) \\ $$

Question Number 196160    Answers: 0   Comments: 3

Question Number 196155    Answers: 1   Comments: 1

Question Number 196154    Answers: 2   Comments: 0

Question Number 196145    Answers: 1   Comments: 0

Calculer ∫^( +∞) _( (1/α)) e^(−αt^2 +2t) dt

$$\mathrm{Calculer}\:\underset{\:\frac{\mathrm{1}}{\alpha}} {\int}^{\:+\infty} {e}^{−\alpha{t}^{\mathrm{2}} +\mathrm{2}{t}} {dt} \\ $$

Question Number 196143    Answers: 2   Comments: 0

say you have 3 (different) books about mathematics, 4 (different) books about physics and 5 (different) books about chemistry. in how many ways can you arrange them in a shelf such that no two books from the same subject are adjacent?

$${say}\:{you}\:{have}\:\mathrm{3}\:\left({different}\right)\:{books} \\ $$$${about}\:{mathematics},\:\mathrm{4}\:\left({different}\right) \\ $$$${books}\:{about}\:{physics}\:{and}\:\mathrm{5}\:\left({different}\right) \\ $$$${books}\:{about}\:{chemistry}.\:{in}\:{how}\:{many} \\ $$$${ways}\:{can}\:{you}\:{arrange}\:{them}\:{in}\:{a}\:{shelf} \\ $$$${such}\:{that}\:{no}\:{two}\:{books}\:{from}\:{the}\:{same} \\ $$$${subject}\:{are}\:{adjacent}? \\ $$

Question Number 196135    Answers: 0   Comments: 0

Question Number 196132    Answers: 0   Comments: 0

Question Number 196121    Answers: 0   Comments: 0

In forest , a hunter obstains that Every morning a snake eats a mouse Every afternoon a scorpion kills a snake Every night a mouse corrodes a scorpion The 8^(th) day morning , there remains Only one of them , a mouse How many were they, in each species?

$${In}\:{forest}\:,\:{a}\:{hunter}\:{obstains}\:{that} \\ $$$${Every}\:{morning}\:{a}\:{snake}\:{eats}\:{a}\:{mouse} \\ $$$${Every}\:{afternoon}\:{a}\:{scorpion}\:{kills}\:{a}\:{snake} \\ $$$${Every}\:{night}\:{a}\:{mouse}\:{corrodes}\:{a}\:{scorpion} \\ $$$${The}\:\mathrm{8}^{{th}} \:{day}\:{morning}\:,\:{there}\:{remains} \\ $$$${Only}\:{one}\:{of}\:{them}\:,\:{a}\:{mouse} \\ $$$$ \\ $$$${How}\:{many}\:{were}\:{they},\:{in}\:{each}\:{species}? \\ $$

Question Number 196119    Answers: 5   Comments: 0

solve (√(100−x^2 ))+(√(64−x^2 ))=12

$${solve} \\ $$$$\sqrt{\mathrm{100}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{64}−{x}^{\mathrm{2}} }=\mathrm{12} \\ $$

Question Number 196117    Answers: 2   Comments: 0

Question Number 196115    Answers: 1   Comments: 0

Question Number 196111    Answers: 1   Comments: 0

Question Number 196107    Answers: 1   Comments: 0

help, please ! lim_(x → (𝛑/4)) ((cos((𝛑/4) −x)−tan x)/(1−sin((𝛑/4)+x))) = ???

$$ \\ $$$$\mathrm{help},\:\mathrm{please}\:! \\ $$$$\underset{{x}\:\rightarrow\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} {{lim}}\:\frac{{cos}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\:−{x}\right)−{tan}\:{x}}{\mathrm{1}−{sin}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}+{x}\right)}\:=\:???\: \\ $$

Question Number 196103    Answers: 0   Comments: 0

(d^n /dx^n )[𝚪(x)] Hence what is (d^(1/2) /dx^(1/2) )[𝚪(x)]

$$\frac{\boldsymbol{\mathrm{d}}^{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{dx}}^{\boldsymbol{\mathrm{n}}} }\left[\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{x}}\right)\right]\:\:\boldsymbol{\mathrm{Hence}}\:\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{is}}\:\frac{\boldsymbol{\mathrm{d}}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\boldsymbol{\mathrm{dx}}^{\frac{\mathrm{1}}{\mathrm{2}}} }\left[\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{x}}\right)\right] \\ $$

Question Number 196102    Answers: 0   Comments: 0

Question Number 196099    Answers: 0   Comments: 0

Question Number 196098    Answers: 1   Comments: 0

Question Number 196097    Answers: 0   Comments: 0

Question Number 196094    Answers: 1   Comments: 0

Question Number 196134    Answers: 3   Comments: 0

Question Number 196086    Answers: 0   Comments: 0

Answer to the question “196008” Σ_(k=1) ^n tan^2 (((kπ)/(2n+1)))=n(2n+1) Ans) according “de moivre” sin(2n+1)α= (((2n+1)),(( 1)) )(cosα)^(2n) sinα− (((2n+1)),(( 3)) )(cosα)^(2n−2) (sinα)^3 +.... =(cosα)^(2n) (sinα)[ (((2n+1)),(( 1)) )− (((2n+1)),(( 3)) ) tan^2 α+...] for “α_k =((kπ)/(2n+1)) ; 1≤k≤n ⇒sin(2n+1)α_k =0 ⇒∀ 1≤k≤n→ (((2n+1)),(( 1)) )− (((2n+1)),(( 3)) ) tan^2 α_k +...=0 therefore “ x_k =tan^2 α_k ” thr roots of the equation are blowe x^n − (((2n+1)),((2n−1)) ) x^n + (((2n+1)),((2n−3)) )x^(n−1) −...=0 the sume of the roots of the equation is “ s= (((2n+1)),((2n−1)) ) ” ⇒s=Σ_(k=1) ^n tan^2 (((kπ)/(2n+1)))=n(2n+1)✓ the proof of the seconf part is done similarly

$${Answer}\:{to}\:{the}\:{question}\:``\mathrm{196008}'' \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{tan}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)={n}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$\left.{Ans}\right) \\ $$$${according}\:\:``{de}\:{moivre}'' \\ $$$${sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha=\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\left({cos}\alpha\right)^{\mathrm{2}{n}} {sin}\alpha−\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\mathrm{3}}\end{pmatrix}\left({cos}\alpha\right)^{\mathrm{2}{n}−\mathrm{2}} \left({sin}\alpha\right)^{\mathrm{3}} +.... \\ $$$$=\left({cos}\alpha\right)^{\mathrm{2}{n}} \left({sin}\alpha\right)\left[\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}−\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:{tan}^{\mathrm{2}} \alpha+...\right] \\ $$$${for}\:\:``\alpha_{{k}} =\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\:\:\:\:\:;\:\:\mathrm{1}\leqslant{k}\leqslant{n}\:\Rightarrow{sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha_{{k}} =\mathrm{0} \\ $$$$\Rightarrow\forall\:\:\mathrm{1}\leqslant{k}\leqslant{n}\rightarrow\:\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\mathrm{1}}\end{pmatrix}−\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:{tan}^{\mathrm{2}} \alpha_{{k}} +...=\mathrm{0} \\ $$$${therefore}\:\:``\:{x}_{{k}} ={tan}^{\mathrm{2}} \alpha_{{k}} \:''\:{thr}\:{roots}\:{of}\:\:{the}\:{equation}\:{are}\:{blowe} \\ $$$${x}^{{n}} −\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\mathrm{2}{n}−\mathrm{1}}\end{pmatrix}\:{x}^{{n}} +\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\mathrm{2}{n}−\mathrm{3}}\end{pmatrix}{x}^{{n}−\mathrm{1}} −...=\mathrm{0} \\ $$$${the}\:{sume}\:{of}\:{the}\:{roots}\:{of}\:{the}\:{equation}\:{is}\:``\:{s}=\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\mathrm{2}{n}−\mathrm{1}}\end{pmatrix}\:'' \\ $$$$\Rightarrow{s}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:{tan}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)={n}\left(\mathrm{2}{n}+\mathrm{1}\right)\checkmark \\ $$$${the}\:{proof}\:{of}\:{the}\:{seconf}\:{part}\:{is}\:{done}\:{similarly} \\ $$$$ \\ $$$$ \\ $$

Question Number 196085    Answers: 2   Comments: 0

Question Number 196084    Answers: 1   Comments: 0

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