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Question Number 196902    Answers: 1   Comments: 0

Question Number 196893    Answers: 1   Comments: 0

Question Number 196886    Answers: 2   Comments: 0

Question Number 196885    Answers: 2   Comments: 0

Question Number 196883    Answers: 1   Comments: 0

Question Number 196881    Answers: 1   Comments: 1

Question Number 196880    Answers: 1   Comments: 0

1. A container of milk is (4/5) full. When 10L of milk is poured into it the container becomes (9/(10)) full. What is the capacity of the container?

$$\mathrm{1}.\:{A}\:{container}\:{of}\:{milk}\:{is}\:\frac{\mathrm{4}}{\mathrm{5}}\:{full}.\:{When}\:\mathrm{10}{L}\:{of}\:{milk}\:{is}\:{poured}\:{into}\:{it}\:{the}\:{container}\:{becomes}\:\:\frac{\mathrm{9}}{\mathrm{10}}\:{full}.\:{What}\:{is}\:{the}\:{capacity}\:{of}\:{the}\:{container}? \\ $$

Question Number 196872    Answers: 1   Comments: 0

Let ξ be a positive Root of x^2 −2023x−1 Define a sequence ϕ_i such That ϕ_0 =1 ϕ_(n+1) =⌊ϕ_n ξ⌋, find The Remainder When ϕ_(2023 ) is divided by (√ϕ_2 )

$${Let}\:\xi\:{be}\:{a}\:{positive}\:{Root}\:{of}\:{x}^{\mathrm{2}} −\mathrm{2023}{x}−\mathrm{1} \\ $$$${Define}\:{a}\:{sequence}\:\varphi_{{i}} \:{such}\:{That}\:\varphi_{\mathrm{0}} =\mathrm{1} \\ $$$$\varphi_{{n}+\mathrm{1}} =\lfloor\varphi_{{n}} \xi\rfloor,\:{find}\:{The}\:{Remainder}\:{When}\:\varphi_{\mathrm{2023}\:} {is}\:{divided}\:{by}\:\sqrt{\varphi_{\mathrm{2}} } \\ $$

Question Number 196870    Answers: 1   Comments: 0

let b_i ∧ a_i >0 where i∈{1,2,3,...,n}& Σ_(i=1) ^n (b_i )=λ Prove that ((λ−(b_1 +b_2 ))/((b_1 +b_2 )))(a_1 +a_2 )+((λ−(b_1 +b_3 ))/((b_1 +b_3 )))(a_1 +a_3 )+....+((λ−(b_2 +b_3 ))/((b_2 +b_3 )))(a_2 +a_3 )+...((λ−(b_(n−1) +b_n ))/((b_(n−1) +b_n )))(a_(n−1) +a_n ) ≥(√(((n(n−1)(n−2)^2 )/4)×ΣΣ_(1≤i<j≤n) (a_i a_j )))

$${let}\:{b}_{{i}} \wedge\:{a}_{{i}} >\mathrm{0}\:{where}\:{i}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3},...,{n}\right\}\&\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({b}_{{i}} \right)=\lambda\:{Prove}\:{that} \\ $$$$\frac{\lambda−\left({b}_{\mathrm{1}} +{b}_{\mathrm{2}} \right)}{\left({b}_{\mathrm{1}} +{b}_{\mathrm{2}} \right)}\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} \right)+\frac{\lambda−\left({b}_{\mathrm{1}} +{b}_{\mathrm{3}} \right)}{\left({b}_{\mathrm{1}} +{b}_{\mathrm{3}} \right)}\left({a}_{\mathrm{1}} +{a}_{\mathrm{3}} \right)+....+\frac{\lambda−\left({b}_{\mathrm{2}} +{b}_{\mathrm{3}} \right)}{\left({b}_{\mathrm{2}} +{b}_{\mathrm{3}} \right)}\left({a}_{\mathrm{2}} +{a}_{\mathrm{3}} \right)+...\frac{\lambda−\left({b}_{{n}−\mathrm{1}} +{b}_{{n}} \right)}{\left({b}_{{n}−\mathrm{1}} +{b}_{{n}} \right)}\left({a}_{{n}−\mathrm{1}} +{a}_{{n}} \right) \\ $$$$\geqslant\sqrt{\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}×\underset{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} {\Sigma\Sigma}\left({a}_{{i}} {a}_{{j}} \right)} \\ $$$$ \\ $$

Question Number 196861    Answers: 0   Comments: 0

Question Number 196860    Answers: 0   Comments: 0

Question Number 196852    Answers: 1   Comments: 0

Question Number 196850    Answers: 0   Comments: 0

Question Number 196848    Answers: 0   Comments: 1

Question Number 196846    Answers: 1   Comments: 0

2xy′′ + (1−4x)y′ + (2x−1)y = y

$$\mathrm{2}{xy}''\:+\:\left(\mathrm{1}−\mathrm{4}{x}\right){y}'\:+\:\left(\mathrm{2}{x}−\mathrm{1}\right){y}\:=\:{y} \\ $$

Question Number 196845    Answers: 0   Comments: 0

Question Number 196843    Answers: 1   Comments: 0

Question Number 196832    Answers: 0   Comments: 1

∫xe^(1/(2x)) dx=?

$$\int{x}\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}{x}}} {dx}=? \\ $$

Question Number 196829    Answers: 1   Comments: 0

Question Number 196828    Answers: 2   Comments: 0

Question Number 196817    Answers: 1   Comments: 0

Question Number 196816    Answers: 1   Comments: 0

Question Number 196815    Answers: 1   Comments: 0

!6×(((!5+9!!!!!+7!!!−16))^(1/4) /(!10))=?

$$!\mathrm{6}×\frac{\sqrt[{\mathrm{4}}]{!\mathrm{5}+\mathrm{9}!!!!!+\mathrm{7}!!!−\mathrm{16}}}{!\mathrm{10}}=? \\ $$

Question Number 196836    Answers: 1   Comments: 1

Question Number 196806    Answers: 2   Comments: 0

lim_(x→0) ( ((((√(2x−2x^2 )))^(1/3) − x)/(2x+ ((1−((x^3 +1))^(1/3) ))^(1/3) )) ) .

$$\:\:\:\: \\ $$$$ \underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\:\frac{\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{2x}−\mathrm{2x}^{\mathrm{2}} }}\:−\:\mathrm{x}}{\mathrm{2x}+\:\sqrt[{\mathrm{3}}]{\mathrm{1}−\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}}}\:\right)\:. \\ $$

Question Number 196793    Answers: 1   Comments: 0

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