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AlgebraQuestion and Answers: Page 64

Question Number 197419 Answers: 1 Comments: 0

Question Number 197409 Answers: 1 Comments: 4

Question Number 197393 Answers: 1 Comments: 0

Question Number 197360 Answers: 1 Comments: 0

Find: ∫_0 ^( ∞) sin^2 ( (√x) ) e^(−x) dx = ?

$$\mathrm{Find}: \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{sin}^{\mathrm{2}} \:\left(\:\sqrt{\mathrm{x}}\:\right)\:\mathrm{e}^{−\boldsymbol{\mathrm{x}}} \:\mathrm{dx}\:=\:? \\ $$

Question Number 197312 Answers: 1 Comments: 0

(((log_2 20)^2 −(log_2 5)^2 )/(log_2 10))=?

$$\frac{\left({log}_{\mathrm{2}} \mathrm{20}\right)^{\mathrm{2}} −\left({log}_{\mathrm{2}} \mathrm{5}\right)^{\mathrm{2}} }{{log}_{\mathrm{2}} \mathrm{10}}=? \\ $$

Question Number 197275 Answers: 1 Comments: 0

how do i prove this, help please. ∣((x^2 −2x−3)/(x^2 +2x+4))∣≤(5/4),∣x∣≤2

$$ \\ $$$$\:{how}\:{do}\:{i}\:{prove}\:{this},\:{help}\:{please}. \\ $$$$\:\mid\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}\mid\leqslant\frac{\mathrm{5}}{\mathrm{4}},\mid{x}\mid\leqslant\mathrm{2} \\ $$$$ \\ $$$$ \\ $$

Question Number 197248 Answers: 0 Comments: 3

a,b,c∈R ((a + 2b − 3ac)/(3ac)) = ((a + 4b − bc)/b) Find: ((2b)/a) − ((3a)/b)

$$\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{R} \\ $$$$\frac{\mathrm{a}\:+\:\mathrm{2b}\:−\:\mathrm{3ac}}{\mathrm{3ac}}\:\:=\:\:\frac{\mathrm{a}\:+\:\mathrm{4b}\:−\:\mathrm{bc}}{\mathrm{b}} \\ $$$$\mathrm{Find}:\:\:\:\frac{\mathrm{2b}}{\mathrm{a}}\:−\:\frac{\mathrm{3a}}{\mathrm{b}} \\ $$

Question Number 197171 Answers: 1 Comments: 0

2^(log_3 (x^2 +1)) +2×(x^2 +1)^(log_3 2) =12 ⇒x=?

$$\mathrm{2}^{{log}_{\mathrm{3}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)} +\mathrm{2}×\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{log}_{\mathrm{3}} \mathrm{2}} \:=\mathrm{12} \\ $$$$\Rightarrow{x}=? \\ $$$$ \\ $$

Question Number 197125 Answers: 1 Comments: 1

Question Number 197064 Answers: 2 Comments: 0

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Question Number 196964 Answers: 2 Comments: 0

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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