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Question Number 227612 Answers: 0 Comments: 0

m_− R (R/m_− ) ⇔

$$ \\ $$$$\:\:\:\:\underset{−} {{m}}\: {R} \\ $$$$\:\:\:\:\:\:\: \:\:\frac{{R}}{\underset{−} {{m}}}\:\:\:\:\Leftrightarrow \: \\ $$

Question Number 227605 Answers: 0 Comments: 0

the two roots of eq x^2 +2ax−2b^2 =0 are α&β a,b are rational but a^2 +b^2 is not complete square. make a quad eq whose one root is α+β+(√(α^2 +β^2 ))

$${the}\:{two}\:{roots}\:{of}\:{eq}\:{x}^{\mathrm{2}} +\mathrm{2}{ax}−\mathrm{2}{b}^{\mathrm{2}} =\mathrm{0}\:{are} \\ $$$$\alpha\&\beta \\ $$$${a},{b}\:{are}\:{rational}\:{but}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:{is}\:{not}\:{complete} \\ $$$${square}.\:{make}\:{a}\:{quad}\:{eq}\:{whose}\:{one}\:{root}\:{is}\: \\ $$$$\alpha+\beta+\sqrt{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} } \\ $$

Question Number 227607 Answers: 2 Comments: 0

Question Number 227592 Answers: 0 Comments: 8

∀ε>0,∃δ>0 s.t. ∣h∣<δ ⇒ ∣((e^(x_0 +h) −e^x_0 )/h)−e^x_0 ∣<ε... (Don′t use Taylar/Maclaurin Series, and derivation only use inequality properties Damn.....i stuck here again .. e^x_0 ∣((e^h −1)/h)−1∣<𝛆 ..... fuck ∣((e^h −1)/h)−1∣ how can i show that h+1<e^h <(1/(1−h)) , ∣h∣<1

$$\forall\epsilon>\mathrm{0},\exists\delta>\mathrm{0}\:\mathrm{s}.\mathrm{t}.\:\mid{h}\mid<\delta\:\Rightarrow\:\mid\frac{{e}^{{x}_{\mathrm{0}} +{h}} −{e}^{{x}_{\mathrm{0}} } }{{h}}−{e}^{{x}_{\mathrm{0}} } \mid<\epsilon... \\ $$$$\left(\mathrm{Don}'\mathrm{t}\:\mathrm{use}\:\mathrm{Taylar}/\mathrm{Maclaurin}\:\mathrm{Series},\:\mathrm{and}\:\mathrm{derivation}\right. \\ $$$$\mathrm{only}\:\mathrm{use}\:\mathrm{inequality}\:\mathrm{properties} \\ $$$$\mathrm{Damn}.....\mathrm{i}\:\mathrm{stuck}\:\mathrm{here}\:\mathrm{again}\:..\:\: \\ $$$${e}^{{x}_{\mathrm{0}} } \mid\frac{{e}^{{h}} −\mathrm{1}}{{h}}−\mathrm{1}\mid<\boldsymbol{\epsilon}\:\:..... \\ $$$$\mathrm{fuck}\:\mid\frac{{e}^{{h}} −\mathrm{1}}{{h}}−\mathrm{1}\mid\: \\ $$$$\mathrm{how}\:\mathrm{can}\:\mathrm{i}\:\mathrm{show}\:\mathrm{that}\:{h}+\mathrm{1}<{e}^{{h}} <\frac{\mathrm{1}}{\mathrm{1}−{h}}\:,\:\mid{h}\mid<\mathrm{1} \\ $$

Question Number 227582 Answers: 1 Comments: 0

cos^2 (((3x)/2) + ((11π)/2)) - sin^2 (((3x)/2) + ((11π)/2)) = - (1/2) x = ?

$$\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{3x}}{\mathrm{2}}\:+\:\frac{\mathrm{11}\pi}{\mathrm{2}}\right)\:-\:\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{3x}}{\mathrm{2}}\:+\:\frac{\mathrm{11}\pi}{\mathrm{2}}\right)\:=\:-\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{x}\:=\:? \\ $$

Question Number 227580 Answers: 0 Comments: 0

(π/4)arctan(tan((7π)/8))+tan2x=(1/2)cos(arccos(−(1/2))+(π/3)) x = ?

$$\frac{\pi}{\mathrm{4}}\mathrm{arctan}\left(\mathrm{tan}\frac{\mathrm{7}\pi}{\mathrm{8}}\right)+\mathrm{tan2x}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{arccos}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\pi}{\mathrm{3}}\right) \\ $$$$\mathrm{x}\:=\:? \\ $$

Question Number 227579 Answers: 1 Comments: 0

cos6x = ((2tan(π/8))/(1−tan^2 (π/8))) ⇒ x = ?

$$\mathrm{cos6x}\:=\:\frac{\mathrm{2tan}\frac{\pi}{\mathrm{8}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}}\:\:\:\:\:\Rightarrow\:\:\:\:\mathrm{x}\:=\:? \\ $$

Question Number 227578 Answers: 1 Comments: 0

sin3x + sin^2 (π/4) = cos^2 (π/4) + ((√3)/2) x = ?

$$\mathrm{sin3x}\:+\:\mathrm{sin}^{\mathrm{2}} \frac{\pi}{\mathrm{4}}\:=\:\mathrm{cos}^{\mathrm{2}} \frac{\pi}{\mathrm{4}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{x}\:=\:? \\ $$

Question Number 227567 Answers: 0 Comments: 0

Question Number 227568 Answers: 1 Comments: 1

AB and AC are two plane mirrors ∠BAC=α P is a point on AB and a ray falls on AC at i=γ and after n times of reflection the ray becomes parrallel to AC find γ in terms of α & n

$${AB}\:{and}\:{AC}\:{are}\:{two}\:{plane}\:{mirrors} \\ $$$$\angle{BAC}=\alpha \\ $$$${P}\:{is}\:{a}\:{point}\:{on}\:{AB}\:{and}\:{a}\:{ray}\:{falls}\:{on}\:{AC} \\ $$$${at}\:{i}=\gamma\:{and}\:{after}\:{n}\:{times}\:{of}\:{reflection} \\ $$$${the}\:{ray}\:{becomes}\:{parrallel}\:{to}\:{AC} \\ $$$${find}\:\gamma\:{in}\:{terms}\:\:{of}\:\alpha\:\&\:{n} \\ $$

Question Number 227561 Answers: 1 Comments: 1

Question Number 227534 Answers: 0 Comments: 8

Post some of the hardests problem you think fall within thescope of high school mathp cometitions under this thread. :)

$$\mathrm{Post}\:\mathrm{some}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hardests} \\ $$$$\mathrm{problem}\:\mathrm{you}\:\mathrm{think}\:\mathrm{fall}\:\mathrm{within}\: \\ $$$$\mathrm{thescope}\:\mathrm{of}\:\mathrm{high}\:\mathrm{school}\:\mathrm{mathp} \\ $$$$\mathrm{cometitions}\:\mathrm{under}\:\mathrm{this}\:\mathrm{thread}. \\ $$$$\left.:\right) \\ $$

Question Number 227533 Answers: 2 Comments: 0

Question Number 227531 Answers: 0 Comments: 0

Question Number 227530 Answers: 0 Comments: 0

Find all real numbers lambdac suh that there exists a positivet ineger N satisfying that for any integer n≥N and any nonnegative realr numbes x_1 x_2 ,…,x_n with sum 1 we have (Σ_(1≤i≤j≤n) x_(i ) x_(j ) )^2 ≤λΣ_(1≤i≤j≤k≤n) x_i x_j x_(k ) +(1/n)Σ_(1≤i≤j≤n) (x_i −x_j )^2 .

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{lambdac} \\ $$$$\mathrm{suh}\:\mathrm{that}\:\mathrm{there}\:\mathrm{exists}\:\mathrm{a}\:\mathrm{positivet} \\ $$$$\mathrm{ineger}\:{N}\:\mathrm{satisfying}\:\mathrm{that}\:\mathrm{for}\:\mathrm{any}\:\mathrm{integer}\:{n}\geq{N}\: \\ $$$$\mathrm{and}\:\mathrm{any}\:\mathrm{nonnegative}\:\mathrm{realr} \\ $$$$\mathrm{numbes}\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} ,\ldots,{x}_{{n}} \mathrm{with}\:\mathrm{sum}\:\mathrm{1}\:\mathrm{we}\:\mathrm{have} \\ $$$$\left(\underset{\mathrm{1}\leq{i}\leq{j}\leq{n}} {\sum}{x}_{{i}\:} {x}_{{j}\:} \right)^{\mathrm{2}} \leq\lambda\underset{\mathrm{1}\leq{i}\leq{j}\leq{k}\leq{n}} {\sum}{x}_{{i}} {x}_{{j}} {x}_{{k}\:} +\frac{\mathrm{1}}{{n}}\underset{\mathrm{1}\leq{i}\leq{j}\leq{n}} {\sum}\left({x}_{{i}} −{x}_{{j}} \right)^{\mathrm{2}} . \\ $$

Question Number 227529 Answers: 0 Comments: 0

There are cards in 30 different colors, with 70 cards of each color. Initially, 70 cards are selected and laid out in a stack from top to bottom. One operation consists of: choosing an X from the bottom 20 cards, such that its color has not appeared in the top 50 cards; then choosing a Y from the top 50 cards, such that its color has appeared at least twice in the top 50 cards; removing X and placing it right above Y. Repeat this process until no more valid X can be chosen. (1) Prove that the number of operations is finite. (2) For all possible initial configurations and sequences of operations, find the maximum possible number of operations.

$$ \\ $$There are cards in 30 different colors, with 70 cards of each color. Initially, 70 cards are selected and laid out in a stack from top to bottom. One operation consists of: choosing an X from the bottom 20 cards, such that its color has not appeared in the top 50 cards; then choosing a Y from the top 50 cards, such that its color has appeared at least twice in the top 50 cards; removing X and placing it right above Y. Repeat this process until no more valid X can be chosen. (1) Prove that the number of operations is finite. (2) For all possible initial configurations and sequences of operations, find the maximum possible number of operations.

Question Number 227528 Answers: 0 Comments: 0

Question Number 227527 Answers: 0 Comments: 0

Question Number 227526 Answers: 1 Comments: 0

9^(sin^2 x) +9^(cos^2 x) =6 x=?

$$\mathrm{9}^{\mathrm{sin}\:^{\mathrm{2}} {x}} +\mathrm{9}^{\mathrm{cos}\:^{\mathrm{2}} {x}} =\mathrm{6} \\ $$$${x}=? \\ $$

Question Number 227521 Answers: 0 Comments: 10

Question Number 227514 Answers: 0 Comments: 0

e^x

$${e}^{{x}} \\ $$

Question Number 227509 Answers: 2 Comments: 2

PLS HELP!!!!! we want to prove lim_(h→0^± ) (((x+h)^n −x^n )/h)=n∙x^(n−1) by using the 𝛆-𝛅 argument def. ∀𝛆>0 , ∃𝛅>0 s.t. ∣x−α∣<𝛅 ⇒∣f(x)−L∣<𝛆 Now, we want to show that for all 𝛆>0 , Exist 𝛅>0 s.t. ∣x−α∣<𝛅 Implies ∣((x^n −α^n )/(x−α))−n∙α^(n−1) ∣<𝛆 But i having a trouble solving this inequality ∣((x^n −α^n )/(x−α))−n∙α^(n−1) ∣<𝛆..... our purpose is: to simplify the Expression ∣((x^n −α^n )/(x−α))−n∙α^(n−1) ∣into the form of K∙∣x−α∣ (K is Constant) and find 𝛅=(1/K)𝛆

$$\boldsymbol{\mathrm{PLS}}\:\boldsymbol{\mathrm{HELP}}!!!!! \\ $$$$\mathrm{we}\:\mathrm{want}\:\mathrm{to}\:\mathrm{prove}\:\underset{{h}\rightarrow\mathrm{0}^{\pm} } {\mathrm{lim}}\:\frac{\left({x}+{h}\right)^{{n}} −{x}^{{n}} }{{h}}={n}\centerdot{x}^{{n}−\mathrm{1}} \:\mathrm{by}\:\mathrm{using}\:\mathrm{the}\:\boldsymbol{\epsilon}-\boldsymbol{\delta}\:\mathrm{argument} \\ $$$$\mathrm{def}.\:\forall\boldsymbol{\epsilon}>\mathrm{0}\:,\:\exists\boldsymbol{\delta}>\mathrm{0}\:\mathrm{s}.\mathrm{t}.\:\mid{x}−\alpha\mid<\boldsymbol{\delta}\:\Rightarrow\mid{f}\left({x}\right)−{L}\mid<\boldsymbol{\epsilon} \\ $$$$\mathrm{Now},\:\mathrm{we}\:\mathrm{want}\:\mathrm{to}\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\mathrm{for}\:\mathrm{all}\:\boldsymbol{\epsilon}>\mathrm{0}\:,\:\mathrm{Exist}\:\boldsymbol{\delta}>\mathrm{0}\:\mathrm{s}.\mathrm{t}.\:\mid{x}−\alpha\mid<\boldsymbol{\delta}\:\:\mathrm{Implies}\:\mid\frac{{x}^{{n}} −\alpha^{{n}} }{{x}−\alpha}−{n}\centerdot\alpha^{{n}−\mathrm{1}} \mid<\boldsymbol{\epsilon} \\ $$$$\mathrm{But}\:\mathrm{i}\:\mathrm{having}\:\mathrm{a}\:\mathrm{trouble}\:\mathrm{solving}\:\mathrm{this}\:\:\mathrm{inequality}\:\:\mid\frac{{x}^{{n}} −\alpha^{{n}} }{{x}−\alpha}−{n}\centerdot\alpha^{{n}−\mathrm{1}} \mid<\boldsymbol{\epsilon}..... \\ $$$$\mathrm{our}\:\mathrm{purpose}\:\mathrm{is}: \\ $$$$\mathrm{to}\:\mathrm{simplify}\:\mathrm{the}\:\mathrm{Expression}\:\mid\frac{{x}^{{n}} −\alpha^{{n}} }{{x}−\alpha}−{n}\centerdot\alpha^{{n}−\mathrm{1}} \mid\mathrm{into}\:\mathrm{the}\:\mathrm{form}\:\mathrm{of} \\ $$$${K}\centerdot\mid{x}−\alpha\mid\:\left({K}\:\mathrm{is}\:\mathrm{Constant}\right) \\ $$$$\mathrm{and}\:\mathrm{find}\:\boldsymbol{\delta}=\frac{\mathrm{1}}{{K}}\boldsymbol{\epsilon} \\ $$

Question Number 227501 Answers: 1 Comments: 0

Question Number 227500 Answers: 0 Comments: 0

Question Number 227499 Answers: 1 Comments: 0

Question Number 227495 Answers: 0 Comments: 1

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