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Question Number 157053 Answers: 0 Comments: 2

If you want to easily write any quadratic function in vertex form, just use these formulas: f(x)=ax^2 +bx+c if a>0, then: f(x)=(x+(b/2))^2 −((b/2))^2 +c if a<0, then: f(x)=−(x−((b/2)))^2 +((b/2))^2 +c Let′s take some examples: f(x)=x^2 −6x+7 f(x)=(x−(6/2))^2 −((6/2))^2 +7 f(x)=(x−3)^2 −9+7 f(x)=(x−3)^2 −2 Now let′s take the same example, but when a<0: f(x)=−x^2 −6x+7 f(x)=−(x−(−(6/2)))^2 +((6/2))^2 +7 f(x)=−(x+3)^2 +9+7 f(x)=−(x+3)^2 +16 Another example: f(x)=x^2 +4x−5 f(x)=(x+(4/2))^2 −((4/2))^2 −5 f(x)=(x+2)^2 −4−5 f(x)=(x+2)^2 −9 Now let′s also see what happens when a<0: f(x)=−x^2 +4x−5 f(x)=−(x−((4/2)))^2 +((4/2))^2 −5 f(x)=−(x−2)^2 +4−5 f(x)=−(x−2)^2 −1

$$\mathrm{If}\:\mathrm{you}\:\mathrm{want}\:\mathrm{to}\:\mathrm{easily}\:\mathrm{write}\:\mathrm{any}\:\mathrm{quadratic}\:\mathrm{function}\:\mathrm{in}\:\mathrm{vertex}\:\mathrm{form},\:\mathrm{just}\:\mathrm{use}\:\mathrm{these}\:\mathrm{formulas}: \\ $$$$\: \\ $$$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$\: \\ $$$$\mathrm{if}\:{a}>\mathrm{0},\:\mathrm{then}: \\ $$$${f}\left({x}\right)=\left({x}+\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} +{c} \\ $$$$\: \\ $$$$\mathrm{if}\:{a}<\mathrm{0},\:\mathrm{then}: \\ $$$${f}\left({x}\right)=−\left({x}−\left(\frac{{b}}{\mathrm{2}}\right)\right)^{\mathrm{2}} +\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} +{c} \\ $$$$\: \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{take}\:\mathrm{some}\:\mathrm{examples}: \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{7} \\ $$$${f}\left({x}\right)=\left({x}−\frac{\mathrm{6}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{6}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{7} \\ $$$${f}\left({x}\right)=\left({x}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{9}+\mathrm{7} \\ $$$${f}\left({x}\right)=\left({x}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{2} \\ $$$$\: \\ $$$$\mathrm{Now}\:\mathrm{let}'\mathrm{s}\:\mathrm{take}\:\mathrm{the}\:\mathrm{same}\:\mathrm{example},\:\mathrm{but}\:\mathrm{when}\:{a}<\mathrm{0}: \\ $$$$\: \\ $$$${f}\left({x}\right)=−{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{7} \\ $$$${f}\left({x}\right)=−\left({x}−\left(−\frac{\mathrm{6}}{\mathrm{2}}\right)\right)^{\mathrm{2}} +\left(\frac{\mathrm{6}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{7} \\ $$$${f}\left({x}\right)=−\left({x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{9}+\mathrm{7} \\ $$$${f}\left({x}\right)=−\left({x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{16} \\ $$$$\: \\ $$$$\mathrm{Another}\:\mathrm{example}: \\ $$$$\: \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{5} \\ $$$${f}\left({x}\right)=\left({x}+\frac{\mathrm{4}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{4}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{5} \\ $$$${f}\left({x}\right)=\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}−\mathrm{5} \\ $$$${f}\left({x}\right)=\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{9} \\ $$$$\: \\ $$$$\mathrm{Now}\:\mathrm{let}'\mathrm{s}\:\mathrm{also}\:\mathrm{see}\:\mathrm{what}\:\mathrm{happens}\:\mathrm{when}\:{a}<\mathrm{0}: \\ $$$$\: \\ $$$${f}\left({x}\right)=−{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{5} \\ $$$${f}\left({x}\right)=−\left({x}−\left(\frac{\mathrm{4}}{\mathrm{2}}\right)\right)^{\mathrm{2}} +\left(\frac{\mathrm{4}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{5} \\ $$$${f}\left({x}\right)=−\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{4}−\mathrm{5} \\ $$$${f}\left({x}\right)=−\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$\: \\ $$

Question Number 157046 Answers: 0 Comments: 1

q={a+((a/9)−(1/(108))−a^3 )^(1/3) }^(1/3) +{b+((b/9)−(1/(108))−b^3 )^(1/3) }^(1/3) a=((9+(√(37)))/(72)) , b=((9−(√(37)))/(72)) find q correct to 5 decimal places.

$${q}=\left\{{a}+\left(\frac{{a}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{108}}−{a}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{3}} \right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:+\left\{{b}+\left(\frac{{b}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{108}}−{b}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{3}} \right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:{a}=\frac{\mathrm{9}+\sqrt{\mathrm{37}}}{\mathrm{72}}\:\:\:,\:\:{b}=\frac{\mathrm{9}−\sqrt{\mathrm{37}}}{\mathrm{72}} \\ $$$$\:{find}\:\boldsymbol{{q}}\:{correct}\:{to}\:\mathrm{5}\:{decimal} \\ $$$$\:{places}. \\ $$

Question Number 157045 Answers: 1 Comments: 1

Question Number 157044 Answers: 1 Comments: 0

Solve for real numbers: (sin2x + 4cos^2 x + 1)(cos5x - cosx)<0

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\left(\mathrm{sin2}\boldsymbol{\mathrm{x}}\:+\:\mathrm{4cos}^{\mathrm{2}} \boldsymbol{\mathrm{x}}\:+\:\mathrm{1}\right)\left(\mathrm{cos5}\boldsymbol{\mathrm{x}}\:-\:\mathrm{cos}\boldsymbol{\mathrm{x}}\right)<\mathrm{0} \\ $$

Question Number 157035 Answers: 1 Comments: 4

Question Number 157033 Answers: 3 Comments: 1

y′′=−y

$${y}''=−{y} \\ $$

Question Number 157031 Answers: 0 Comments: 0

If xcos θ+ycos ∅+zcos ψ=0, xsin θ+ysin ∅+zsin ψ=0 and xsec θ+ysec ∅+zsec ψ=0 then prove that (x^2 +y^2 −z^2 )^2 = 4x^2 y^2

$$\mathrm{If}\:\:\mathrm{xcos}\:\theta+\mathrm{ycos}\:\emptyset+\mathrm{zcos}\:\psi=\mathrm{0}, \\ $$$$\:\:\:\:\:\:\mathrm{xsin}\:\theta+\mathrm{ysin}\:\emptyset+\mathrm{zsin}\:\psi=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{xsec}\:\theta+\mathrm{ysec}\:\emptyset+\mathrm{zsec}\:\psi=\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{z}^{\mathrm{2}} \right)^{\mathrm{2}} =\:\mathrm{4x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \\ $$

Question Number 157057 Answers: 1 Comments: 2

(1/7)+(1/(13))+(1/(19))=a find (2/7)+(4/(13))+(6/(19))=?

$$\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{13}}+\frac{\mathrm{1}}{\mathrm{19}}={a} \\ $$$${find}\:\frac{\mathrm{2}}{\mathrm{7}}+\frac{\mathrm{4}}{\mathrm{13}}+\frac{\mathrm{6}}{\mathrm{19}}=? \\ $$

Question Number 157021 Answers: 1 Comments: 0

Find: 𝛀(n) =∫_( 1) ^( n) ([x]^2 ∙{x} + [x]∙{x}^2 )dx n∈N ; [∗]-GIF ; {x}=x-[x]

$$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\left(\boldsymbol{\mathrm{n}}\right)\:=\underset{\:\mathrm{1}} {\overset{\:\boldsymbol{\mathrm{n}}} {\int}}\left(\left[\mathrm{x}\right]^{\mathrm{2}} \centerdot\left\{\mathrm{x}\right\}\:+\:\left[\mathrm{x}\right]\centerdot\left\{\mathrm{x}\right\}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$\mathrm{n}\in\mathbb{N}\:\:;\:\:\left[\ast\right]-\mathrm{GIF}\:\:;\:\:\left\{\mathrm{x}\right\}=\mathrm{x}-\left[\mathrm{x}\right] \\ $$

Question Number 157016 Answers: 1 Comments: 0

f(x)=(1/(1+2^x ))+(1/(1+3^x ))+(1/(4^x +1)) find ∫_1 ^5 f(x)dx+∫_(−5) ^(−1) f(x)dx

$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}^{{x}} }+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{3}^{{x}} }+\frac{\mathrm{1}}{\mathrm{4}^{{x}} +\mathrm{1}}\:\: \\ $$$${find}\:\:\:\int_{\mathrm{1}} ^{\mathrm{5}} {f}\left({x}\right){dx}+\int_{−\mathrm{5}} ^{−\mathrm{1}} {f}\left({x}\right){dx} \\ $$

Question Number 157010 Answers: 0 Comments: 3

Question Number 157007 Answers: 1 Comments: 1

Question Number 157006 Answers: 1 Comments: 0

(√(4+27(√(4+29(√(4+31(√(4+…))))))))=?

$$\sqrt{\mathrm{4}+\mathrm{27}\sqrt{\mathrm{4}+\mathrm{29}\sqrt{\mathrm{4}+\mathrm{31}\sqrt{\mathrm{4}+\ldots}}}}=? \\ $$$$ \\ $$

Question Number 156992 Answers: 1 Comments: 1

Question Number 156991 Answers: 1 Comments: 0

Question Number 156979 Answers: 2 Comments: 2

Question Number 156977 Answers: 0 Comments: 0

{ ((a_1 =((√3)/2))),((a_(n+1) =4a_n ^3 −3a_n ; ∀n≥1)) :} a_n =?

$$\:\begin{cases}{{a}_{\mathrm{1}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\\{{a}_{{n}+\mathrm{1}} =\mathrm{4}{a}_{{n}} ^{\mathrm{3}} −\mathrm{3}{a}_{{n}} \:;\:\forall{n}\geqslant\mathrm{1}}\end{cases} \\ $$$$\:{a}_{{n}} =? \\ $$

Question Number 156973 Answers: 0 Comments: 1

Question Number 156993 Answers: 0 Comments: 3

lim_(n→∞) (sin(sin(sin…(sin(x))…)_(n) (√n)=? 0<x<π

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{n}} {\underbrace{\left({sin}\left({sin}\left({sin}\ldots\left({sin}\left({x}\right)\right)\ldots\right)}}\:\sqrt{{n}}=?\right.\right. \\ $$$$\mathrm{0}<{x}<\pi \\ $$

Question Number 157003 Answers: 1 Comments: 0

let n∈Z^+ shov that ∫_( 0) ^( ∞) ((sin(x^(-n) )ln(x))/x) dx = ((π𝛄)/(2n^2 )) where 𝛄 is the Euler-Mascheroni constan

$$\mathrm{let}\:\:\boldsymbol{\mathrm{n}}\in\mathbb{Z}^{+} \\ $$$$\mathrm{shov}\:\mathrm{that}\:\:\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{sin}\left(\mathrm{x}^{-\boldsymbol{\mathrm{n}}} \right)\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}}\:\mathrm{dx}\:=\:\frac{\pi\boldsymbol{\gamma}}{\mathrm{2n}^{\mathrm{2}} }\: \\ $$$$\mathrm{where}\:\:\boldsymbol{\gamma}\:\:\mathrm{is}\:\mathrm{the}\:\mathrm{Euler}-\mathrm{Mascheroni}\:\mathrm{constan}\: \\ $$

Question Number 156969 Answers: 0 Comments: 0

Question Number 156968 Answers: 0 Comments: 0

Question Number 156966 Answers: 2 Comments: 1

Question Number 156962 Answers: 1 Comments: 0

Solve for real numbers: (3/( ((1 + x))^(1/3) )) + (x/( ((1 + x^3 ))^(1/3) )) = 2 (4)^(1/3)

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}\:+\:\mathrm{x}}}\:+\:\frac{\mathrm{x}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{3}} }}\:=\:\mathrm{2}\:\sqrt[{\mathrm{3}}]{\mathrm{4}} \\ $$

Question Number 156961 Answers: 0 Comments: 0

𝛀 =∫_( 0) ^( ∞) ((cos^2 (x) - sin^2 (x))/((1 + x^4 )^3 )) dx = ?

$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)\:-\:\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)}{\left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{4}} \right)^{\mathrm{3}} }\:\mathrm{dx}\:=\:? \\ $$

Question Number 156951 Answers: 1 Comments: 7

solve for n∈N (n−1)!+1=n^2

$${solve}\:{for}\:{n}\in{N} \\ $$$$\left({n}−\mathrm{1}\right)!+\mathrm{1}={n}^{\mathrm{2}} \\ $$

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