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Question Number 222697    Answers: 1   Comments: 0

Question Number 222685    Answers: 2   Comments: 0

Question Number 222684    Answers: 1   Comments: 1

Question Number 222679    Answers: 2   Comments: 0

If x=Π_(x=1) ^(10) x then (1/(log _2 x))+(1/(log _3 x))+(1/(log _4 x))...+(1/(log _(10) x))=??

$${If}\:{x}=\underset{{x}=\mathrm{1}} {\overset{\mathrm{10}} {\prod}}{x}\:{then}\:\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{2}} {x}}+\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{3}} {x}}+\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{4}} {x}}...+\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{10}} {x}}=?? \\ $$

Question Number 222673    Answers: 1   Comments: 1

Question Number 222662    Answers: 1   Comments: 4

Question Number 222652    Answers: 0   Comments: 1

Question Number 222646    Answers: 2   Comments: 0

Question Number 222639    Answers: 2   Comments: 0

Question Number 222638    Answers: 1   Comments: 0

Question Number 222636    Answers: 3   Comments: 0

Question Number 222635    Answers: 1   Comments: 0

Question Number 222622    Answers: 0   Comments: 0

Question Number 222619    Answers: 2   Comments: 0

Question Number 222614    Answers: 0   Comments: 0

complicated solution of the integral ; ∫ tan(((cos x)/x))^(13) dx

$$ \\ $$$$\:\:\mathrm{complicated}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{integral}\:; \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\mathrm{tan}\left(\frac{\mathrm{cos}\:{x}}{{x}}\right)^{\mathrm{13}} \:\mathrm{d}{x} \\ $$$$ \\ $$

Question Number 222613    Answers: 0   Comments: 0

what is I = ∫ tan (((cos x)/x)) dx

$$ \\ $$$$\:\:\:\:\:\:\mathrm{what}\:\mathrm{is}\:\:{I}\:=\:\int\:\mathrm{tan}\:\left(\frac{\mathrm{cos}\:{x}}{{x}}\right)\:\mathrm{d}{x}\: \\ $$$$ \\ $$

Question Number 222616    Answers: 1   Comments: 0

Question Number 222610    Answers: 0   Comments: 0

Σ_(n=1) ^∞ (1/(n(e^(2πn) − 1)))

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\left({e}^{\mathrm{2}\pi{n}} \:−\:\mathrm{1}\right)} \\ $$$$ \\ $$

Question Number 222624    Answers: 1   Comments: 0

Question Number 222607    Answers: 1   Comments: 0

Prove:∫_0 ^1 (x^3 /(5−x^3 ))∙(1/( ((1−x^3 ))^(1/3) ))dx=((((10))^(1/3) −2)/(3(√3)))π

$$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{3}} }{\mathrm{5}−{x}^{\mathrm{3}} }\centerdot\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}−{x}^{\mathrm{3}} }}{dx}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{10}}−\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\pi \\ $$

Question Number 222601    Answers: 0   Comments: 1

L 60 1. y=x^2 +5x Find the equation of a line with the slope of 7 that touches y=x^2 +5x. [Sol.] Let f(x)=x^2 +5x Then f′(x)=2x+5 Since 2x+5=7⇒x=1 then the point is (1, 1^2 +5∙1)=(1, 6) So the equation of a line is y−6=7(x−1)⇒y=7x−1 2. y=ax^2 +bx (2, 2) a, b Find the values of constants a, b that the slope of the line that touches (2, 2) and y=ax^2 +bx is 5. [Sol.] Let f(x)=ax^2 +bx Then f′(x)=2ax+b and build two equations to solve for a and b { ((f(2)=a∙2^2 +b∙2=4a+2b=2)),((f′(2)=2a∙2+b=4a+b=5)) :} Solving for a, b gives a=2, b=−3 3. y=x^3 −3x^2 −1 Find the equation of a line that is drawn, touches y=x^3 −3x^2 −1. [Sol.] The line of the equation is y−(a^3 −3a^2 −1)=(3a^2 −6a)(x−a) Calculating gives y=(3a^2 −6a)x−(3a^2 −6a)a+(a^3 −3a^2 −1) y=(3a^2 −6a)x+(−3a^3 +6a^2 )+(a^3 −3a^2 −1) y=(3a^2 −6a)x+(−2a^3 +3a^2 −1) −2a^3 +3a^2 −1=0 a=−(1/2) or a=2 ...a=−3x, a=((15)/4)x

$$\mathrm{L}\:\mathrm{60} \\ $$$$ \\ $$$$\mathrm{1}.\: {y}={x}^{\mathrm{2}} +\mathrm{5}{x} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{line}\:\mathrm{with}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{of}\:\mathrm{7}\:\mathrm{that}\:\mathrm{touches}\:{y}={x}^{\mathrm{2}} +\mathrm{5}{x}. \\ $$$$\left[\mathrm{Sol}.\right]\:\mathrm{Let}\:{f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{5}{x}\:\mathrm{Then}\:{f}'\left({x}\right)=\mathrm{2}{x}+\mathrm{5} \\ $$$$\mathrm{Since}\:\mathrm{2}{x}+\mathrm{5}=\mathrm{7}\Rightarrow{x}=\mathrm{1}\:\mathrm{then}\:\mathrm{the}\:\mathrm{point}\:\mathrm{is}\:\left(\mathrm{1},\:\mathrm{1}^{\mathrm{2}} +\mathrm{5}\centerdot\mathrm{1}\right)=\left(\mathrm{1},\:\mathrm{6}\right) \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{line}\:\mathrm{is}\:{y}−\mathrm{6}=\mathrm{7}\left({x}−\mathrm{1}\right)\Rightarrow{y}=\mathrm{7}{x}−\mathrm{1} \\ $$$$ \\ $$$$\mathrm{2}.\: \:{y}={ax}^{\mathrm{2}} +{bx}\: \:\left(\mathrm{2},\:\mathrm{2}\right) \:{a},\:{b} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{constants}\:{a},\:{b}\:\mathrm{that}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line}\:\mathrm{that}\:\mathrm{touches}\:\left(\mathrm{2},\:\mathrm{2}\right)\:\mathrm{and}\:{y}={ax}^{\mathrm{2}} +{bx}\:\mathrm{is}\:\mathrm{5}. \\ $$$$\left[\mathrm{Sol}.\right]\:\mathrm{Let}\:{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}\:\mathrm{Then}\:{f}'\left({x}\right)=\mathrm{2}{ax}+{b}\:\mathrm{and}\:\mathrm{build}\:\mathrm{two}\:\mathrm{equations}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{for}\:{a}\:\mathrm{and}\:{b} \\ $$$$\begin{cases}{{f}\left(\mathrm{2}\right)={a}\centerdot\mathrm{2}^{\mathrm{2}} +{b}\centerdot\mathrm{2}=\mathrm{4}{a}+\mathrm{2}{b}=\mathrm{2}}\\{{f}'\left(\mathrm{2}\right)=\mathrm{2}{a}\centerdot\mathrm{2}+{b}=\mathrm{4}{a}+{b}=\mathrm{5}}\end{cases} \\ $$$$\mathrm{Solving}\:\mathrm{for}\:{a},\:{b}\:\mathrm{gives}\:{a}=\mathrm{2},\:{b}=−\mathrm{3} \\ $$$$ \\ $$$$\mathrm{3}.\: \:{y}={x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{line}\:\mathrm{that}\:\mathrm{is}\:\mathrm{drawn},\:\mathrm{touches}\:{y}={x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}. \\ $$$$\left[\mathrm{Sol}.\right]\:\mathrm{The}\:\mathrm{line}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{is}\:{y}−\left({a}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} −\mathrm{1}\right)=\left(\mathrm{3}{a}^{\mathrm{2}} −\mathrm{6}{a}\right)\left({x}−{a}\right) \\ $$$$\mathrm{Calculating}\:\mathrm{gives}\:{y}=\left(\mathrm{3}{a}^{\mathrm{2}} −\mathrm{6}{a}\right){x}−\left(\mathrm{3}{a}^{\mathrm{2}} −\mathrm{6}{a}\right){a}+\left({a}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${y}=\left(\mathrm{3}{a}^{\mathrm{2}} −\mathrm{6}{a}\right){x}+\left(−\mathrm{3}{a}^{\mathrm{3}} +\mathrm{6}{a}^{\mathrm{2}} \right)+\left({a}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${y}=\left(\mathrm{3}{a}^{\mathrm{2}} −\mathrm{6}{a}\right){x}+\left(−\mathrm{2}{a}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$−\mathrm{2}{a}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${a}=−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{or}\:{a}=\mathrm{2} \\ $$$$...{a}=−\mathrm{3}{x},\:{a}=\frac{\mathrm{15}}{\mathrm{4}}{x} \\ $$

Question Number 222599    Answers: 0   Comments: 1

Σ_(k=1) ^∞ (−1)^(k+1) (1/((2k−1)^(2n+1) ))=(π^(2n+1) /(2^(2n+2) (2n)!))∣E_(2m) ∣

$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }=\frac{\pi^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} \left(\mathrm{2}{n}\right)!}\mid{E}_{\mathrm{2}{m}} \mid \\ $$

Question Number 222592    Answers: 0   Comments: 0

Σ_(n = 1) ^∞ (1/(cosh^4 [n ln((√2) + 1)]))

$$\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{cosh}^{\mathrm{4}} \left[\mathrm{n}\:\mathrm{ln}\left(\sqrt{\mathrm{2}}\:\:+\:\:\mathrm{1}\right)\right]} \\ $$

Question Number 222585    Answers: 3   Comments: 2

Question Number 222584    Answers: 0   Comments: 0

find the correct const. to preconst. “ x^n −2=−x ”

$${find}\:{the}\:{correct}\:{const}.\:{to}\:{preconst}.\:``\:{x}^{{n}} −\mathrm{2}=−{x}\:'' \\ $$

Question Number 222582    Answers: 2   Comments: 0

If: a_i > 0 , b_i > 0 , i = 1,...,n^(−) Prove that: (√(a_1 ^2 + b_1 ^2 )) + (√(a_2 ^2 + b_2 ^2 )) +...+ (√(a_n ^2 + b_n ^2 )) ≥ ≥ (√((a_1 +a_2 +...+a_n )^2 + (b_1 +b_2 +...+b_n )^2 ))

$$\mathrm{If}:\:\:\:\mathrm{a}_{\boldsymbol{\mathrm{i}}} \:>\:\mathrm{0}\:\:\:,\:\:\:\mathrm{b}_{\boldsymbol{\mathrm{i}}} \:>\:\mathrm{0}\:\:\:,\:\:\:\mathrm{i}\:=\:\overline {\mathrm{1},...,\mathrm{n}} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\sqrt{\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\mathrm{1}} ^{\mathrm{2}} }\:+\:\sqrt{\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\mathrm{2}} ^{\mathrm{2}} }\:+...+\:\sqrt{\mathrm{a}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} }\:\:\:\geqslant \\ $$$$\geqslant\:\sqrt{\left(\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} +...+\mathrm{a}_{\boldsymbol{\mathrm{n}}} \right)^{\mathrm{2}} \:+\:\left(\mathrm{b}_{\mathrm{1}} +\mathrm{b}_{\mathrm{2}} +...+\mathrm{b}_{\boldsymbol{\mathrm{n}}} \right)^{\mathrm{2}} } \\ $$

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