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Question Number 220110    Answers: 2   Comments: 0

Question Number 220108    Answers: 7   Comments: 0

Question Number 220104    Answers: 1   Comments: 0

Question Number 220103    Answers: 8   Comments: 0

Question Number 220101    Answers: 0   Comments: 0

Question Number 220097    Answers: 1   Comments: 0

x ∈ Q ; x ≠ 1 (7/(x − 1)) + (6/x) − (4/(x + 1)) + ((3x + 5)/(x^2 − 1)) = (1/x)

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:\in\:\mathbb{Q}\:\:\:;\:\:\:\:{x}\:\neq\:\mathrm{1} \\ $$$$\:\frac{\mathrm{7}}{{x}\:−\:\mathrm{1}}\:+\:\frac{\mathrm{6}}{{x}}\:−\:\frac{\mathrm{4}}{{x}\:+\:\mathrm{1}}\:+\:\frac{\mathrm{3}{x}\:+\:\mathrm{5}}{{x}^{\mathrm{2}} \:−\:\mathrm{1}}\:=\:\frac{\mathrm{1}}{{x}} \\ $$$$\:\:\:\:\:\:\: \\ $$

Question Number 220096    Answers: 1   Comments: 0

let a, b, c, d, e is a positive real numbers and K = a + b + c + d + e +1 . prove that; Σ_(cyc) (1/(k−a)) < (1/4) ((((e^3 d^3 c))^(1/(4 )) /(c^(3/4) d^(1/2) e^(1/4) (√a))) + (((d^( 3) c^2 b))^(1/(4 )) /(d^( 3/4) c^(1/2) b^(1/4) (√e))) + (((c^3 b^2 a))^(1/(4 )) /(c^(3/4) b^(1/2) a^(1/4) (√d))) + (((b^3 a^2 e))^(1/(4 )) /(b^(3/4) a^(1/2) e^(1/4) (√c))) + (((a^3 e^2 d))^(1/(4 )) /(a^(3/4) e^(1/2) d^(1/4) (√b))))

$$ \\ $$$$\:\:\:\mathrm{let}\:{a},\:{b},\:{c},\:{d},\:{e}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{and} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{K}\:=\:{a}\:+\:{b}\:+\:{c}\:+\:{d}\:+\:{e}\:+\mathrm{1}\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\:\:\underset{\boldsymbol{{cyc}}} {\sum}\:\frac{\mathrm{1}}{\boldsymbol{{k}}−\boldsymbol{{a}}}\:<\:\frac{\mathrm{1}}{\mathrm{4}}\:\left(\frac{\sqrt[{\mathrm{4}\:\:\:\:}]{\boldsymbol{{e}}^{\mathrm{3}} \boldsymbol{{d}}^{\mathrm{3}} \boldsymbol{{c}}}}{\boldsymbol{{c}}^{\mathrm{3}/\mathrm{4}} \boldsymbol{{d}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{e}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{a}}}}\:+\:\frac{\sqrt[{\mathrm{4}\:\:\:}]{\boldsymbol{{d}}^{\:\mathrm{3}} \boldsymbol{{c}}^{\mathrm{2}} \boldsymbol{{b}}}}{\boldsymbol{{d}}^{\:\mathrm{3}/\mathrm{4}} \boldsymbol{{c}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{b}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{e}}}}\:+\:\frac{\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{c}}^{\mathrm{3}} \boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{a}}}}{\boldsymbol{{c}}^{\mathrm{3}/\mathrm{4}} \boldsymbol{{b}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{a}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{d}}}}\:+\:\frac{\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{b}}^{\mathrm{3}} \boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{e}}}}{\boldsymbol{{b}}^{\mathrm{3}/\mathrm{4}} \boldsymbol{{a}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{e}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{c}}}}\:+\:\frac{\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{a}}^{\mathrm{3}} \boldsymbol{{e}}^{\mathrm{2}} \boldsymbol{{d}}}}{\boldsymbol{{a}}^{\mathrm{3}/\mathrm{4}} \boldsymbol{{e}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{d}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{b}}}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\: \\ $$

Question Number 220094    Answers: 1   Comments: 0

let n ≥ 2 ∈ Z and x_1 , x_2 , ..., x_n are a positive real numbers such that Σ_(i=1) ^n x_i = n , prove that Σ_(i=1) ^n (x_i ^n /(x_1 + ∙∙∙ + x_i ^ + ∙∙∙ + x_n )) ≥ (n/(n − 1))

$$ \\ $$$$\:\:\:\:\mathrm{let}\:{n}\:\geqslant\:\mathrm{2}\:\in\:\mathbb{Z}\:\mathrm{and}\:{x}_{\mathrm{1}} ,\:{x}_{\mathrm{2}} ,\:...,\:{x}_{{n}} \:\mathrm{are}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{such}\:\mathrm{that}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:{x}_{{i}} \:=\:{n}\:,\:\mathrm{prove}\:\mathrm{that}\:\:\:\: \\ $$$$\:\:\:\:\:\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{{x}_{{i}} ^{{n}} }{{x}_{\mathrm{1}} +\:\centerdot\centerdot\centerdot\:+\:\hat {{x}}_{{i}} \:+\:\centerdot\centerdot\centerdot\:+\:{x}_{{n}} }\:\geqslant\:\frac{{n}}{{n}\:−\:\mathrm{1}} \\ $$$$ \\ $$

Question Number 220081    Answers: 2   Comments: 3

Question Number 220074    Answers: 0   Comments: 0

Question Number 220072    Answers: 1   Comments: 0

If x,y∈(0,(π/2)) Then prove that: log_(sinx) ^2 (((sin2x)/(sinx + cosx))) + log_(cosx) ^2 (((sin2x)/(sinx + cosx))) ≥ 2

$$\mathrm{If}\:\:\:\mathrm{x},\mathrm{y}\in\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{log}_{\boldsymbol{\mathrm{sinx}}} ^{\mathrm{2}} \:\left(\frac{\mathrm{sin2x}}{\mathrm{sinx}\:+\:\mathrm{cosx}}\right)\:+\:\mathrm{log}_{\boldsymbol{\mathrm{cosx}}} ^{\mathrm{2}} \:\left(\frac{\mathrm{sin2x}}{\mathrm{sinx}\:+\:\mathrm{cosx}}\right)\:\geqslant\:\mathrm{2} \\ $$

Question Number 220069    Answers: 1   Comments: 0

Let be (H_n )_(n≥1) H_n = Σ_(k=1) ^n (1/k) Find: lim_(n→∞) e^(2H_n ) ((((n+1)!))^(1/(n+1)) − ((n!))^(1/n) ) sin (π/n^2 ) = ?

$$\mathrm{Let}\:\mathrm{be}\:\:\:\left(\mathrm{H}_{\boldsymbol{\mathrm{n}}} \right)_{\boldsymbol{\mathrm{n}}\geqslant\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{H}_{\boldsymbol{\mathrm{n}}} \:=\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{k}} \\ $$$$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{e}^{\mathrm{2}\boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} } \:\left(\sqrt[{\boldsymbol{\mathrm{n}}+\mathrm{1}}]{\left(\mathrm{n}+\mathrm{1}\right)!}\:−\:\sqrt[{\boldsymbol{\mathrm{n}}}]{\mathrm{n}!}\:\right)\:\mathrm{sin}\:\frac{\pi}{\mathrm{n}^{\mathrm{2}} }\:=\:? \\ $$

Question Number 220066    Answers: 0   Comments: 0

evaluate −((csc(πs))/(iπ))∫_( C) (−t)^(s−1) e^(−t) dt , path C;(−∞,∞) −((𝚪(1−s))/(2πi)) ∫_( C) (((−t)^(s−1) )/(e^t −1)) dt , path C;(−∞,∞)

$$\mathrm{evaluate} \\ $$$$−\frac{\mathrm{csc}\left(\pi{s}\right)}{\boldsymbol{{i}}\pi}\int_{\:\boldsymbol{\mathcal{C}}} \:\left(−{t}\right)^{{s}−\mathrm{1}} {e}^{−{t}} \:\mathrm{d}{t}\:,\:\mathrm{path}\:\boldsymbol{\mathcal{C}};\left(−\infty,\infty\right) \\ $$$$−\frac{\boldsymbol{\Gamma}\left(\mathrm{1}−{s}\right)}{\mathrm{2}\pi\boldsymbol{{i}}}\:\int_{\:\boldsymbol{\mathcal{C}}} \:\frac{\left(−{t}\right)^{{s}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}\:\mathrm{d}{t}\:,\:\mathrm{path}\:\boldsymbol{\mathcal{C}};\left(−\infty,\infty\right) \\ $$

Question Number 220065    Answers: 2   Comments: 0

α∈R ; ω∈R^+ I(α) = ∫_(−∞) ^( ∞) ((x^2 ln(1+x^2 ))/((x^4 +1)^α )) e^(−x^2 ) cos(ωx) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha\in\mathbb{R}\:\:\:;\:\:\:\:\omega\in\mathbb{R}^{+} \\ $$$$\:\:\:\:\:{I}\left(\alpha\right)\:=\:\int_{−\infty} ^{\:\infty} \:\frac{{x}^{\mathrm{2}} \:\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\alpha} }\:{e}^{−{x}^{\mathrm{2}} } \:\mathrm{cos}\left(\omega{x}\right)\:{dx} \\ $$$$ \\ $$

Question Number 220040    Answers: 1   Comments: 0

∫^( ∞) _0 (1/( (√(x^8 + 1)))) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{0}} {\int}^{\:\infty} \:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{8}} \:+\:\mathrm{1}}}\:{dx} \\ $$$$ \\ $$

Question Number 220038    Answers: 1   Comments: 0

∫^( 1) _( 0) (1/( (√(x^8 + 1)))) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{8}} \:+\:\mathrm{1}}}\:{dx} \\ $$$$\: \\ $$

Question Number 220037    Answers: 1   Comments: 0

∫ (1/( (√(x^8 + 1)))) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{8}} \:+\:\mathrm{1}}}\:{dx} \\ $$$$ \\ $$

Question Number 220034    Answers: 5   Comments: 0

Question Number 220022    Answers: 1   Comments: 0

If 0 ≤ a ≤ b ≤ 1 Then prove that: ∫_a ^( b) ∫_a ^( b) ∫_a ^( b) ((dxdydz)/( (√(1 + xyz)))) ≥ (b−a)^2 ∫_a ^( b) (dx/( (√(1 + x^3 ))))

$$\mathrm{If}\:\:\:\mathrm{0}\:\leqslant\:\mathrm{a}\:\leqslant\:\mathrm{b}\:\leqslant\:\mathrm{1} \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \:\frac{\mathrm{dxdydz}}{\:\sqrt{\mathrm{1}\:+\:\mathrm{xyz}}}\:\geqslant\:\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{2}} \:\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{3}} }}\: \\ $$

Question Number 220020    Answers: 2   Comments: 0

If f:[a,b]→[−1,∞) a,b∈R a ≤ b f-continuous Then prove that: (∫_a ^( b) (1+f(x))dx)^3 ≥ (b−a)^3 + 3(b−a)^2 ∫_a ^( b) f(x)dx

$$\mathrm{If}\:\:\:\mathrm{f}:\left[\mathrm{a},\mathrm{b}\right]\rightarrow\left[−\mathrm{1},\infty\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{a},\mathrm{b}\in\mathbb{R} \\ $$$$\:\:\:\:\:\:\:\mathrm{a}\:\leqslant\:\mathrm{b} \\ $$$$\:\:\:\:\:\:\:\mathrm{f}-\mathrm{continuous} \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\left(\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \:\left(\mathrm{1}+\mathrm{f}\left(\mathrm{x}\right)\right)\mathrm{dx}\right)^{\mathrm{3}} \geqslant\:\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{3}} +\:\mathrm{3}\left(\mathrm{b}−\mathrm{a}\right)^{\mathrm{2}} \:\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$

Question Number 220019    Answers: 0   Comments: 1

please help me .The photos I upload becomes blurred HELP please

$${please}\:{help}\:{me}\:.{The}\:{photos}\:{I}\:{upload} \\ $$$${becomes}\:{blurred} \\ $$$${HELP}\:{please} \\ $$

Question Number 220016    Answers: 1   Comments: 0

Question Number 220015    Answers: 1   Comments: 0

∫_0 ^( ∞) ∣∣J_ν (r)∣∣e^(−rt) dr=Σ_(h=1) ^∞ ∫_z_h ^( z_(h+1) ) J_ν (r)e^(−rt) dr z_j is point of J_ν (z)=0 , z_1 =0 Σ_(h=1) ^∞ [F(r,t)]_(r=z_h ) ^(r=z_(h+1) ) i can′t solve anymore

$$\int_{\mathrm{0}} ^{\:\infty} \:\mid\mid{J}_{\nu} \left({r}\right)\mid\mid{e}^{−{rt}} \:\mathrm{d}{r}=\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\int_{{z}_{{h}} } ^{\:{z}_{{h}+\mathrm{1}} } \:{J}_{\nu} \left({r}\right){e}^{−{rt}} \mathrm{d}{r} \\ $$$${z}_{{j}} \:\mathrm{is}\:\mathrm{point}\:\mathrm{of}\:\:{J}_{\nu} \left({z}\right)=\mathrm{0}\:,\:{z}_{\mathrm{1}} =\mathrm{0} \\ $$$$\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left[{F}\left({r},{t}\right)\right]_{{r}={z}_{{h}} } ^{{r}={z}_{{h}+\mathrm{1}} } \: \\ $$$$\mathrm{i}\:\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{anymore} \\ $$

Question Number 220007    Answers: 2   Comments: 1

Question Number 219998    Answers: 1   Comments: 0

Question Number 219996    Answers: 1   Comments: 0

Solve Equation ((dx(t))/dt)=2x(t)+y(t) ((dy(t))/dt)=−3y(t) (((x^((1)) (t))),((y^((1)) (t))) )= ((2,( 1)),(0,(−3)) ) (((x(t))),((y(t))) ) A= ((2,( 1)),(0,(−3)) ) det{A−𝛌E}=0 det{ ((2,( 1)),(0,(−3)) )− ((𝛌,0),(0,𝛌) )}=0 𝛌=2,−3 v_1 = (((−1)),(( 5)) ) v_2 = ((1),(0) ) ((x),(y) )=C_1 e^(λt) v_1 +C_2 e^(λt) v_2 = (((−1),1),(( 5),0) ) (((C_1 e^(2t) )),((C_2 e^(−3t) )) ) ((x),(y) )= (((−C_1 e^(2t) +C_2 e^(−3t) )),(( 5C_1 e^(2t) )) ) ((dx(t))/dt)=−2C_1 e^(2t) −3C_2 e^(−3t) ≠ 2x(t)+y(t)=−2C_1 e^(2t) −2C_2 e^(−3t) +5C_1 e^(2t) ((dy(t))/dt)=10e^(2t) ≠ −3y(t)=−15C_2 e^(2t) Wrong..... Help

$$\mathrm{Solve}\:\mathrm{Equation} \\ $$$$\frac{\mathrm{d}{x}\left({t}\right)}{\mathrm{d}{t}}=\mathrm{2}{x}\left({t}\right)+{y}\left({t}\right) \\ $$$$\frac{\mathrm{d}{y}\left({t}\right)}{\mathrm{d}{t}}=−\mathrm{3}{y}\left({t}\right) \\ $$$$\begin{pmatrix}{{x}^{\left(\mathrm{1}\right)} \left({t}\right)}\\{{y}^{\left(\mathrm{1}\right)} \left({t}\right)}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}}&{\:\:\:\:\mathrm{1}}\\{\mathrm{0}}&{−\mathrm{3}}\end{pmatrix}\begin{pmatrix}{{x}\left({t}\right)}\\{{y}\left({t}\right)}\end{pmatrix} \\ $$$$\mathrm{A}=\begin{pmatrix}{\mathrm{2}}&{\:\:\:\:\mathrm{1}}\\{\mathrm{0}}&{−\mathrm{3}}\end{pmatrix} \\ $$$$\mathrm{det}\left\{\mathrm{A}−\boldsymbol{\lambda}\mathrm{E}\right\}=\mathrm{0} \\ $$$$\mathrm{det}\left\{\begin{pmatrix}{\mathrm{2}}&{\:\:\:\:\mathrm{1}}\\{\mathrm{0}}&{−\mathrm{3}}\end{pmatrix}−\begin{pmatrix}{\boldsymbol{\lambda}}&{\mathrm{0}}\\{\mathrm{0}}&{\boldsymbol{\lambda}}\end{pmatrix}\right\}=\mathrm{0} \\ $$$$\boldsymbol{\lambda}=\mathrm{2},−\mathrm{3} \\ $$$$\boldsymbol{\mathrm{v}}_{\mathrm{1}} =\begin{pmatrix}{−\mathrm{1}}\\{\:\:\:\:\mathrm{5}}\end{pmatrix} \\ $$$$\boldsymbol{\mathrm{v}}_{\mathrm{2}} =\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}={C}_{\mathrm{1}} {e}^{\lambda{t}} \boldsymbol{\mathrm{v}}_{\mathrm{1}} +{C}_{\mathrm{2}} {e}^{\lambda{t}} \boldsymbol{\mathrm{v}}_{\mathrm{2}} =\begin{pmatrix}{−\mathrm{1}}&{\mathrm{1}}\\{\:\:\:\:\mathrm{5}}&{\mathrm{0}}\end{pmatrix}\begin{pmatrix}{{C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} }\\{{C}_{\mathrm{2}} {e}^{−\mathrm{3}{t}} }\end{pmatrix} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{−{C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} +{C}_{\mathrm{2}} {e}^{−\mathrm{3}{t}} }\\{\:\:\:\:\:\:\:\mathrm{5C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} }\end{pmatrix} \\ $$$$\frac{\mathrm{d}{x}\left({t}\right)}{\mathrm{d}{t}}=−\mathrm{2}{C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} −\mathrm{3}{C}_{\mathrm{2}} {e}^{−\mathrm{3}{t}} \neq \\ $$$$\mathrm{2}{x}\left({t}\right)+{y}\left({t}\right)=−\mathrm{2}{C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} −\mathrm{2}{C}_{\mathrm{2}} {e}^{−\mathrm{3}{t}} +\mathrm{5}{C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} \\ $$$$\frac{\mathrm{d}{y}\left({t}\right)}{\mathrm{d}{t}}=\mathrm{10}{e}^{\mathrm{2}{t}} \neq \\ $$$$−\mathrm{3}{y}\left({t}\right)=−\mathrm{15}{C}_{\mathrm{2}} {e}^{\mathrm{2}{t}} \\ $$$$\:\mathrm{Wrong}.....\:\mathrm{Help} \\ $$

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