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Question Number 220877 Answers: 4 Comments: 0

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Find the maximum value of x^2 y^3 z^4 subject to the condition x+y+z=18

$${Find}\:{the}\:{maximum}\:{value}\:{of}\:{x}^{\mathrm{2}} {y}^{\mathrm{3}} {z}^{\mathrm{4}} \:{subject}\:{to}\:{the}\:{condition}\:{x}+{y}+{z}=\mathrm{18} \\ $$

Question Number 220863 Answers: 1 Comments: 0

(211) Find the derivative of Δx, where Δx= determinant (((f_1 (x)),(φ_1 (x)),(Ψ_1 (x))),((f_2 (x)),(φ_2 (x)),(Ψ_2 (x))),((f_3 (x)),(φ_3 (x)),(Ψ_3 (x)))) and f_1 (x) ,f_2 (x), f_3 (x),φ_1 (x), etc. are different functions of x.

$$\left(\mathrm{211}\right) \\ $$$$\:\: \\ $$$${Find}\:{the}\:{derivative}\:{of}\:\Delta{x},\:{where} \\ $$$$\Delta{x}=\begin{vmatrix}{{f}_{\mathrm{1}} \left({x}\right)}&{\phi_{\mathrm{1}} \left({x}\right)}&{\Psi_{\mathrm{1}} \left({x}\right)}\\{{f}_{\mathrm{2}} \left({x}\right)}&{\phi_{\mathrm{2}} \left({x}\right)}&{\Psi_{\mathrm{2}} \left({x}\right)}\\{{f}_{\mathrm{3}} \left({x}\right)}&{\phi_{\mathrm{3}} \left({x}\right)}&{\Psi_{\mathrm{3}} \left({x}\right)}\end{vmatrix} \\ $$$${and}\:{f}_{\mathrm{1}} \left({x}\right)\:,{f}_{\mathrm{2}} \left({x}\right),\:{f}_{\mathrm{3}} \left({x}\right),\phi_{\mathrm{1}} \left({x}\right),\:{etc}.\:{are}\:{different}\:{functions}\:{of}\:{x}. \\ $$

Question Number 220858 Answers: 1 Comments: 2

Question Number 220857 Answers: 1 Comments: 0

Prove that tan 20^0 tan40^0 tan 80^0 =tan 60^0

$${Prove}\:{that}\:\mathrm{tan}\:\mathrm{20}^{\mathrm{0}} \mathrm{tan40}^{\mathrm{0}} \:\mathrm{tan}\:\mathrm{80}^{\mathrm{0}} =\mathrm{tan}\:\mathrm{60}^{\mathrm{0}} \\ $$

Question Number 220855 Answers: 1 Comments: 0

If b cos(θ+120^0 )=c cos (θ+240^0 ) then prove that b−c=−(b+c)(√3) tan θ

$${If}\:\:{b}\:\mathrm{cos}\left(\theta+\mathrm{120}^{\mathrm{0}} \right)={c}\:\mathrm{cos}\:\left(\theta+\mathrm{240}^{\mathrm{0}} \right)\:{then}\:{prove}\:{that} \\ $$$${b}−{c}=−\left({b}+{c}\right)\sqrt{\mathrm{3}}\:\mathrm{tan}\:\theta \\ $$

Question Number 220854 Answers: 2 Comments: 0

Solve for x and y 3^x +3^y =4, 3^(−x) +3^(−y ) =(4/3)

$${Solve}\:{for}\:{x}\:\:\:{and}\:\:\:\:{y} \\ $$$$\mathrm{3}^{{x}} +\mathrm{3}^{{y}} =\mathrm{4},\:\:\mathrm{3}^{−{x}} +\mathrm{3}^{−{y}\:} =\frac{\mathrm{4}}{\mathrm{3}} \\ $$

Question Number 220853 Answers: 1 Comments: 0

The two solutions of the equation are the same a(b−c)x^(2 ) +b(c−a)x+c(a−b)=0 Prove that (1/a)+(1/c)=(2/b)

$${The}\:{two}\:{solutions}\:{of}\:{the}\:{equation}\:{are}\:{the}\:{same} \\ $$$${a}\left({b}−{c}\right){x}^{\mathrm{2}\:} +{b}\left({c}−{a}\right){x}+{c}\left({a}−{b}\right)=\mathrm{0} \\ $$$${Prove}\:{that}\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{2}}{{b}} \\ $$

Question Number 220852 Answers: 1 Comments: 0

Lim_(x→0) {((xe^x −log(1+x))/x^2 )}

$$\underset{{x}\rightarrow\mathrm{0}} {{Lim}}\left\{\frac{{xe}^{{x}} −{log}\left(\mathrm{1}+{x}\right)}{{x}^{\mathrm{2}} }\right\} \\ $$

Question Number 220851 Answers: 0 Comments: 0

Question Number 220850 Answers: 1 Comments: 1

∫(√((x+1)/(x+2))).(1/(x+3))dx

$$\int\sqrt{\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}}.\frac{\mathrm{1}}{{x}+\mathrm{3}}{dx} \\ $$

Question Number 220848 Answers: 1 Comments: 2

∫_( 0) ^( 1) (1/2) (√(((2 − 6x + 3x^2 )^2 + 4(2x − 3x^2 + x^3 ))/(2x − 3x^2 + x^3 ))) dx

$$ \\ $$$$\:\:\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{2}}\:\sqrt{\frac{\left(\mathrm{2}\:−\:\mathrm{6}{x}\:+\:\mathrm{3}{x}^{\mathrm{2}} \right)^{\mathrm{2}} +\:\mathrm{4}\left(\mathrm{2}{x}\:−\:\mathrm{3}{x}^{\mathrm{2}} \:+\:{x}^{\mathrm{3}} \right)}{\mathrm{2}{x}\:−\:\mathrm{3}{x}^{\mathrm{2}} \:+\:{x}^{\mathrm{3}} }}\:{dx}\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 220843 Answers: 0 Comments: 1

α ∈ R lim_(x→1) (((1 − x)^α )/(^3 (√(1 − x^4 )))) ∈(0,∞)

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha\:\in\:\mathbb{R} \\ $$$$\:\:\:\:\:\mathrm{lim}_{{x}\rightarrow\mathrm{1}} \:\frac{\left(\mathrm{1}\:−\:{x}\right)^{\alpha} }{\:^{\mathrm{3}} \sqrt{\mathrm{1}\:−\:{x}^{\mathrm{4}} }}\:\:\:\:\:\:\:\:\in\left(\mathrm{0},\infty\right) \\ $$$$ \\ $$

Question Number 220842 Answers: 0 Comments: 2

J=∫_0 ^( ∞) e^(−t) u_π (t)cos(t)dt=? note: u_c (t)= { (( 0 t<c)),(( 1 t>c )) :} ; c≥0

$$ \\ $$$$\:\:\:\:\:\:\mathrm{J}=\int_{\mathrm{0}} ^{\:\infty} {e}^{−{t}} {u}_{\pi} \left({t}\right){cos}\left({t}\right){dt}=? \\ $$$$ \\ $$$$\:{note}:\:\:{u}_{{c}} \left({t}\right)=\:\begin{cases}{\:\mathrm{0}\:\:\:\:\:\:\:\:{t}<{c}}\\{\:\mathrm{1}\:\:\:\:\:\:\:\:\:{t}>{c}\:\:}\end{cases}\:\:;\:\:\:\:{c}\geqslant\mathrm{0} \\ $$

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To Tinkutara [(a),(b) ], determinant ((a),(b)), ((a),(b) ) , { (a),(b) :} , {: (a),(b) } , determinant (((abcdefg)),((pqrstvw))) is work well but invisible line matrix(?) dosen′t work pls Fix bug

$$\mathrm{To}\:\mathrm{Tinkutara} \\ $$$$\begin{bmatrix}{\mathrm{a}}\\{\mathrm{b}}\end{bmatrix},\begin{vmatrix}{\mathrm{a}}\\{\mathrm{b}}\end{vmatrix},\begin{pmatrix}{\mathrm{a}}\\{\mathrm{b}}\end{pmatrix}\:,\begin{cases}{\mathrm{a}}\\{\mathrm{b}}\end{cases}\:,\:\:\left.\begin{matrix}{\mathrm{a}}\\{\mathrm{b}}\end{matrix}\right\}\:,\begin{array}{|c|c|}{\mathrm{abcdefg}}\\{\mathrm{pqrstvw}}\\\hline\end{array}\:\mathrm{is}\:\mathrm{work}\:\mathrm{well} \\ $$$$\mathrm{but}\:\mathrm{invisible}\:\mathrm{line}\:\mathrm{matrix}\left(?\right)\:\mathrm{dosen}'\mathrm{t}\:\mathrm{work} \\ $$$$\mathrm{pls}\:\mathrm{Fix}\:\mathrm{bug} \\ $$

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