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AllQuestion and Answers: Page 34

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

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

Question Number 220830 Answers: 0 Comments: 2

Question Number 220832 Answers: 3 Comments: 0

Question Number 220831 Answers: 1 Comments: 0

Question Number 220825 Answers: 1 Comments: 0

Question Number 220820 Answers: 1 Comments: 1

Question Number 220811 Answers: 0 Comments: 0

Question Number 220810 Answers: 1 Comments: 3

Question Number 220800 Answers: 0 Comments: 0

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

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

Question Number 220791 Answers: 0 Comments: 0

∫_( 0) ^( 1) (x^2 /(sin x + 1)) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{x}^{\mathrm{2}} }{\boldsymbol{\mathrm{sin}}\:{x}\:+\:\mathrm{1}}\:{dx} \\ $$$$ \\ $$

Question Number 220790 Answers: 1 Comments: 0

Complex integral ∮_( C) (dz/(z^3 +1))=?? , C;x^2 +y^2 =4 ∮_( C) (1/z)e^z dz, C; { ((y=1 , −1≤x≤1)),((y=−1 , −1≤x≤1)),((x=1 , −1≤y≤1)),((x=−1 , −1≤y≤1)) :}

$$\mathrm{Complex}\:\mathrm{integral} \\ $$$$\oint_{\:\mathrm{C}} \:\frac{\mathrm{d}{z}}{{z}^{\mathrm{3}} +\mathrm{1}}=??\:,\:{C};{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4} \\ $$$$\oint_{\:{C}} \:\frac{\mathrm{1}}{{z}}{e}^{{z}} \:\mathrm{d}{z},\:{C};\begin{cases}{{y}=\mathrm{1}\:,\:−\mathrm{1}\leq{x}\leq\mathrm{1}}\\{{y}=−\mathrm{1}\:,\:−\mathrm{1}\leq{x}\leq\mathrm{1}}\\{{x}=\mathrm{1}\:,\:−\mathrm{1}\leq{y}\leq\mathrm{1}}\\{{x}=−\mathrm{1}\:,\:−\mathrm{1}\leq{y}\leq\mathrm{1}}\end{cases}\: \\ $$

Question Number 220770 Answers: 1 Comments: 0

∫_0 ^( (π/2)) (( sin(x))/( (√(1 +(√(sin(2x)))))))dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\:{sin}\left({x}\right)}{\:\sqrt{\mathrm{1}\:+\sqrt{{sin}\left(\mathrm{2}{x}\right)}}}{dx}\: \\ $$

Question Number 220769 Answers: 2 Comments: 0

Question Number 220764 Answers: 1 Comments: 0

L= lim _( n→∞) (Σ_(k=1) ^n (k/(n^2 +k^2 ))).(∫^( 1) _( 0) e^(−x^2 ) dx)^(−1) .(Σ_(m=0) ^∞ (((−1)^m )/((2m+1)3^m )))

$$ \\ $$$$\:\:\boldsymbol{\mathrm{L}}=\:\boldsymbol{\mathrm{lim}}\underset{\:\boldsymbol{{n}}\rightarrow\infty} {\:}\left(\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\sum}}\:\frac{\boldsymbol{{k}}}{\boldsymbol{{n}}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} }\right).\left(\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \boldsymbol{{e}}^{−\boldsymbol{{x}}^{\mathrm{2}} } \boldsymbol{{dx}}\overset{−\mathrm{1}} {\right)}.\left(\underset{\boldsymbol{{m}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{m}}} }{\left(\mathrm{2}\boldsymbol{{m}}+\mathrm{1}\right)\mathrm{3}^{\boldsymbol{{m}}} }\right)\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 220755 Answers: 0 Comments: 0

Hmm....i need a help... generalize L_z ^(−1) {(1/(z^n +1))}= ∮_( C) (e^(zt) /(z^n +1)) dz

$$ \\ $$$$\mathrm{Hmm}....\mathrm{i}\:\mathrm{need}\:\mathrm{a}\:\mathrm{help}... \\ $$$$\mathrm{generalize}\: \\ $$$$\mathcal{L}_{{z}} ^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{{z}^{{n}} +\mathrm{1}}\right\}=\:\oint_{\:{C}} \:\:\frac{{e}^{{zt}} }{{z}^{{n}} +\mathrm{1}}\:\mathrm{d}{z} \\ $$

Question Number 220745 Answers: 3 Comments: 0

Question Number 220744 Answers: 1 Comments: 0

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