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Question Number 170116    Answers: 1   Comments: 1

solve the D.E. (x+2y−4)dx+(2x+y−5)dy=0

$${solve}\:{the}\:{D}.{E}. \\ $$$$\left({x}+\mathrm{2}{y}−\mathrm{4}\right){dx}+\left(\mathrm{2}{x}+{y}−\mathrm{5}\right){dy}=\mathrm{0} \\ $$

Question Number 170115    Answers: 0   Comments: 0

In △ABC then: ((cos (A/2) ∙ cos (B/2) ∙ cos (C/2))/((1 - sin (A/2))∙(1 - sin (B/2))∙(1 - sin (C/2)))) ≥ 3(√3)

$$\mathrm{In}\:\:\bigtriangleup\mathrm{ABC}\:\:\mathrm{then}: \\ $$$$\frac{\mathrm{cos}\:\frac{\mathrm{A}}{\mathrm{2}}\:\centerdot\:\mathrm{cos}\:\frac{\mathrm{B}}{\mathrm{2}}\:\centerdot\:\mathrm{cos}\:\frac{\mathrm{C}}{\mathrm{2}}}{\left(\mathrm{1}\:-\:\mathrm{sin}\:\frac{\mathrm{A}}{\mathrm{2}}\right)\centerdot\left(\mathrm{1}\:-\:\mathrm{sin}\:\frac{\mathrm{B}}{\mathrm{2}}\right)\centerdot\left(\mathrm{1}\:-\:\mathrm{sin}\:\frac{\mathrm{C}}{\mathrm{2}}\right)}\:\geqslant\:\mathrm{3}\sqrt{\mathrm{3}} \\ $$

Question Number 170113    Answers: 0   Comments: 0

Question Number 170109    Answers: 0   Comments: 0

Question Number 170106    Answers: 1   Comments: 0

Question Number 170110    Answers: 2   Comments: 1

prove that e^(iθ) =cosθ+isinθ

$${prove}\:{that} \\ $$$${e}^{{i}\theta} ={cos}\theta+{isin}\theta \\ $$

Question Number 170091    Answers: 0   Comments: 0

Question Number 170100    Answers: 0   Comments: 1

Question Number 170099    Answers: 0   Comments: 0

Question Number 170084    Answers: 0   Comments: 2

Question Number 170079    Answers: 2   Comments: 0

∣a^→ ∣=1 ∣b^→ ∣=2 ∢(a^→ , b^→ )=(π/3) ∣a^→ +b^→ ∣=?

$$\mid\overset{\rightarrow} {{a}}\mid=\mathrm{1} \\ $$$$\mid\overset{\rightarrow} {{b}}\mid=\mathrm{2} \\ $$$$\sphericalangle\left(\overset{\rightarrow} {{a}},\:\overset{\rightarrow} {{b}}\right)=\frac{\pi}{\mathrm{3}} \\ $$$$\mid\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}\mid=? \\ $$

Question Number 170077    Answers: 1   Comments: 0

How do I find for the accurate measure of sine from 1 to 30

$$ \\ $$How do I find for the accurate measure of sine from 1 to 30

Question Number 170076    Answers: 0   Comments: 0

Question Number 170075    Answers: 0   Comments: 0

Question Number 170073    Answers: 3   Comments: 3

Question Number 170065    Answers: 1   Comments: 0

a:b:c=5:6:7 3a+4b−5c=16 faind volue of c=?

$${a}:{b}:{c}=\mathrm{5}:\mathrm{6}:\mathrm{7} \\ $$$$\mathrm{3}{a}+\mathrm{4}{b}−\mathrm{5}{c}=\mathrm{16}\:\:\:\:\:\: \\ $$$${faind}\:{volue}\:{of}\:\:\:\:{c}=? \\ $$

Question Number 170064    Answers: 2   Comments: 0

(((a−2))/2)=(((b−3))/3)=(((c−4))/4) and a∙b∙c=192 faind a+b+c=?

$$\frac{\left({a}−\mathrm{2}\right)}{\mathrm{2}}=\frac{\left({b}−\mathrm{3}\right)}{\mathrm{3}}=\frac{\left({c}−\mathrm{4}\right)}{\mathrm{4}} \\ $$$${and}\:\:\:\:{a}\centerdot{b}\centerdot{c}=\mathrm{192} \\ $$$${faind}\:\:\:{a}+{b}+{c}=? \\ $$

Question Number 170057    Answers: 1   Comments: 0

lim_(x→0) (x/a)⌊(b/x)⌋ where a,b>0 please....

$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{{x}}{{a}}\lfloor\frac{{b}}{{x}}\rfloor\:\:{where}\:{a},{b}>\mathrm{0} \\ $$$${please}.... \\ $$

Question Number 170054    Answers: 1   Comments: 0

Question Number 170053    Answers: 1   Comments: 0

Question Number 170050    Answers: 1   Comments: 0

Show that,∀a,b,c∈R :a^2 +b^2 +c^2 +12≥4(a+b+c)

$$\mathrm{Show}\:\mathrm{that},\forall\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{R}\::\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{12}\geqslant\mathrm{4}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right) \\ $$

Question Number 170046    Answers: 2   Comments: 0

PROVE THAT 6 ∣ (8^x −2^x )

$$\mathrm{PROVE}\:\mathrm{THAT} \\ $$$$\:\:\:\:\mathrm{6}\:\mid\:\left(\mathrm{8}^{{x}} −\mathrm{2}^{{x}} \right) \\ $$

Question Number 170032    Answers: 1   Comments: 2

((8^x −2^x )/(6^x −3^x ))=2 , find x ? Mastermind

$$\frac{\mathrm{8}^{{x}} −\mathrm{2}^{{x}} }{\mathrm{6}^{{x}} −\mathrm{3}^{{x}} }=\mathrm{2}\:,\:{find}\:{x}\:? \\ $$$$ \\ $$$${Mastermind} \\ $$

Question Number 170028    Answers: 0   Comments: 1

verify that ⟨p_x ∣x^ ∣Ψ⟩=ih(d/dp_x )<p_x ∣Ψ>

$${verify}\:{that}\:\:\langle{p}_{{x}} \mid\hat {{x}}\mid\Psi\rangle={ih}\frac{{d}}{{dp}_{{x}} }<{p}_{{x}} \mid\Psi> \\ $$

Question Number 170025    Answers: 1   Comments: 0

Question Number 170022    Answers: 0   Comments: 0

2. [((1 2)),((2 −1)) ] Soln: [((1 2)),((2 −1)) ]= [((1 0)),((0 1)) ].A ⇒ [((1 2)),((0 −5)) ]= [(( 1 0)),((−2 1)) ].A [R_2 →R_2 −2R_1 ] ⇒ [((1 2)),((0 1)) ]= [(( 1 0)),(((2/5) −(1/5))) ].A [R_2 →(−(1/5))R_2 ] ⇒ [((1 0)),((0 1)) ]= [(((1/5) (2/5))),(((2/5) −(1/5))) ].A [R_1 →R_1 −2R_2 ] ∴A^(−1) =(1/5) [((1 2)),((2 −1)) ]

$$\mathrm{2}.\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\:\:−\mathrm{1}}\end{bmatrix} \\ $$$${Soln}:\:\:\:\:\:\:\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\:\:−\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{bmatrix}.{A} \\ $$$$\Rightarrow\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:−\mathrm{5}}\end{bmatrix}=\begin{bmatrix}{\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{0}}\\{−\mathrm{2}\:\:\:\:\:\:\mathrm{1}}\end{bmatrix}.{A}\:\:\:\:\:\:\left[{R}_{\mathrm{2}} \rightarrow{R}_{\mathrm{2}} −\mathrm{2}{R}_{\mathrm{1}} \right] \\ $$$$\Rightarrow\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\frac{\mathrm{2}}{\mathrm{5}}\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{5}}}\end{bmatrix}.{A}\:\:\:\:\:\left[{R}_{\mathrm{2}} \rightarrow\left(−\frac{\mathrm{1}}{\mathrm{5}}\right){R}_{\mathrm{2}} \right] \\ $$$$\Rightarrow\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\frac{\mathrm{1}}{\mathrm{5}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{5}}}\\{\frac{\mathrm{2}}{\mathrm{5}}\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{5}}}\end{bmatrix}.{A}\:\:\:\:\:\:\left[{R}_{\mathrm{1}} \rightarrow{R}_{\mathrm{1}} −\mathrm{2}{R}_{\mathrm{2}} \right] \\ $$$$\therefore{A}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{5}}\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\:\:\:\:−\mathrm{1}}\end{bmatrix} \\ $$

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