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Question Number 222300    Answers: 0   Comments: 5

How do you put a box around something?? please tell me

$${How}\:{do}\:{you}\:{put}\:{a}\:{box}\:{around}\:{something}??\:{please}\:{tell}\:{me} \\ $$

Question Number 222299    Answers: 1   Comments: 0

For what value of k the roots of the equation ((x^2 −2x)/(4x−1))=((k−1)/(k+1)) will have same value but with opposite symbol(like x=a and −a) i mean the two valuea of x will be this type x=2 and −2(both 2 but opposite symbols)

$${For}\:{what}\:{value}\:{of}\:\:{k}\:\:{the}\:{roots}\:{of}\:{the}\:{equation} \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{4}{x}−\mathrm{1}}=\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}} \\ $$$${will}\:{have}\:{same}\:{value}\:{but}\:\:{with}\:{opposite}\:{symbol}\left({like}\:{x}={a}\:{and}\:−{a}\right) \\ $$$${i}\:{mean}\:{the}\:{two}\:{valuea}\:{of}\:{x}\:{will}\:{be}\:{this}\:{type} \\ $$$${x}=\mathrm{2}\:{and}\:−\mathrm{2}\left({both}\:\mathrm{2}\:{but}\:{opposite}\:{symbols}\right) \\ $$

Question Number 222295    Answers: 1   Comments: 0

Question Number 222292    Answers: 1   Comments: 0

∫_(−∞) ^∞ sech(z) sech(z−a) dz

$$ \\ $$$$\:\:\:\int_{−\infty} ^{\infty} \mathrm{sech}\left({z}\right)\:\mathrm{sech}\left({z}−{a}\right)\:{dz} \\ $$$$ \\ $$

Question Number 222288    Answers: 1   Comments: 0

Prove that: 1 + 2 + 3 + ... + n = ((n∙(n + 1))/2)

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:...\:+\:\boldsymbol{\mathrm{n}}\:=\:\frac{\mathrm{n}\centerdot\left(\mathrm{n}\:+\:\mathrm{1}\right)}{\mathrm{2}} \\ $$

Question Number 222296    Answers: 0   Comments: 0

(1+x^4 )y′−x^3 y = x^5 −x^3 +2x+1

$$\left(\mathrm{1}+{x}^{\mathrm{4}} \right){y}'−{x}^{\mathrm{3}} {y}\:=\:{x}^{\mathrm{5}} −{x}^{\mathrm{3}} +\mathrm{2}{x}+\mathrm{1} \\ $$

Question Number 222284    Answers: 1   Comments: 0

y=(8^x /((in8)^3 )) find (d^6 y/dx^6 )

$$\boldsymbol{\mathrm{y}}=\frac{\mathrm{8}^{\boldsymbol{\mathrm{x}}} }{\left(\boldsymbol{\mathrm{in}}\mathrm{8}\right)^{\mathrm{3}} } \\ $$$$\boldsymbol{\mathrm{find}}\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{6}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{6}} } \\ $$

Question Number 222280    Answers: 0   Comments: 1

y=3x^(2024) −18x^(2020) +5x^(47) −8 find (d^(2025) y/dx^(2025) )

$$\boldsymbol{\mathrm{y}}=\mathrm{3}\boldsymbol{\mathrm{x}}^{\mathrm{2024}} −\mathrm{18}\boldsymbol{\mathrm{x}}^{\mathrm{2020}} +\mathrm{5}\boldsymbol{\mathrm{x}}^{\mathrm{47}} −\mathrm{8} \\ $$$$\boldsymbol{\mathrm{find}}\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2025}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2025}} } \\ $$

Question Number 222279    Answers: 2   Comments: 0

(a^2 −b^2 )sin θ+2abcos θ=a^2 +b^2 tan θ=??

$$\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\mathrm{sin}\:\theta+\mathrm{2}{ab}\mathrm{cos}\:\theta={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\mathrm{tan}\:\theta=?? \\ $$

Question Number 222276    Answers: 0   Comments: 0

Prove; Σ_(k = 1) ^∞ (k/(sinh(πk))) = ((Γ((1/4))^4 − 8π^2 )/(32π^3 ))

$$ \\ $$$$\:\mathrm{Prove};\:\underset{{k}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{k}}{\mathrm{sinh}\left(\pi{k}\right)}\:=\:\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{4}} −\:\mathrm{8}\pi^{\mathrm{2}} }{\mathrm{32}\pi^{\mathrm{3}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 222275    Answers: 1   Comments: 0

Prove ; ∫_(−π) ^( π) ((z sin(z) )/((1 + z + (√(1 + z^2 )))(√(3 + sin^2 (z))))) dz = ζ(2)

$$ \\ $$$$\:\:\mathrm{Prove}\:;\:\int_{−\pi} ^{\:\pi} \:\frac{{z}\:\mathrm{sin}\left({z}\right)\:}{\left(\mathrm{1}\:+\:{z}\:+\:\sqrt{\mathrm{1}\:+\:{z}^{\mathrm{2}} }\right)\sqrt{\mathrm{3}\:+\:\mathrm{sin}^{\mathrm{2}} \left({z}\right)}}\:{dz}\:=\:\zeta\left(\mathrm{2}\right)\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 222271    Answers: 2   Comments: 0

y=(((1+(√5))/2))^(10) ,Prove:y=((123+55(√5))/2)

$${y}=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{10}} ,\mathrm{Prove}:{y}=\frac{\mathrm{123}+\mathrm{55}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

Question Number 222261    Answers: 1   Comments: 0

lim_(x→0) ((tan(x^2 +4x))/(sin(9x^2 +x))) No L′ho^ pital′s rule allowed!

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\left({x}^{\mathrm{2}} +\mathrm{4}{x}\right)}{\mathrm{sin}\left(\mathrm{9}{x}^{\mathrm{2}} +{x}\right)} \\ $$$$\mathrm{No}\:\mathrm{L}'\mathrm{h}\hat {\mathrm{o}pital}'\mathrm{s}\:\mathrm{rule}\:\mathrm{allowed}! \\ $$

Question Number 222249    Answers: 3   Comments: 0

Prove:∀n∈Z^+ ,1^3 +2^3 +…+n^3 =(1+2+…+n)^2

$$\mathrm{Prove}:\forall{n}\in\mathbb{Z}^{+} ,\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\ldots+{n}^{\mathrm{3}} =\left(\mathrm{1}+\mathrm{2}+\ldots+{n}\right)^{\mathrm{2}} \\ $$

Question Number 222245    Answers: 2   Comments: 0

∫_0 ^( (π/2)) ((cos^(−1) (((√(1 − sin^2 (x) cos^2 (x)))/(1 + sin^2 (x))))∙ln(((1 + sin(x))/(1 + cos(x)))))/( (√(1 + cos^2 (x) − sin^2 (x))))) dx

$$ \\ $$$$\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{cos}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}\:−\:\mathrm{sin}^{\mathrm{2}} \left({x}\right)\:\mathrm{cos}^{\mathrm{2}} \left({x}\right)}}{\mathrm{1}\:+\:\mathrm{sin}^{\mathrm{2}} \left({x}\right)}\right)\centerdot\mathrm{ln}\left(\frac{\mathrm{1}\:+\:\mathrm{sin}\left({x}\right)}{\mathrm{1}\:+\:\mathrm{cos}\left({x}\right)}\right)}{\:\sqrt{\mathrm{1}\:+\:\mathrm{cos}^{\mathrm{2}} \left({x}\right)\:−\:\mathrm{sin}^{\mathrm{2}} \left({x}\right)}}\:\:\mathrm{d}{x}\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 222225    Answers: 0   Comments: 9

a+b+c=−D ab+bc+ac=E abc=−1 (a^2 −bc)(b^2 −ac)(c^2 −ab)=40 D−E=4 (D+E)^2 =...

$${a}+{b}+{c}=−{D} \\ $$$${ab}+{bc}+{ac}={E} \\ $$$${abc}=−\mathrm{1} \\ $$$$\left({a}^{\mathrm{2}} −{bc}\right)\left({b}^{\mathrm{2}} −{ac}\right)\left({c}^{\mathrm{2}} −{ab}\right)=\mathrm{40} \\ $$$${D}−{E}=\mathrm{4} \\ $$$$\left({D}+{E}\right)^{\mathrm{2}} =... \\ $$

Question Number 222224    Answers: 0   Comments: 1

Prove:∫_0 ^1 ((ln(1−x^2 ))/x)cos(ln x)dx=1−(π/2)cosh(π/2)

$$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}}\mathrm{cos}\left(\mathrm{ln}\:{x}\right){dx}=\mathrm{1}−\frac{\pi}{\mathrm{2}}\mathrm{cosh}\frac{\pi}{\mathrm{2}} \\ $$

Question Number 222218    Answers: 1   Comments: 0

∫_0 ^( π) tan^(−1) (((ln sin(x))/x)) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\pi} \:\mathrm{tan}^{−\mathrm{1}} \:\left(\frac{\mathrm{ln}\:\mathrm{sin}\left({x}\right)}{{x}}\right)\:{dx} \\ $$$$ \\ $$

Question Number 222217    Answers: 0   Comments: 0

∫_0 ^∞ ∫_0 ^∞ ((ln x ln y)/( (√(xy))))cos(x+y)=π^2 (γ+2 ln 2) Sol:∫_0 ^∞ ∫_0 ^∞ ((ln x ln y)/( (√(xy))))cos(x+y)dxdy=Re((∫_0 ^∞ x^(−(1/2)) e^(ix) ln xdx)(∫_0 ^∞ y^(−(1/2)) e^(iy) ln ydy)) ∫_0 ^∞ x^a e^(ix) dx=e^(iπ(a+1)) Γ(a+1),−1<Re a<0 ∫_0 ^∞ x^a e^(ix) dx=∫_0 ^∞ x^a e^(ix) ln xdx=(∂/∂u)[e^(iπ(a+1)/2) Γ(a+1)] =e^(iπ(a+1)) Γ(a+1)(((iπ)/2)+ψ(a+1)) a=−(1/2): ∫_0 ^∞ x^(−(1/2)) e^(ix) ln xdx=e^(iπ/1) (√π)(((iπ)/2)+ψ((1/2))) c=e^(iπ/4) (√π)(((iπ)/2)−γ−2 ln 2) e^(iπ/4) =((√2)/2)(1+i) c=(√π)∙((√2)/2)(1+i)(−γ−2 ln 2+i(π/2))=((√(2π))/2)[(γ−ln 2−(π/2))+i(−γ−2 ln 2+(π/2))] p=−γ−2 ln 2−(π/2),q=−γ−2 ln 2+(π/2) c^2 =(((√(2π))/2))^2 (p+ip)^2 =((2π)/4)(p^2 −q^2 +2ipq)=(π/2)(p^2 −q^2 +2ipq) p+q+2=(−γ−2 ln 2) p−q=π p^2 −q^2 =(p−q)(p+q)=(−π)∙2(−γ−2 ln 2)=2π(γ+2 ln 2) [2π(γ+2 ln 2)+2ipq]=π^2 (γ+2 ln 2)+iπpq Re(c^2 )=π^2 (γ+2 ln 2)

$$\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\:{x}\:\mathrm{ln}\:{y}}{\:\sqrt{{xy}}}\mathrm{cos}\left({x}+{y}\right)=\pi^{\mathrm{2}} \left(\gamma+\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$$$\mathrm{Sol}:\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\:{x}\:\mathrm{ln}\:{y}}{\:\sqrt{{xy}}}\mathrm{cos}\left({x}+{y}\right){dxdy}=\mathrm{Re}\left(\left(\int_{\mathrm{0}} ^{\infty} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{{ix}} \mathrm{ln}\:{xdx}\right)\left(\int_{\mathrm{0}} ^{\infty} {y}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{{iy}} \mathrm{ln}\:{ydy}\right)\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {e}^{{ix}} {dx}={e}^{{i}\pi\left({a}+\mathrm{1}\right)} \Gamma\left({a}+\mathrm{1}\right),−\mathrm{1}<\mathrm{Re}\:{a}<\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {e}^{{ix}} {dx}=\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {e}^{{ix}} \mathrm{ln}\:{xdx}=\frac{\partial}{\partial{u}}\left[{e}^{{i}\pi\left({a}+\mathrm{1}\right)/\mathrm{2}} \Gamma\left({a}+\mathrm{1}\right)\right] \\ $$$$={e}^{{i}\pi\left({a}+\mathrm{1}\right)} \Gamma\left({a}+\mathrm{1}\right)\left(\frac{{i}\pi}{\mathrm{2}}+\psi\left({a}+\mathrm{1}\right)\right) \\ $$$${a}=−\frac{\mathrm{1}}{\mathrm{2}}: \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{{ix}} \mathrm{ln}\:{xdx}={e}^{{i}\pi/\mathrm{1}} \sqrt{\pi}\left(\frac{{i}\pi}{\mathrm{2}}+\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$${c}={e}^{{i}\pi/\mathrm{4}} \sqrt{\pi}\left(\frac{{i}\pi}{\mathrm{2}}−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$$${e}^{{i}\pi/\mathrm{4}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+{i}\right) \\ $$$${c}=\sqrt{\pi}\centerdot\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+{i}\right)\left(−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}+{i}\frac{\pi}{\mathrm{2}}\right)=\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{2}}\left[\left(\gamma−\mathrm{ln}\:\mathrm{2}−\frac{\pi}{\mathrm{2}}\right)+{i}\left(−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}+\frac{\pi}{\mathrm{2}}\right)\right] \\ $$$${p}=−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}−\frac{\pi}{\mathrm{2}},{q}=−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}+\frac{\pi}{\mathrm{2}} \\ $$$${c}^{\mathrm{2}} =\left(\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{2}}\right)^{\mathrm{2}} \left({p}+{ip}\right)^{\mathrm{2}} =\frac{\mathrm{2}\pi}{\mathrm{4}}\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} +\mathrm{2}{ipq}\right)=\frac{\pi}{\mathrm{2}}\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} +\mathrm{2}{ipq}\right) \\ $$$${p}+{q}+\mathrm{2}=\left(−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$$${p}−{q}=\pi \\ $$$${p}^{\mathrm{2}} −{q}^{\mathrm{2}} =\left({p}−{q}\right)\left({p}+{q}\right)=\left(−\pi\right)\centerdot\mathrm{2}\left(−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right)=\mathrm{2}\pi\left(\gamma+\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$$$\left[\mathrm{2}\pi\left(\gamma+\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right)+\mathrm{2}{ipq}\right]=\pi^{\mathrm{2}} \left(\gamma+\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right)+{i}\pi{pq} \\ $$$$\mathrm{Re}\left({c}^{\mathrm{2}} \right)=\pi^{\mathrm{2}} \left(\gamma+\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$

Question Number 222195    Answers: 1   Comments: 0

f(x)=x^3 +(√x)+sin x f=?

$${f}\left({x}\right)={x}^{\mathrm{3}} +\sqrt{{x}}+\mathrm{sin}\:{x} \\ $$$${f}=? \\ $$

Question Number 222193    Answers: 1   Comments: 0

Prove: (1/(sin(((3π)/8))))=2(cos((π/8))−sin((π/8))) (2/(sin(((8π)/9))))−(1/(sin(((5π)/9))))=2(√3)+4sin((π/9)) (2/(sin(((6π)/7))))+(1/(sin(((4π)/7))))=4(sin((π/7))+cos((π/(14))))

$$\mathrm{Prove}: \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)}=\mathrm{2}\left(\mathrm{cos}\left(\frac{\pi}{\mathrm{8}}\right)−\mathrm{sin}\left(\frac{\pi}{\mathrm{8}}\right)\right) \\ $$$$\frac{\mathrm{2}}{\mathrm{sin}\left(\frac{\mathrm{8}\pi}{\mathrm{9}}\right)}−\frac{\mathrm{1}}{\mathrm{sin}\left(\frac{\mathrm{5}\pi}{\mathrm{9}}\right)}=\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4sin}\left(\frac{\pi}{\mathrm{9}}\right) \\ $$$$\frac{\mathrm{2}}{\mathrm{sin}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)}+\frac{\mathrm{1}}{\mathrm{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)}=\mathrm{4}\left(\mathrm{sin}\left(\frac{\pi}{\mathrm{7}}\right)+\mathrm{cos}\left(\frac{\pi}{\mathrm{14}}\right)\right) \\ $$

Question Number 222192    Answers: 1   Comments: 0

there are 32 students in a class. for each competition in a sport event in the school each class can send a team with three students. if no two students may be in the same team for more than one time, in how many different competitions can this class participate?

$${there}\:{are}\:\mathrm{32}\:{students}\:{in}\:{a}\:{class}.\:{for} \\ $$$${each}\:{competition}\:{in}\:{a}\:{sport}\:{event}\: \\ $$$${in}\:{the}\:{school}\:{each}\:{class}\:{can}\:{send} \\ $$$${a}\:{team}\:{with}\:{three}\:{students}.\:{if}\:{no} \\ $$$${two}\:{students}\:{may}\:{be}\:{in}\:{the}\:{same} \\ $$$${team}\:{for}\:{more}\:{than}\:{one}\:{time},\:{in} \\ $$$${how}\:{many}\:{different}\:{competitions}\: \\ $$$${can}\:{this}\:{class}\:{participate}? \\ $$

Question Number 222191    Answers: 0   Comments: 0

((d )/dt) ∫_( V^( 3) ) ρ_q (r,t)dV=−∮_( ∂V) J_q (r,t)∙da+∫_( V^( 3) ) S_q (r,t)dV ∫_( V^( 3) ) ((∂ρ_q (r,t))/∂t) dV=−∫_V^( 3) ▽^→ ∙J_q (r,t)dV+∫_( V^( 3) ) S_q (r,t)dV ∫_( V^( 3) ) [ (∂ρ/∂t)+▽^→ ∙J^ (r,t)−S(r,t)]dV=0 ∴((∂ρ_q (r,t))/∂t)+▽^→ ∙J_q (r,t)=S_q (r,t)

$$\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\:\int_{\:{V}^{\:\mathrm{3}} } \rho_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)\mathrm{d}{V}=−\oint_{\:\partial{V}} \:\boldsymbol{\mathrm{J}}_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)\centerdot\mathrm{d}\boldsymbol{\mathrm{a}}+\int_{\:{V}^{\:\mathrm{3}} } \:{S}_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)\mathrm{d}{V} \\ $$$$\int_{\:{V}^{\:\mathrm{3}} } \:\frac{\partial\rho_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)}{\partial{t}}\:\mathrm{dV}=−\int_{{V}^{\:\mathrm{3}} } \overset{\rightarrow} {\bigtriangledown}\centerdot\boldsymbol{\mathrm{J}}_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)\mathrm{d}{V}+\int_{\:{V}^{\:\mathrm{3}} } {S}_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)\mathrm{d}{V} \\ $$$$\int_{\:\mathcal{V}^{\:\mathrm{3}} } \left[\:\frac{\partial\rho}{\partial{t}}+\overset{\rightarrow} {\bigtriangledown}\centerdot\boldsymbol{\mathrm{J}}^{\:} \left(\boldsymbol{\mathrm{r}},{t}\right)−{S}\left(\boldsymbol{\mathrm{r}},{t}\right)\right]\mathrm{d}{V}=\mathrm{0} \\ $$$$\therefore\frac{\partial\rho_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)}{\partial{t}}+\overset{\rightarrow} {\bigtriangledown}\centerdot\boldsymbol{\mathrm{J}}_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)={S}_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right) \\ $$

Question Number 222184    Answers: 2   Comments: 0

f(2) = 8 f(3) = 5 ∫_2 ^( 3) (f(x) + f^′ (x)∙x) dx = ?

$$\mathrm{f}\left(\mathrm{2}\right)\:=\:\mathrm{8} \\ $$$$\mathrm{f}\left(\mathrm{3}\right)\:=\:\mathrm{5} \\ $$$$\int_{\mathrm{2}} ^{\:\mathrm{3}} \:\left(\mathrm{f}\left(\mathrm{x}\right)\:+\:\mathrm{f}\:^{'} \left(\mathrm{x}\right)\centerdot\mathrm{x}\right)\:\mathrm{dx}\:=\:? \\ $$

Question Number 222176    Answers: 1   Comments: 0

Question Number 222175    Answers: 1   Comments: 0

solve (e^(2y) −y)cosx(dy/dx)=e^y sin2x klipto−quanta

$$\boldsymbol{\mathrm{solve}} \\ $$$$\left(\boldsymbol{\mathrm{e}}^{\mathrm{2}\boldsymbol{\mathrm{y}}} −\boldsymbol{\mathrm{y}}\right)\boldsymbol{\mathrm{cosx}}\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}=\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{y}}} \boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}} \\ $$$$\boldsymbol{\mathrm{klipto}}−\boldsymbol{\mathrm{quanta}} \\ $$

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