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Question Number 107171    Answers: 0   Comments: 0

Question Number 107170    Answers: 0   Comments: 0

Question Number 107169    Answers: 2   Comments: 1

If a,b,c,d∈R a+b=8 ab+c+d=23 ad+bc=28 cd=12 Find a^2 +b^2 +c^2 +d^2 .

$$\mathrm{If}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\in\mathbb{R} \\ $$$$\mathrm{a}+\mathrm{b}=\mathrm{8} \\ $$$$\mathrm{ab}+\mathrm{c}+\mathrm{d}=\mathrm{23} \\ $$$$\mathrm{ad}+\mathrm{bc}=\mathrm{28} \\ $$$$\mathrm{cd}=\mathrm{12} \\ $$$$\mathrm{Find}\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} . \\ $$$$ \\ $$

Question Number 107163    Answers: 2   Comments: 0

Question Number 107162    Answers: 1   Comments: 1

Question Number 107157    Answers: 0   Comments: 0

Question Number 107155    Answers: 2   Comments: 0

Question Number 107153    Answers: 1   Comments: 0

@bemath@ (((14)/5))^(((28)/(√x))−5) = ((5/(14)))^((5/(√x))−160)

$$\:\:\:\:\:\:@{bemath}@ \\ $$$$\left(\frac{\mathrm{14}}{\mathrm{5}}\right)^{\frac{\mathrm{28}}{\sqrt{{x}}}−\mathrm{5}} =\:\left(\frac{\mathrm{5}}{\mathrm{14}}\right)^{\frac{\mathrm{5}}{\sqrt{{x}}}−\mathrm{160}} \\ $$

Question Number 107151    Answers: 0   Comments: 0

@JS@ (D^2 +7D+12)y = e^x cos 2x

$$\:\:\:\:\:\:\:\:\:@\mathrm{JS}@ \\ $$$$\left(\mathrm{D}^{\mathrm{2}} +\mathrm{7D}+\mathrm{12}\right)\mathrm{y}\:=\:\mathrm{e}^{\mathrm{x}} \:\mathrm{cos}\:\mathrm{2x}\: \\ $$

Question Number 107147    Answers: 0   Comments: 0

find lim_(x→0^+ ) ((∫_0 ^x e^t ln(t)dt)/(e^x lnx))

$$\mathrm{find}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}^{+} } \:\:\:\:\frac{\int_{\mathrm{0}} ^{\mathrm{x}} \:\mathrm{e}^{\mathrm{t}} \mathrm{ln}\left(\mathrm{t}\right)\mathrm{dt}}{\mathrm{e}^{\mathrm{x}} \mathrm{lnx}} \\ $$

Question Number 107144    Answers: 1   Comments: 0

solve x^2 y^(′′) −xy^′ +2y =xe^(−x) sin(2x)

$$\mathrm{solve}\:\mathrm{x}^{\mathrm{2}} \mathrm{y}^{''} −\mathrm{xy}^{'} \:+\mathrm{2y}\:=\mathrm{xe}^{−\mathrm{x}} \mathrm{sin}\left(\mathrm{2x}\right) \\ $$

Question Number 107141    Answers: 0   Comments: 0

∫(x^3 +x^6 )(((x^3 +2))^(1/3) )dx

$$\int\left({x}^{\mathrm{3}} +{x}^{\mathrm{6}} \right)\left(\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{2}}\right){dx} \\ $$

Question Number 107133    Answers: 0   Comments: 0

Question Number 107124    Answers: 2   Comments: 0

Question Number 107123    Answers: 0   Comments: 0

Question Number 107122    Answers: 1   Comments: 0

Question Number 107121    Answers: 0   Comments: 0

Question Number 107117    Answers: 2   Comments: 0

Question Number 107113    Answers: 0   Comments: 0

Question Number 107108    Answers: 1   Comments: 3

y′′+y=(1/(cosx))

$$\mathrm{y}''+\mathrm{y}=\frac{\mathrm{1}}{\mathrm{cosx}} \\ $$

Question Number 107107    Answers: 1   Comments: 2

Question Number 107105    Answers: 1   Comments: 0

Question Number 107102    Answers: 0   Comments: 0

Question Number 107101    Answers: 0   Comments: 0

Fun time 1+2x+3x^2 +4x^3 +....=(1/((1−x)^2 )) 1+4+12+32+...=(1/((1−2)^2 )) 4+12+32+....=0 (No 1 fun) 5+11+17+23+...=0 Σ_(n=1) ^∞ 6n−1=6Σ_(n=1) ^∞ n−Σ^∞ 1=6.(−(1/(12)))−(−(1/2))=0 Σ^∞ n=−(1/(12)) (Ramanujan sum) Σ^∞ 1=1+1+1+1+1+...=−(1/2) Σ^∞ n^2 .Σ^∞ (1/n^2 )≥(Σ^∞ 1)^2 (Cauchy schwarz ineqality) Σ^∞ n^2 .(π^2 /6)≥(1/4) Σ^∞ n^2 ≥(3/(2π^2 ))

$$\mathrm{Fun}\:\mathrm{time} \\ $$$$ \\ $$$$\mathrm{1}+\mathrm{2x}+\mathrm{3x}^{\mathrm{2}} +\mathrm{4x}^{\mathrm{3}} +....=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$\mathrm{1}+\mathrm{4}+\mathrm{12}+\mathrm{32}+...=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\mathrm{4}+\mathrm{12}+\mathrm{32}+....=\mathrm{0}\:\:\left(\mathrm{No}\:\mathrm{1}\:\mathrm{fun}\right) \\ $$$$ \\ $$$$\mathrm{5}+\mathrm{11}+\mathrm{17}+\mathrm{23}+...=\mathrm{0}\:\:\: \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{6n}−\mathrm{1}=\mathrm{6}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{n}−\overset{\infty} {\sum}\mathrm{1}=\mathrm{6}.\left(−\frac{\mathrm{1}}{\mathrm{12}}\right)−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\overset{\infty} {\sum}\mathrm{n}=−\frac{\mathrm{1}}{\mathrm{12}}\:\:\:\:\left(\mathrm{Ramanujan}\:\mathrm{sum}\right) \\ $$$$\overset{\infty} {\sum}\mathrm{1}=\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+...=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\overset{\infty} {\sum}\mathrm{n}^{\mathrm{2}} .\overset{\infty} {\sum}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\geqslant\left(\overset{\infty} {\sum}\mathrm{1}\right)^{\mathrm{2}} \:\:\:\left(\mathrm{Cauchy}\:\mathrm{schwarz}\:\mathrm{ineqality}\right) \\ $$$$\overset{\infty} {\sum}\mathrm{n}^{\mathrm{2}} .\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\geqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\overset{\infty} {\sum}\mathrm{n}^{\mathrm{2}} \geqslant\frac{\mathrm{3}}{\mathrm{2}\pi^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 107092    Answers: 1   Comments: 2

Question Number 107088    Answers: 3   Comments: 1

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