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Question Number 186998    Answers: 1   Comments: 1

Question Number 186996    Answers: 1   Comments: 0

Question Number 186989    Answers: 0   Comments: 0

Question Number 186983    Answers: 3   Comments: 0

Question Number 186960    Answers: 1   Comments: 0

Question Number 186958    Answers: 0   Comments: 0

Question Number 186952    Answers: 2   Comments: 0

Question Number 186951    Answers: 2   Comments: 0

Question Number 186950    Answers: 0   Comments: 0

Question Number 186941    Answers: 0   Comments: 1

((sinx)/x^b )=((Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!))x^(2n+1) )/x^b ) prove that

$$\frac{{sinx}}{{x}^{{b}} }=\frac{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{x}^{\mathrm{2}{n}+\mathrm{1}} }{{x}^{{b}} } \\ $$$${prove}\:{that} \\ $$

Question Number 186929    Answers: 3   Comments: 0

If { ((B+R+P=−1)),((B^2 +R^2 +P^2 =17)),((B^3 +R^3 +P^3 =11)) :} then B^5 +R^5 +P^5 =?

$$\:\:{If}\:\begin{cases}{{B}+{R}+{P}=−\mathrm{1}}\\{{B}^{\mathrm{2}} +{R}^{\mathrm{2}} +{P}^{\mathrm{2}} =\mathrm{17}}\\{{B}^{\mathrm{3}} +{R}^{\mathrm{3}} +{P}^{\mathrm{3}} =\mathrm{11}}\end{cases} \\ $$$$\:{then}\:{B}^{\mathrm{5}} +{R}^{\mathrm{5}} +{P}^{\mathrm{5}} \:=? \\ $$

Question Number 186926    Answers: 1   Comments: 1

Question Number 186924    Answers: 3   Comments: 0

Question Number 186923    Answers: 0   Comments: 0

Question Number 186925    Answers: 0   Comments: 1

Question Number 186921    Answers: 0   Comments: 0

Question Number 186912    Answers: 0   Comments: 2

Question Number 186911    Answers: 1   Comments: 0

∫_0 ^∞ (1/(1+a^x +a^(x/2) ))dx = (1/(ln a))[ln 3−(π/(3(√3)))]

$$ \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{a}^{{x}} +{a}^{\frac{{x}}{\mathrm{2}}} }{dx}\:=\:\frac{\mathrm{1}}{\mathrm{ln}\:{a}}\left[\mathrm{ln}\:\mathrm{3}−\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\right] \\ $$

Question Number 186910    Answers: 1   Comments: 0

∫_0 ^a (√((cos 2x−cos 2a)/(cos 2x+1))) dx=(π/2)(1−cos a)

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

Question Number 186967    Answers: 1   Comments: 1

Question Number 186966    Answers: 1   Comments: 2

Question Number 186965    Answers: 1   Comments: 1

Question Number 186964    Answers: 1   Comments: 2

Question Number 186963    Answers: 1   Comments: 1

Question Number 186962    Answers: 1   Comments: 0

Question Number 186976    Answers: 1   Comments: 1

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