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Question Number 37037    Answers: 4   Comments: 0

If ((a+b−c)/(a+b)) = ((b+c−a)/(b+c)) =((c+a−b)/(c+a)) and a+b+c ≠ 0 then which of the following is true 1) a=−b=c 2)−a=−b=c 3) a=b=c 4) a=b≠c

$${If}\:\frac{{a}+{b}−{c}}{{a}+{b}}\:=\:\frac{{b}+{c}−{a}}{{b}+{c}}\:=\frac{{c}+{a}−{b}}{{c}+{a}}\:{and}\: \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}\:\neq\:\mathrm{0}\:{then}\:{which}\:{of}\:{the}\:{following} \\ $$$$\left.{i}\left.{s}\:{true}\:\mathrm{1}\right)\:{a}=−{b}={c}\:\mathrm{2}\right)−{a}=−{b}={c}\: \\ $$$$\left.\mathrm{3}\left.\right)\:\mathrm{a}=\mathrm{b}=\mathrm{c}\:\mathrm{4}\right)\:\mathrm{a}=\mathrm{b}\neq{c} \\ $$

Question Number 37036    Answers: 0   Comments: 3

∫_0 ^( a) (1−((b−x)/(√((b−x)^2 +cx)))) dx = ?

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

Question Number 37032    Answers: 1   Comments: 1

Question Number 37028    Answers: 1   Comments: 0

∫((2x−1)/(5x^2 −x+2)) dx = ?

$$\int\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{5}{x}^{\mathrm{2}} −{x}+\mathrm{2}}\:{dx}\:\:=\:\:? \\ $$

Question Number 37026    Answers: 0   Comments: 0

Question Number 37023    Answers: 0   Comments: 0

Question Number 37022    Answers: 0   Comments: 0

Question Number 37021    Answers: 0   Comments: 0

Question Number 37020    Answers: 0   Comments: 0

Question Number 37018    Answers: 3   Comments: 3

Question Number 37010    Answers: 0   Comments: 0

Question Number 37009    Answers: 0   Comments: 0

Question Number 37008    Answers: 0   Comments: 0

Question Number 37005    Answers: 0   Comments: 0

Question Number 37004    Answers: 3   Comments: 0

Question Number 36997    Answers: 2   Comments: 1

∫ (√(((1+x)/x) ))dx = ?

$$\int\:\sqrt{\frac{\mathrm{1}+{x}}{{x}}\:}{dx}\:=\:? \\ $$

Question Number 36990    Answers: 0   Comments: 0

Question Number 36986    Answers: 0   Comments: 0

Question Number 36985    Answers: 0   Comments: 0

Question Number 36978    Answers: 1   Comments: 1

if (a^2 /(b+c)) = (b^2 /(c+a)) = (c^2 /(a+b)) = 1 then find the value of (1/(1+a)) + (1/(1+b)) + (1/(1+c))

$${if}\:\:\frac{{a}^{\mathrm{2}} }{{b}+{c}}\:=\:\frac{{b}^{\mathrm{2}} }{{c}+{a}}\:=\:\frac{{c}^{\mathrm{2}} }{{a}+{b}}\:=\:\mathrm{1}\:{then}\:{find}\:{the}\:{value}\:{of}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{a}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+{b}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+{c}} \\ $$

Question Number 36965    Answers: 2   Comments: 3

∫((x^5 −x^4 +x^3 −1)/((x^2 −x+1)^3 ))dx=

$$\int\frac{{x}^{\mathrm{5}} −{x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{1}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{3}} }{dx}= \\ $$

Question Number 36957    Answers: 2   Comments: 0

Interval in which given function is decreasing. f(x)= (2^x −1)(2^x −2)^2

$$\mathrm{Interval}\:\mathrm{in}\:\mathrm{which}\:\mathrm{given}\:\mathrm{function}\:\mathrm{is} \\ $$$${decreasing}. \\ $$$$\mathrm{f}\left({x}\right)=\:\left(\mathrm{2}^{{x}} −\mathrm{1}\right)\left(\mathrm{2}^{{x}} −\mathrm{2}\right)^{\mathrm{2}} \\ $$

Question Number 36953    Answers: 1   Comments: 3

lim_(n→∞) nsin (2π(√(1+n^2 )) ) ,( n∈N).

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{nsin}\:\left(\mathrm{2}\pi\sqrt{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\:\right)\:,\left(\:\mathrm{n}\in\mathbb{N}\right). \\ $$

Question Number 36948    Answers: 0   Comments: 2

calculate ∫∫_D (√(xy)) dxdy with D={(x,y)∈R^2 / (x+y)^2 ≥2x and xy≥0}

$${calculate}\:\:\:\int\int_{{D}} \sqrt{{xy}}\:{dxdy}\:\:{with} \\ $$$${D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} \:/\:\left({x}+{y}\right)^{\mathrm{2}} \:\geqslant\mathrm{2}{x}\:\:{and}\:{xy}\geqslant\mathrm{0}\right\} \\ $$

Question Number 36947    Answers: 1   Comments: 1

integrate the d.equation xy^′ +y = ((2x)/(√(1−x^4 ))) .

$${integrate}\:{the}\:{d}.{equation}\:\:{xy}^{'} \:+{y}\:=\:\frac{\mathrm{2}{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\:. \\ $$

Question Number 36946    Answers: 0   Comments: 2

calculateϕ(λ)= ∫_0 ^π ((cos(t))/(1−2λ cost +λ^2 )) dt

$${calculate}\varphi\left(\lambda\right)=\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\:\:\frac{{cos}\left({t}\right)}{\mathrm{1}−\mathrm{2}\lambda\:{cost}\:+\lambda^{\mathrm{2}} }\:{dt} \\ $$

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