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Question Number 88889    Answers: 0   Comments: 1

Question Number 88886    Answers: 1   Comments: 5

Question Number 88882    Answers: 1   Comments: 1

The coefficient of x^3 in ((√x^5 ) +(3/(√x^3 )))^6 is

$$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:\:{x}^{\mathrm{3}} \:\mathrm{in}\:\left(\sqrt{{x}^{\mathrm{5}} }\:+\frac{\mathrm{3}}{\sqrt{{x}^{\mathrm{3}} }}\right)^{\mathrm{6}} \:\:\mathrm{is} \\ $$

Question Number 88881    Answers: 1   Comments: 0

Question Number 88876    Answers: 0   Comments: 2

Given a 10−digit number X = 1345789026 How many 10−digit number that can be made using every digit from X, with condition: If a number n is located in k^(th) position of X, then the new created number must not contain number n in k^(th) position Example: • Number 1 is located in 1^(st) position of X, hence 1234567890 is not valid, but 2134567890 is valid • Number 5 and 0 are located in 4^(th) and 8^(th) position of X, hence 9435162087 is not valid, but 9431506287 is valid. • 1345026789 is not valid • and so on...

$$\mathrm{Given}\:\mathrm{a}\:\mathrm{10}−\mathrm{digit}\:\mathrm{number}\:{X}\:=\:\mathrm{1345789026} \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{10}−\mathrm{digit}\:\mathrm{number}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{made} \\ $$$$\mathrm{using}\:\mathrm{every}\:\mathrm{digit}\:\mathrm{from}\:{X},\:\mathrm{with}\:\mathrm{condition}: \\ $$$$\mathrm{If}\:\mathrm{a}\:\mathrm{number}\:{n}\:\:\mathrm{is}\:\mathrm{located}\:\mathrm{in}\:{k}^{{th}} \:\mathrm{position}\:\mathrm{of}\:{X},\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{new}\:\mathrm{created}\:\mathrm{number}\:\mathrm{must}\:\mathrm{not}\:\mathrm{contain} \\ $$$$\mathrm{number}\:{n}\:\mathrm{in}\:{k}^{{th}} \:\mathrm{position} \\ $$$$ \\ $$$$\mathrm{Example}: \\ $$$$\bullet\:\mathrm{Number}\:\mathrm{1}\:\mathrm{is}\:\mathrm{located}\:\mathrm{in}\:\mathrm{1}^{{st}} \:\mathrm{position}\:\mathrm{of}\:{X},\:\mathrm{hence} \\ $$$$\mathrm{1234567890}\:\mathrm{is}\:\mathrm{not}\:\mathrm{valid},\:\mathrm{but}\:\mathrm{2134567890} \\ $$$$\mathrm{is}\:\mathrm{valid} \\ $$$$\bullet\:\mathrm{Number}\:\mathrm{5}\:\mathrm{and}\:\mathrm{0}\:\mathrm{are}\:\mathrm{located}\:\mathrm{in}\:\mathrm{4}^{{th}} \:\mathrm{and}\:\mathrm{8}^{{th}} \:\mathrm{position} \\ $$$$\mathrm{of}\:{X},\:\mathrm{hence}\:\mathrm{9435162087}\:\mathrm{is}\:\mathrm{not}\:\mathrm{valid},\:\mathrm{but} \\ $$$$\mathrm{9431506287}\:\mathrm{is}\:\mathrm{valid}. \\ $$$$\bullet\:\mathrm{1345026789}\:\mathrm{is}\:\mathrm{not}\:\mathrm{valid} \\ $$$$\bullet\:\mathrm{and}\:\mathrm{so}\:\mathrm{on}... \\ $$

Question Number 88873    Answers: 0   Comments: 1

3+(√(x^2 −5)) > ∣x−1∣

$$\mathrm{3}+\sqrt{{x}^{\mathrm{2}} −\mathrm{5}}\:>\:\mid{x}−\mathrm{1}\mid\: \\ $$

Question Number 88872    Answers: 0   Comments: 1

The system of equations kx+y+z=1, x+ky+z=k and x+y+kz=k^2 have no solution if k equals

$$\mathrm{The}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations}\:\:\:{kx}+{y}+{z}=\mathrm{1}, \\ $$$${x}+{ky}+{z}={k}\:\:\:\mathrm{and}\:\:\:{x}+{y}+{kz}={k}^{\mathrm{2}} \:\mathrm{have} \\ $$$$\mathrm{no}\:\mathrm{solution}\:\mathrm{if}\:\:\:{k}\:\:\mathrm{equals} \\ $$

Question Number 88871    Answers: 0   Comments: 0

Question Number 88867    Answers: 1   Comments: 1

Question Number 88865    Answers: 0   Comments: 1

Question Number 88863    Answers: 0   Comments: 1

∫((8cos^3 (x))/(8+sin^3 2x))dx

$$\int\frac{\mathrm{8}{cos}^{\mathrm{3}} \left({x}\right)}{\mathrm{8}+{sin}^{\mathrm{3}} \mathrm{2}{x}}{dx} \\ $$$$ \\ $$

Question Number 88860    Answers: 0   Comments: 3

Question Number 88859    Answers: 0   Comments: 1

∫ e^(ax) cos bx dx ∫x^2 e^(2x) ln3x^2 dx

$$\int\:\boldsymbol{{e}}^{\boldsymbol{{ax}}} \mathrm{cos}\:\boldsymbol{{bx}}\:\boldsymbol{{dx}} \\ $$$$\int\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{x}}} \boldsymbol{{ln}}\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} \:\boldsymbol{{dx}} \\ $$

Question Number 88858    Answers: 0   Comments: 0

Question Number 88852    Answers: 1   Comments: 0

prove that ∫_0 ^n ⌈x⌉dx= ((n(n+1))/2) and ∫_0 ^n ⌊x⌋dx=((n(n−1))/2) when ⌊..⌋ is floor and ⌈..⌉ is ceil

$${prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{{n}} \lceil{x}\rceil{dx}=\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:{and}\:\int_{\mathrm{0}} ^{{n}} \lfloor{x}\rfloor{dx}=\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${when}\:\lfloor..\rfloor\:{is}\:{floor}\:{and}\:\lceil..\rceil\:{is}\:{ceil} \\ $$

Question Number 88851    Answers: 0   Comments: 1

Question Number 88820    Answers: 2   Comments: 5

Question Number 88811    Answers: 1   Comments: 0

{ (((x+1)^2 (y+1)^2 =27xy)),(((x^2 +1)(y^2 +1) =10xy)) :}

$$\begin{cases}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({y}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{27}{xy}}\\{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{1}\right)\:=\mathrm{10}{xy}}\end{cases} \\ $$

Question Number 88789    Answers: 0   Comments: 7

∫∫ln(x+1) dx dy

$$\int\int{ln}\left({x}+\mathrm{1}\right)\:{dx}\:{dy} \\ $$

Question Number 88778    Answers: 0   Comments: 3

Question Number 88794    Answers: 1   Comments: 0

Question Number 88771    Answers: 0   Comments: 4

solve x^x^4 =64

$${solve} \\ $$$${x}^{{x}^{\mathrm{4}} } =\mathrm{64} \\ $$

Question Number 88761    Answers: 0   Comments: 4

if a_n =((n!)/(n^n e^(−n) (√(2πn)))) and b_n =(((2n)!(√n))/(4^n (n!)^2 )) lim_(n→∞) a_n =1 find lim_(n→∞) b_n =?

$${if}\:{a}_{{n}} =\frac{{n}!}{{n}^{{n}} \:{e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}} \\ $$$${and}\:{b}_{{n}} =\frac{\left(\mathrm{2}{n}\right)!\sqrt{{n}}}{\mathrm{4}^{{n}} \:\left({n}!\right)^{\mathrm{2}} } \\ $$$$\underset{{n}\rightarrow\infty} {{lim}a}_{{n}} =\mathrm{1} \\ $$$${find}\:\underset{{n}\rightarrow\infty} {{lim}b}_{{n}} =? \\ $$

Question Number 88758    Answers: 2   Comments: 22

Some people may have noticed that i usually calculate areas concerning parabola directly, without applying complicated integral calculus. Here i am giving you the backgroud. Actually you know all these things and you are able to prove them. Maybe you just forget to apply them.

$${Some}\:{people}\:{may}\:{have}\:{noticed}\:{that}\:{i} \\ $$$${usually}\:{calculate}\:{areas}\:{concerning} \\ $$$${parabola}\:{directly},\:{without}\:{applying} \\ $$$${complicated}\:{integral}\:{calculus}. \\ $$$${Here}\:{i}\:{am}\:{giving}\:{you}\:{the}\:{backgroud}.\: \\ $$$${Actually}\:{you}\:{know}\:{all}\:{these}\:{things}\:{and} \\ $$$${you}\:{are}\:{able}\:{to}\:{prove}\:{them}.\:{Maybe} \\ $$$${you}\:{just}\:{forget}\:{to}\:{apply}\:{them}. \\ $$

Question Number 88756    Answers: 0   Comments: 0

E is reported in (i^→ ;j^→ ) base. e_1 ^→ =2i^→ +3j^→ ; e_2 ^→ =i^→ −2j^→ and e_3 ^→ =4i^→ −5j^→ belong to E. 1)Determinate the cordonnates of e_3 ^→ in the base B(e_1 ^→ ;e_2 ^→ ).

$$\mathrm{E}\:\mathrm{is}\:\mathrm{reported}\:\mathrm{in}\:\left(\overset{\rightarrow} {\mathrm{i}};\overset{\rightarrow} {\mathrm{j}}\right)\:\mathrm{base}. \\ $$$$\overset{\rightarrow} {\mathrm{e}}_{\mathrm{1}} =\mathrm{2}\overset{\rightarrow} {\mathrm{i}}+\mathrm{3}\overset{\rightarrow} {\mathrm{j}}\:;\:\:\:\overset{\rightarrow} {\mathrm{e}}_{\mathrm{2}} =\overset{\rightarrow} {\mathrm{i}}−\mathrm{2}\overset{\rightarrow} {\mathrm{j}}\:\mathrm{and}\:\:\:\overset{\rightarrow} {\mathrm{e}}_{\mathrm{3}} =\mathrm{4}\overset{\rightarrow} {\mathrm{i}}−\mathrm{5}\overset{\rightarrow} {\mathrm{j}}\: \\ $$$$\mathrm{belong}\:\mathrm{to}\:\mathrm{E}. \\ $$$$\left.\mathrm{1}\right)\mathrm{Determinate}\:\mathrm{the}\:\mathrm{cordonnates}\:\mathrm{of}\:\overset{\rightarrow} {\mathrm{e}}_{\mathrm{3}} \:\mathrm{in}\:\mathrm{the}\: \\ $$$$\mathrm{base}\:\mathrm{B}\left(\overset{\rightarrow} {\mathrm{e}}_{\mathrm{1}} ;\overset{\rightarrow} {\mathrm{e}}_{\mathrm{2}} \right). \\ $$

Question Number 88754    Answers: 1   Comments: 3

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