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Question Number 66462 Answers: 0 Comments: 1

1)simplify S_n (x)=Σ_(k=0) ^n C_n ^k cos^k (x)cos(kx) 2)find the value of A_n =Σ_(k=0) ^n C_n ^k cos^k ((π/n))cos(((kπ)/n))

$$\left.\mathrm{1}\right){simplify}\:{S}_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{cos}^{{k}} \left({x}\right){cos}\left({kx}\right) \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:{of}\:{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{cos}^{{k}} \left(\frac{\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right) \\ $$

Question Number 66461 Answers: 0 Comments: 0

x(n)=3n^2 −2n+7 find even and odd component

$${x}\left({n}\right)=\mathrm{3}{n}^{\mathrm{2}} −\mathrm{2}{n}+\mathrm{7} \\ $$$${find}\:{even}\:{and}\:{odd}\:{component} \\ $$

Question Number 66459 Answers: 0 Comments: 1

1) calculate by residus method ∫_0 ^∞ (dx/((1+x^2 )^3 )) 2) find the value of ∫_0 ^1 ((1+x^4 )/((1+x^2 )^3 ))dx

$$\left.\mathrm{1}\right)\:{calculate}\:{by}\:{residus}\:{method}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+{x}^{\mathrm{4}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx} \\ $$

Question Number 66446 Answers: 0 Comments: 1

Find ∫_1 ^∞ ((1/(E(x))) −(1/x))dx

$$\:\:{Find}\:\:\:\:\int_{\mathrm{1}} ^{\infty} \:\left(\frac{\mathrm{1}}{{E}\left({x}\right)}\:−\frac{\mathrm{1}}{{x}}\right){dx} \\ $$

Question Number 66444 Answers: 0 Comments: 1

calculate lim_(x→0) (x!)^(1/x) if x!=Π(x)=∫_0 ^∞ t^x e^(−t) dt

$$\:{calculate}\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left({x}!\right)^{\frac{\mathrm{1}}{{x}}} \:\:\:\:\:\:\:{if}\:\:\:\:\:{x}!=\Pi\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {t}^{{x}} \:{e}^{−{t}} {dt} \\ $$

Question Number 66439 Answers: 0 Comments: 4

for a geometric series. can the sun to infinty use the two formulas S_∞ = (a/(1−r)) ∣r∣ <1 and S_∞ = (a/(r−1)) ∣r∣ > 1 ?? please i am getting confused on this.

$${for}\:{a}\:{geometric}\:{series}. \\ $$$${can}\:{the}\:{sun}\:{to}\:{infinty}\:{use}\:{the}\:{two}\:{formulas} \\ $$$${S}_{\infty} =\:\frac{{a}}{\mathrm{1}−{r}}\:\:\mid{r}\mid\:\:<\mathrm{1}\:\:{and}\:{S}_{\infty} \:=\:\frac{{a}}{{r}−\mathrm{1}}\:\mid{r}\mid\:>\:\mathrm{1}\:??\:{please}\:{i}\:{am}\:{getting}\:{confused}\:{on}\:{this}. \\ $$

Question Number 66434 Answers: 0 Comments: 2

lim_(x→∞) ((cos^2 x−x)/(1−2x))

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{x}}{\mathrm{1}−\mathrm{2x}} \\ $$

Question Number 66433 Answers: 0 Comments: 0

A 2kg is attached to the end of a vertical wire of length 2m with a diameter of 0.64mm and having an extension of 0.60m.Calculate the tensile strain on the wire(take g=9.8)

$$\mathrm{A}\:\mathrm{2kg}\:\mathrm{is}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{the}\:\mathrm{end}\:\mathrm{of}\:\mathrm{a}\:\mathrm{vertical}\:\mathrm{wire}\:\mathrm{of}\:\mathrm{length}\:\mathrm{2m}\:\mathrm{with}\:\mathrm{a}\:\mathrm{diameter}\:\mathrm{of}\:\mathrm{0}.\mathrm{64mm}\:\mathrm{and}\:\mathrm{having}\:\mathrm{an}\:\mathrm{extension}\:\mathrm{of}\:\mathrm{0}.\mathrm{60m}.\mathrm{Calculate}\:\mathrm{the}\:\mathrm{tensile}\:\mathrm{strain}\:\mathrm{on}\:\mathrm{the}\:\mathrm{wire}\left(\mathrm{take}\:\mathrm{g}=\mathrm{9}.\mathrm{8}\right) \\ $$

Question Number 66431 Answers: 1 Comments: 0

Determine x e y: { ((x^(1/(√i)) + (1/y^(i(√i)) ) = 10)),(((1/((xy)^(i(√i)) )) = 21)) :}

Question Number 66421 Answers: 1 Comments: 0

show that for a given complex number z z^n = r^n (cosnθ + isinnθ)

$${show}\:{that}\:{for}\:{a}\:{given}\:{complex}\:{number}\:{z} \\ $$$$\:{z}^{{n}} \:=\:{r}^{{n}} \:\left({cosn}\theta\:+\:{isinn}\theta\right)\: \\ $$

Question Number 66420 Answers: 0 Comments: 3

solve the differential equation 2(d^2 y/dx^2 ) + (dy/dx) − e^(−x) = 4

$${solve}\:{the}\:{differential}\:{equation} \\ $$$$\:\mathrm{2}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\frac{{dy}}{{dx}}\:−\:{e}^{−{x}} \:=\:\mathrm{4} \\ $$

Question Number 66418 Answers: 0 Comments: 0

what operation on interger used in 9(7.8)=(9.7).8??

$${what}\:{operation}\:{on}\:{interger}\:{used}\:{in}\:\mathrm{9}\left(\mathrm{7}.\mathrm{8}\right)=\left(\mathrm{9}.\mathrm{7}\right).\mathrm{8}?? \\ $$

Question Number 66406 Answers: 0 Comments: 0

Question Number 66405 Answers: 0 Comments: 0

Question Number 66404 Answers: 0 Comments: 3

Question Number 66403 Answers: 0 Comments: 0

Question Number 66401 Answers: 0 Comments: 0

Question Number 66399 Answers: 0 Comments: 1

Show that for all real values of x; x^(2/3) + 6x^(1/3) + 10 >0

$${Show}\:{that}\:{for}\:{all}\:{real} \\ $$$${values}\:{of}\:{x};\: \\ $$$$\:\:{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \:+\:\mathrm{6}{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \:+\:\mathrm{10}\:>\mathrm{0} \\ $$

Question Number 66396 Answers: 1 Comments: 0

Seja 53^(log_(1/(√e^𝛑 )) [(((x+11)!))^(1/(9999999)) ]) = 1. Calcule (x_1 /x_2 )+0,9.

$$\: \\ $$$$\:\boldsymbol{\mathrm{Seja}}\:\:\mathrm{53}^{\boldsymbol{\mathrm{log}}_{\frac{\mathrm{1}}{\sqrt{\boldsymbol{{e}}^{\boldsymbol{\pi}} }}} \left[\sqrt[{\mathrm{9999999}}]{\left(\boldsymbol{{x}}+\mathrm{11}\right)!}\right]} \:=\:\mathrm{1}. \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{Calcule}}\:\:\frac{\boldsymbol{\mathrm{x}}_{\mathrm{1}} }{\boldsymbol{\mathrm{x}}_{\mathrm{2}} }+\mathrm{0},\mathrm{9}. \\ $$

Question Number 66413 Answers: 1 Comments: 0

(√(8+log_6 (x!)))+(√(17−log_(x!) (6))) = 7

$$\: \\ $$$$\:\:\sqrt{\mathrm{8}+\boldsymbol{\mathrm{log}}_{\mathrm{6}} \left(\boldsymbol{\mathrm{x}}!\right)}+\sqrt{\mathrm{17}−\boldsymbol{\mathrm{log}}_{\boldsymbol{\mathrm{x}}!} \left(\mathrm{6}\right)}\:=\:\mathrm{7} \\ $$$$\: \\ $$

Question Number 66412 Answers: 1 Comments: 3

if f(x)=ln(x+(√(x^2 +1))) find f^(−1) (x)=?

$${if} \\ $$$$ \\ $$$${f}\left({x}\right)={ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$ \\ $$$${find} \\ $$$$ \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=? \\ $$

Question Number 66382 Answers: 0 Comments: 14

Give me any Quintic, i shall solve it. For sure! At^5 +Bt^4 +Ct^3 +Dt^2 +Et+F=0 wont even assume A=1, or B=0. but if A+C+E=B+D+F then my formula dont work but then obviously t=−1 is a root!

$${Give}\:{me}\:{any}\:{Quintic},\:{i}\:{shall}\:{solve} \\ $$$${it}.\:{For}\:{sure}! \\ $$$${At}^{\mathrm{5}} +{Bt}^{\mathrm{4}} +{Ct}^{\mathrm{3}} +{Dt}^{\mathrm{2}} +{Et}+{F}=\mathrm{0} \\ $$$${wont}\:{even}\:{assume}\:{A}=\mathrm{1},\:{or}\:{B}=\mathrm{0}. \\ $$$${but}\:{if}\:{A}+{C}+{E}={B}+{D}+{F}\: \\ $$$${then}\:{my}\:{formula}\:{dont}\:{work} \\ $$$${but}\:{then}\:{obviously}\:{t}=−\mathrm{1}\:{is}\:{a}\:{root}! \\ $$

Question Number 66379 Answers: 0 Comments: 2

Question Number 66381 Answers: 0 Comments: 0

Question Number 66356 Answers: 1 Comments: 0

Question Number 66355 Answers: 0 Comments: 1

Value of x satiesfied y=((log_4 (x^2 −1))/(4x^2 +2x+1)) negative value is... a. −1<x<(√2) b. −(√2)<x<1 c. −(√2)<x<(√2) d. −(√2)<x<−1 e. x<−2

$${V}\mathrm{alue}\:\mathrm{of}\:{x}\:\mathrm{satiesfied}\:{y}=\frac{\mathrm{log}_{\mathrm{4}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}} \\ $$$${negative}\:{value}\:\mathrm{is}... \\ $$$$\mathrm{a}.\:−\mathrm{1}<{x}<\sqrt{\mathrm{2}} \\ $$$${b}.\:−\sqrt{\mathrm{2}}<{x}<\mathrm{1} \\ $$$${c}.\:−\sqrt{\mathrm{2}}<{x}<\sqrt{\mathrm{2}} \\ $$$${d}.\:−\sqrt{\mathrm{2}}<{x}<−\mathrm{1} \\ $$$${e}.\:{x}<−\mathrm{2} \\ $$

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