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Question Number 192630    Answers: 0   Comments: 0

z=xy−5x+2y. find (dz/dx) and (dz/dy) at(2,4)

$${z}={xy}−\mathrm{5}{x}+\mathrm{2}{y}.\:{find}\:\frac{{dz}}{{dx}}\:{and}\:\frac{{dz}}{{dy}}\:{at}\left(\mathrm{2},\mathrm{4}\right) \\ $$

Question Number 192629    Answers: 0   Comments: 0

Z=f(x_(1,) x_(2,) x_3 )=x_1 x_2 +x_1 ^5 −x_2 ^2 x_3 find f_1 ,f_(11) ,and f_(21)

$${Z}={f}\left({x}_{\mathrm{1},} {x}_{\mathrm{2},} {x}_{\mathrm{3}} \right)={x}_{\mathrm{1}} {x}_{\mathrm{2}} +{x}_{\mathrm{1}} ^{\mathrm{5}} −{x}_{\mathrm{2}} ^{\mathrm{2}} {x}_{\mathrm{3}} \:{find}\:{f}_{\mathrm{1}} ,{f}_{\mathrm{11}} ,{and}\:{f}_{\mathrm{21}} \\ $$

Question Number 192625    Answers: 3   Comments: 1

lim_(x→0) ((sin^4 (πcos(x)))/(1−cos(1−cos(1−cos(x)))))

$${lim}_{{x}\rightarrow\mathrm{0}} \frac{{sin}^{\mathrm{4}} \left(\pi{cos}\left({x}\right)\right)}{\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left({x}\right)\right)\right)} \\ $$

Question Number 192624    Answers: 1   Comments: 0

cos36−cos72=?

$${cos}\mathrm{36}−{cos}\mathrm{72}=? \\ $$

Question Number 192623    Answers: 1   Comments: 0

cot70+4cos70=?

$${cot}\mathrm{70}+\mathrm{4}{cos}\mathrm{70}=? \\ $$

Question Number 192622    Answers: 1   Comments: 0

(3/2)cos^(−1) (√(2/(2+π^2 )))+(1/4)sin^(−1) ((2(√(2π)))/(2+π^2 ))+tan^(−1) ((√2)/π)

$$\frac{\mathrm{3}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \sqrt{\frac{\mathrm{2}}{\mathrm{2}+\pi^{\mathrm{2}} }}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}\sqrt{\mathrm{2}\pi}}{\mathrm{2}+\pi^{\mathrm{2}} }+\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}}{\pi} \\ $$

Question Number 192617    Answers: 2   Comments: 0

Question Number 192616    Answers: 0   Comments: 0

Question Number 192612    Answers: 1   Comments: 3

(1/x) + (1/y) + (1/z) = 1 find the minimum value of x^2 + y^2 + z^2

$$\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{z}}\:=\:\mathrm{1}\: \\ $$$${find}\:{the}\:{minimum}\:{value}\:{of}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \\ $$

Question Number 192608    Answers: 2   Comments: 0

let k be natural number. defined s_k as the sum of the infinite series s_k =((k^2 −1)/k^0 ) + ((k^2 −1)/k^1 ) + ((k^2 −1)/k^2 ) +... find the value of Σ_(k=1) ^∞ [(s_k /2^(k−1) )] .

$${let}\:{k}\:{be}\:{natural}\:{number}.\:{defined}\:{s}_{{k}} \:{as}\:{the} \\ $$$${sum}\:{of}\:{the}\:{infinite}\:{series}\:{s}_{{k}} =\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{0}} }\:+\:\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{1}} }\:+\:\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} }\:+... \\ $$$${find}\:{the}\:{value}\:{of}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left[\frac{{s}_{{k}} }{\mathrm{2}^{{k}−\mathrm{1}} }\right]\:\:. \\ $$

Question Number 192606    Answers: 0   Comments: 1

proof that 0^0 =1 without limit!

$${proof}\:{that}\:\mathrm{0}^{\mathrm{0}} =\mathrm{1}\:{without}\:{limit}! \\ $$

Question Number 192604    Answers: 0   Comments: 1

many people say that the 0^(0 ) is an uninfinity ones of them say that 0^0 is infinity and equal to 1! what do you think wich ones of them say right?

$${many}\:{people}\:{say}\:{that}\:{the}\:\mathrm{0}^{\mathrm{0}\:} {is}\:{an}\:{uninfinity} \\ $$$${ones}\:{of}\:{them}\:{say}\:{that}\:\mathrm{0}^{\mathrm{0}} \:{is}\:{infinity}\:{and}\:{equal}\:{to}\:\mathrm{1}! \\ $$$${what}\:{do}\:{you}\:{think}\:{wich}\:{ones}\:{of}\:{them} \\ $$$${say}\:{right}? \\ $$

Question Number 192597    Answers: 1   Comments: 0

a_(1 ) , a_2 , a_(3 ) , .... , a_n is a sequence satifies that a_(n+2) =a_(n+1) −a_n for n ≥ 1. suppose the sum of the first 999 terms = 1003 and the sum of the first 1003 terms = −999 find the sum of the first 2002 terms.

$${a}_{\mathrm{1}\:} \:,\:{a}_{\mathrm{2}} \:,\:{a}_{\mathrm{3}\:} \:,\:....\:,\:{a}_{{n}} \:{is}\:{a}\:{sequence}\:{satifies}\:{that} \\ $$$${a}_{{n}+\mathrm{2}} ={a}_{{n}+\mathrm{1}} −{a}_{{n}} \:{for}\:{n}\:\geq\:\mathrm{1}.\:{suppose}\:{the}\:{sum}\: \\ $$$${of}\:{the}\:{first}\:\mathrm{999}\:{terms}\:=\:\mathrm{1003}\:{and}\:{the}\:{sum} \\ $$$${of}\:{the}\:{first}\:\mathrm{1003}\:{terms}\:=\:−\mathrm{999}\:{find}\:{the}\: \\ $$$${sum}\:{of}\:{the}\:{first}\:\mathrm{2002}\:{terms}. \\ $$

Question Number 192596    Answers: 2   Comments: 0

Question Number 192594    Answers: 0   Comments: 0

Question Number 192593    Answers: 0   Comments: 0

Question Number 192590    Answers: 1   Comments: 0

Σ_(n=1) ^∞ Σ_(m=1) ^∞ [((m^2 n)/(3^m (3^n .m+3^m .n)))]=λ find λ.

$$ \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{{m}^{\mathrm{2}} {n}}{\mathrm{3}^{{m}} \left(\mathrm{3}^{{n}} .{m}+\mathrm{3}^{{m}} .{n}\right)}\right]=\lambda\: \\ $$$${find}\:\lambda. \\ $$$$ \\ $$

Question Number 192584    Answers: 0   Comments: 0

Question Number 192582    Answers: 1   Comments: 5

Question Number 192580    Answers: 0   Comments: 0

This might be helpful: How to easily calculate z_1 ^z_2 with z_1 , z_2 ∈C: Transform z_1 =re^(iθ) and z_2 =a+bi ⇒ z_1 ^z_2 =(re^(iθ) )^(a+bi) z_1 ^z_2 =r^(a+bi) e^(−bθ+iaθ) z_1 ^z_2 =r^a e^(−bθ) r^(bi) e^(iaθ) z_1 ^z_2 =(r^a /e^(bθ) )e^(i(aθ+bln r))

$$\mathrm{This}\:\mathrm{might}\:\mathrm{be}\:\mathrm{helpful}: \\ $$$$\mathrm{How}\:\mathrm{to}\:\mathrm{easily}\:\mathrm{calculate}\:{z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } \:\mathrm{with}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\in\mathbb{C}: \\ $$$$\mathrm{Transform}\:{z}_{\mathrm{1}} ={r}\mathrm{e}^{\mathrm{i}\theta} \:\mathrm{and}\:{z}_{\mathrm{2}} ={a}+{b}\mathrm{i}\:\Rightarrow \\ $$$${z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } =\left({r}\mathrm{e}^{\mathrm{i}\theta} \right)^{{a}+{b}\mathrm{i}} \\ $$$${z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } ={r}^{{a}+{b}\mathrm{i}} \mathrm{e}^{−{b}\theta+\mathrm{i}{a}\theta} \\ $$$${z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } ={r}^{{a}} \mathrm{e}^{−{b}\theta} {r}^{{b}\mathrm{i}} \mathrm{e}^{\mathrm{i}{a}\theta} \\ $$$${z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } =\frac{{r}^{{a}} }{\mathrm{e}^{{b}\theta} }\mathrm{e}^{\mathrm{i}\left({a}\theta+{b}\mathrm{ln}\:{r}\right)} \\ $$

Question Number 192560    Answers: 1   Comments: 3

Question Number 192572    Answers: 0   Comments: 0

Let ABCD be a rectangle having an area of 290. Let E be on BC such that BE : BC = 3 : 2. Let F be on CD such that CF : FD = 3 : 1. If G is the intersection of AE and BF, compute the area of △BEG.

$$\mathrm{Let}\:\:{ABCD}\:\:\mathrm{be}\:\:\mathrm{a}\:\:\mathrm{rectangle}\:\:\mathrm{having}\:\:\mathrm{an}\:\mathrm{area}\:\:\mathrm{of}\:\:\mathrm{290}. \\ $$$$\mathrm{Let}\:\:{E}\:\:\mathrm{be}\:\:\mathrm{on}\:\:{BC}\:\:\mathrm{such}\:\:\mathrm{that}\:\:{BE}\::\:{BC}\:=\:\mathrm{3}\::\:\mathrm{2}. \\ $$$$\mathrm{Let}\:\:{F}\:\:\mathrm{be}\:\:\mathrm{on}\:\:{CD}\:\:\mathrm{such}\:\:\mathrm{that}\:\:{CF}\::\:{FD}\:=\:\mathrm{3}\::\:\mathrm{1}. \\ $$$$\mathrm{If}\:\:{G}\:\:\mathrm{is}\:\:\mathrm{the}\:\:\mathrm{intersection}\:\:\mathrm{of}\:\:{AE}\:\:\mathrm{and}\:\:{BF},\:\:\mathrm{compute} \\ $$$$\mathrm{the}\:\:\mathrm{area}\:\:\mathrm{of}\:\:\bigtriangleup{BEG}. \\ $$

Question Number 192570    Answers: 2   Comments: 0

Question Number 192573    Answers: 0   Comments: 0

solve; lim_(x→0) x^2 tan(((sinπx)/(2x))) solution let L=lim_(x→0) x^2 tan(((sinπx)/(2x))) since sinx∼x−(x^3 /6) L=lim_(x→0) x^2 tan(((πx)/(2x))−((π^3 x^3 )/(12x))) L=lim_(x→0) x^2 tan((π/2)−((π^3 x^2 )/(12))) since tan((π/2)−x)=(1/(tanx)) L=lim_(x→0) (x^2 /(tan(((π^3 x^2 )/(12))))) L=lim_(x→0) (((π^3 x^2 )/(12))/(tan(((π^3 x^2 )/(12))))) ((12)/π^3 ) L=((12)/π^3 )lim_(x→0) (((π^3 x^2 )/(12))/(tan(((π^3 x^2 )/(12))))) since lim_(x→0) (x/(tanx))=1 L=((12)/π^3 ) ∙1=((12)/π^3 ) solved by HY a.k.a senestro

$${solve}; \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}^{\mathrm{2}} {tan}\left(\frac{{sin}\pi{x}}{\mathrm{2}{x}}\right) \\ $$$${solution} \\ $$$${let}\:{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}^{\mathrm{2}} {tan}\left(\frac{{sin}\pi{x}}{\mathrm{2}{x}}\right) \\ $$$${since}\:{sinx}\sim{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\: \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}^{\mathrm{2}} {tan}\left(\frac{\pi{x}}{\mathrm{2}{x}}−\frac{\pi^{\mathrm{3}} {x}^{\mathrm{3}} }{\mathrm{12}{x}}\right) \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}^{\mathrm{2}} {tan}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi^{\mathrm{3}} {x}^{\mathrm{2}} }{\mathrm{12}}\right) \\ $$$${since}\:{tan}\left(\frac{\pi}{\mathrm{2}}−{x}\right)=\frac{\mathrm{1}}{{tanx}} \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} }{{tan}\left(\frac{\pi^{\mathrm{3}} {x}^{\mathrm{2}} }{\mathrm{12}}\right)} \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\pi^{\mathrm{3}} {x}^{\mathrm{2}} }{\mathrm{12}}}{{tan}\left(\frac{\pi^{\mathrm{3}} {x}^{\mathrm{2}} }{\mathrm{12}}\right)}\:\frac{\mathrm{12}}{\pi^{\mathrm{3}} } \\ $$$${L}=\frac{\mathrm{12}}{\pi^{\mathrm{3}} }\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\pi^{\mathrm{3}} {x}^{\mathrm{2}} }{\mathrm{12}}}{{tan}\left(\frac{\pi^{\mathrm{3}} {x}^{\mathrm{2}} }{\mathrm{12}}\right)} \\ $$$${since}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}}{{tanx}}=\mathrm{1} \\ $$$${L}=\frac{\mathrm{12}}{\pi^{\mathrm{3}} }\:\centerdot\mathrm{1}=\frac{\mathrm{12}}{\pi^{\mathrm{3}} } \\ $$$${solved}\:{by}\:{HY}\:{a}.{k}.{a}\:{senestro} \\ $$

Question Number 192550    Answers: 1   Comments: 0

how can find the sum Σ_(i=1) ^r (2v_i +1) ?

$${how}\:{can}\:{find}\:{the}\:{sum}\:\underset{{i}=\mathrm{1}} {\overset{{r}} {\sum}}\left(\mathrm{2}{v}_{{i}} +\mathrm{1}\right)\:? \\ $$

Question Number 192549    Answers: 2   Comments: 0

lim_(x→0) x^2 tan (((sin πx)/(2x))) =?

$$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{x}^{\mathrm{2}} \:\mathrm{tan}\:\left(\frac{\mathrm{sin}\:\pi\mathrm{x}}{\mathrm{2x}}\right)\:=? \\ $$

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