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Question Number 54841 Answers: 1 Comments: 0

Question Number 54830 Answers: 0 Comments: 1

let V_n = ∫_0 ^∞ ((cos(nx))/(n +x^2 ))dx with n integr nstural not 0 . 1) calculate V_n 2)calculate lim_(n→+∞) nV_n 3) calculate the sum Σ_(n=0) ^∞ V_n

$${let}\:{V}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({nx}\right)}{{n}\:+{x}^{\mathrm{2}} }{dx}\:\:\:{with}\:{n}\:{integr}\:{nstural}\:{not}\:\mathrm{0}\:. \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{V}_{{n}} \\ $$$$\left.\mathrm{2}\right){calculate}\:{lim}_{{n}\rightarrow+\infty} {nV}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{the}\:{sum}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{V}_{{n}} \\ $$

Question Number 54827 Answers: 0 Comments: 0

logx^2 +log_2 (x−6)=3 solve for x

$$\mathrm{logx}^{\mathrm{2}} +\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}−\mathrm{6}\right)=\mathrm{3} \\ $$$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x} \\ $$

Question Number 54826 Answers: 4 Comments: 5

1) ∫_0 ^( 1) ((1−x^7 )^(1/4) −(1−x^4 )^(1/7) )dx = ? 2) If f(x)=x^3 +3x+4 then the value of ∫_(−1) ^( 1) f(x)dx + ∫_0 ^( 4) f^( −1) (x)dx = ? 3) ∫_(−π) ^( π) (1+cosx+cos2x+....+cos13x)(1+sinx+...+sin13x)dx=? 4) ∫_0 ^( 2) ((√(x^3 +1)) + (x^2 +2x)^(1/3) )dx =? (This time func. are not inverse of each other,right ?)

$$\left.\mathrm{1}\right)\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\left(\left(\mathrm{1}−{x}^{\mathrm{7}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} −\left(\mathrm{1}−{x}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{7}}} \right){dx}\:=\:? \\ $$$$\left.\mathrm{2}\right)\:{If}\:{f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{4}\:{then}\:{the}\:{value}\:{of} \\ $$$$\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} {f}\left({x}\right){dx}\:+\:\int_{\mathrm{0}} ^{\:\mathrm{4}} {f}^{\:−\mathrm{1}} \left({x}\right){dx}\:=\:? \\ $$$$\left.\mathrm{3}\right)\:\int_{−\pi} ^{\:\pi} \left(\mathrm{1}+{cosx}+{cos}\mathrm{2}{x}+....+{cos}\mathrm{13}{x}\right)\left(\mathrm{1}+{sinx}+...+{sin}\mathrm{13}{x}\right){dx}=? \\ $$$$\left.\mathrm{4}\right)\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \:\left(\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}\:+\:\left({x}^{\mathrm{2}} +\mathrm{2}{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\right){dx}\:=? \\ $$$$\left({This}\:{time}\:{func}.\:{are}\:{not}\:{inverse}\:{of}\:{each}\right. \\ $$$$\left.{other},{right}\:?\right) \\ $$

Question Number 54825 Answers: 1 Comments: 0

Question Number 54808 Answers: 1 Comments: 2

let p(x)=(1+x^2 )(1+x^4 )...(1+x^2^n ) with n integr natural 1) find a simple form of p(x) 2) find roots of p(x)and decompose p(x) inside C[x] 3)calculate ∫_0 ^1 p(x)dx 4) decompose the fraction F(x)=(1/(p(x))) .

$${let}\:{p}\left({x}\right)=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)...\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \right) \\ $$$${with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{roots}\:{of}\:{p}\left({x}\right){and}\:{decompose} \\ $$$${p}\left({x}\right)\:{inside}\:{C}\left[{x}\right] \\ $$$$\left.\mathrm{3}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{p}\left({x}\right){dx} \\ $$$$\left.\mathrm{4}\right)\:{decompose}\:{the}\:{fraction} \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{p}\left({x}\right)}\:. \\ $$

Question Number 54797 Answers: 1 Comments: 0

∫ (√( x + (√( x + (√(x + (√( .....)))))))) dx = ?

$$ \\ $$$$ \\ $$$$\:\:\:\int\:\:\sqrt{\:\boldsymbol{{x}}\:+\:\sqrt{\:\boldsymbol{{x}}\:+\:\sqrt{\boldsymbol{{x}}\:+\:\sqrt{\:.....}}}}\:\:\boldsymbol{{dx}}\:\:=\:\:\:? \\ $$$$ \\ $$

Question Number 54821 Answers: 0 Comments: 1

find lim_(n→+∞) ∫_0 ^n ((arctan(nx))/(n^2 +x^2 ))dx

$${find}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\int_{\mathrm{0}} ^{{n}} \:\:\:\frac{{arctan}\left({nx}\right)}{{n}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 54790 Answers: 3 Comments: 2

differenciatethefollowing i)x^x +(sinx)^(lnx) = ii)sin^(−1) (tanhx)= iii)(√(1+x^2 /1−x^2 =))

$${differenciatethefollowing} \\ $$$$\left.{i}\right){x}^{{x}} +\left({sinx}\right)^{{lnx}} = \\ $$$$\left.{ii}\right){sin}^{−\mathrm{1}} \left({tanhx}\right)= \\ $$$$\left.{iii}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} /\mathrm{1}−{x}^{\mathrm{2}} =} \\ $$

Question Number 54788 Answers: 2 Comments: 1

How can cut a right angeled triangle to make a square? how can cut a equilateral triangle to make a rectangle?

Question Number 54787 Answers: 0 Comments: 0

if: (a/((√b)+(√c)))=(b/((√c)+(√d)))=(c/((√d)+(√a)))=(d/((√a)+(√b))) ⇒(a/d)=?

$$\:{if}:\:\:\frac{{a}}{\sqrt{{b}}+\sqrt{{c}}}=\frac{{b}}{\sqrt{{c}}+\sqrt{{d}}}=\frac{{c}}{\sqrt{{d}}+\sqrt{{a}}}=\frac{{d}}{\sqrt{{a}}+\sqrt{{b}}} \\ $$$$\Rightarrow\frac{{a}}{{d}}=? \\ $$

Question Number 54786 Answers: 0 Comments: 5

solve for:a,b,c,d∈R a^2 =b+(√c) b^2 =c+(√d) c^2 =d+(√a) d^2 =a+(√b)

$${solve}\:{for}:{a},{b},{c},{d}\in\boldsymbol{{R}} \\ $$$$\:\:\:\:\:\:{a}^{\mathrm{2}} ={b}+\sqrt{{c}} \\ $$$$\:\:\:\:\:\:{b}^{\mathrm{2}} ={c}+\sqrt{{d}} \\ $$$$\:\:\:\:\:\:{c}^{\mathrm{2}} ={d}+\sqrt{{a}} \\ $$$$\:\:\:\:\:\:{d}^{\mathrm{2}} ={a}+\sqrt{{b}} \\ $$

Question Number 54777 Answers: 0 Comments: 1

let u_n =∫_0 ^∞ ((sin(nx^2 ))/(x^2 +6))dx 1) calculate u_n and lim u_n (n→+∞) 2) find nature of Σ u_n and calaculate it. 3) find nature of Σ u_n ^2

$${let}\:{u}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{sin}\left({nx}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{6}}{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\:{u}_{{n}} \:\:\:{and}\:{lim}\:{u}_{{n}} \left({n}\rightarrow+\infty\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{nature}\:{of}\:\Sigma\:{u}_{{n}} \:\:\:{and}\:{calaculate}\:{it}. \\ $$$$\left.\mathrm{3}\right)\:{find}\:{nature}\:\:{of}\:\Sigma\:{u}_{{n}} ^{\mathrm{2}} \\ $$

Question Number 54775 Answers: 0 Comments: 2

found something interesting, it was published by Tschirnhaus in 1683 we can reduce x^3 +ax^2 +bx+c=0 (1) to y^3 +py+q=0 (2) and further to z^3 =t (1) is the well known linear substitution y=x+(a/3) → x=y−(a/3) ⇒ y^3 −((a^2 −3b)/3)y+((2a^3 −9ab+27c)/(27))=0 p=−((a^2 −3b)/3) and q=((2a^3 −9ab+27c)/(27)) ⇒ y^3 +py+q=0 (2) quadratic substitution z=y^2 +αy+β → y^2 +αy+(β−z)=0 we could solve this for y and then plug in above... (Tschirnhaus did) but there′s an easier way: we calculate the determinant of the Sylvester Matrix we have (a) 1y^3 +0y^2 +py+q=0 (b) 0y^3 +y^2 +αy+(β−z)=0 the matrix is [(1,0,p,q,0),(0,1,0,p,q),(0,1,α,(β−z),0),(0,0,1,α,(β−z)),(1,α,(β−z),0,0) ] the determinant is −z^3 +(3β−2p)z^2 −(pα^2 +3qα+3β^2 −4pβ+p^2 )z −(qα^3 −pα^2 β−3qαβ+pqα−β^3 +2pβ^2 −p^2 β−q^2 )p we want the square and the linear terms to disappear so we set their constants zero to get α and β ⇒ β=((2p)/3); α=−((3q)/(2p))±((√(12p^3 +81q^2 ))/(6p)) this leads to z^3 =((8p^3 )/(27))+((27q^4 )/(2p^3 ))+4q^2 ±((q(√(3(4p^3 +27q^2 )^3 )))/(18p^3 )) Tschirnhaus thought he could solve polynomes of any degree with this method but it′s getting harder to solve because you need a cubic substitution to eliminate 3 constants and so on...

$$\mathrm{found}\:\mathrm{something}\:\mathrm{interesting},\:\mathrm{it}\:\mathrm{was}\:\mathrm{published} \\ $$$$\mathrm{by}\:\mathrm{Tschirnhaus}\:\mathrm{in}\:\mathrm{1683} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{reduce} \\ $$$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{to} \\ $$$${y}^{\mathrm{3}} +{py}+{q}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{and}\:\mathrm{further}\:\mathrm{to} \\ $$$${z}^{\mathrm{3}} ={t} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{well}\:\mathrm{known}\:\mathrm{linear}\:\mathrm{substitution} \\ $$$${y}={x}+\frac{{a}}{\mathrm{3}}\:\rightarrow\:{x}={y}−\frac{{a}}{\mathrm{3}} \\ $$$$\Rightarrow\:{y}^{\mathrm{3}} −\frac{{a}^{\mathrm{2}} −\mathrm{3}{b}}{\mathrm{3}}{y}+\frac{\mathrm{2}{a}^{\mathrm{3}} −\mathrm{9}{ab}+\mathrm{27}{c}}{\mathrm{27}}=\mathrm{0} \\ $$$${p}=−\frac{{a}^{\mathrm{2}} −\mathrm{3}{b}}{\mathrm{3}}\:\mathrm{and}\:{q}=\frac{\mathrm{2}{a}^{\mathrm{3}} −\mathrm{9}{ab}+\mathrm{27}{c}}{\mathrm{27}} \\ $$$$\Rightarrow\:{y}^{\mathrm{3}} +{py}+{q}=\mathrm{0} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\mathrm{quadratic}\:\mathrm{substitution} \\ $$$${z}={y}^{\mathrm{2}} +\alpha{y}+\beta\:\rightarrow\:{y}^{\mathrm{2}} +\alpha{y}+\left(\beta−{z}\right)=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{could}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{for}\:{y}\:\mathrm{and}\:\mathrm{then}\:\mathrm{plug}\:\mathrm{in} \\ $$$$\mathrm{above}...\:\left(\mathrm{Tschirnhaus}\:\mathrm{did}\right)\:\mathrm{but}\:\mathrm{there}'\mathrm{s}\:\mathrm{an} \\ $$$$\mathrm{easier}\:\mathrm{way}:\:\mathrm{we}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{determinant}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{Sylvester}\:\mathrm{Matrix} \\ $$$$\mathrm{we}\:\mathrm{have} \\ $$$$\left({a}\right)\:\mathrm{1}{y}^{\mathrm{3}} +\mathrm{0}{y}^{\mathrm{2}} +{py}+{q}=\mathrm{0} \\ $$$$\left({b}\right)\:\mathrm{0}{y}^{\mathrm{3}} +{y}^{\mathrm{2}} +\alpha{y}+\left(\beta−{z}\right)=\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{matrix}\:\mathrm{is} \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}&{{p}}&{{q}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}&{{p}}&{{q}}\\{\mathrm{0}}&{\mathrm{1}}&{\alpha}&{\beta−{z}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\alpha}&{\beta−{z}}\\{\mathrm{1}}&{\alpha}&{\beta−{z}}&{\mathrm{0}}&{\mathrm{0}}\end{bmatrix} \\ $$$$\mathrm{the}\:\mathrm{determinant}\:\mathrm{is} \\ $$$$−{z}^{\mathrm{3}} \\ $$$$\:\:+\left(\mathrm{3}\beta−\mathrm{2}{p}\right){z}^{\mathrm{2}} \\ $$$$\:\:−\left({p}\alpha^{\mathrm{2}} +\mathrm{3}{q}\alpha+\mathrm{3}\beta^{\mathrm{2}} −\mathrm{4}{p}\beta+{p}^{\mathrm{2}} \right){z} \\ $$$$\:\:−\left({q}\alpha^{\mathrm{3}} −{p}\alpha^{\mathrm{2}} \beta−\mathrm{3}{q}\alpha\beta+{pq}\alpha−\beta^{\mathrm{3}} +\mathrm{2}{p}\beta^{\mathrm{2}} −{p}^{\mathrm{2}} \beta−{q}^{\mathrm{2}} \right){p} \\ $$$$\mathrm{we}\:\mathrm{want}\:\mathrm{the}\:\mathrm{square}\:\mathrm{and}\:\mathrm{the}\:\mathrm{linear}\:\mathrm{terms} \\ $$$$\mathrm{to}\:\mathrm{disappear}\:\mathrm{so}\:\mathrm{we}\:\mathrm{set}\:\mathrm{their}\:\mathrm{constants}\:\mathrm{zero} \\ $$$$\mathrm{to}\:\mathrm{get}\:\alpha\:\mathrm{and}\:\beta \\ $$$$\Rightarrow\:\beta=\frac{\mathrm{2}{p}}{\mathrm{3}};\:\alpha=−\frac{\mathrm{3}{q}}{\mathrm{2}{p}}\pm\frac{\sqrt{\mathrm{12}{p}^{\mathrm{3}} +\mathrm{81}{q}^{\mathrm{2}} }}{\mathrm{6}{p}} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to} \\ $$$${z}^{\mathrm{3}} =\frac{\mathrm{8}{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{\mathrm{27}{q}^{\mathrm{4}} }{\mathrm{2}{p}^{\mathrm{3}} }+\mathrm{4}{q}^{\mathrm{2}} \pm\frac{{q}\sqrt{\mathrm{3}\left(\mathrm{4}{p}^{\mathrm{3}} +\mathrm{27}{q}^{\mathrm{2}} \right)^{\mathrm{3}} }}{\mathrm{18}{p}^{\mathrm{3}} } \\ $$$$\mathrm{Tschirnhaus}\:\mathrm{thought}\:\mathrm{he}\:\mathrm{could}\:\mathrm{solve}\:\mathrm{polynomes} \\ $$$$\mathrm{of}\:\mathrm{any}\:\mathrm{degree}\:\mathrm{with}\:\mathrm{this}\:\mathrm{method}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{getting} \\ $$$$\mathrm{harder}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{because}\:\mathrm{you}\:\mathrm{need}\:\mathrm{a}\:\mathrm{cubic} \\ $$$$\mathrm{substitution}\:\mathrm{to}\:\mathrm{eliminate}\:\mathrm{3}\:\mathrm{constants}\:\mathrm{and} \\ $$$$\mathrm{so}\:\mathrm{on}... \\ $$

Question Number 54770 Answers: 0 Comments: 0

A glass bottle full of mercury has mass 500g. On being heated through 35°C, 2.43g of mercury are expelled. Calculate the mass of mercury remaining in the bottle. (Cubic expansivity of mercury is 1.8 × 10^(−4) K^(−1) , linear expansivity of glass is 8.0 × 10^(−6) K^(−1) .

$$\mathrm{A}\:\mathrm{glass}\:\mathrm{bottle}\:\mathrm{full}\:\mathrm{of}\:\mathrm{mercury}\:\mathrm{has}\:\mathrm{mass}\:\mathrm{500g}.\:\mathrm{On}\:\mathrm{being}\:\mathrm{heated}\:\mathrm{through} \\ $$$$\mathrm{35}°\mathrm{C},\:\:\mathrm{2}.\mathrm{43g}\:\mathrm{of}\:\mathrm{mercury}\:\mathrm{are}\:\mathrm{expelled}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{mercury} \\ $$$$\mathrm{remaining}\:\mathrm{in}\:\mathrm{the}\:\mathrm{bottle}.\:\left(\mathrm{Cubic}\:\mathrm{expansivity}\:\mathrm{of}\:\mathrm{mercury}\:\mathrm{is}\:\:\mathrm{1}.\mathrm{8}\:×\:\mathrm{10}^{−\mathrm{4}} \:\mathrm{K}^{−\mathrm{1}} \:,\right. \\ $$$$\mathrm{linear}\:\mathrm{expansivity}\:\mathrm{of}\:\mathrm{glass}\:\mathrm{is}\:\:\mathrm{8}.\mathrm{0}\:×\:\mathrm{10}^{−\mathrm{6}} \mathrm{K}^{−\mathrm{1}} \:. \\ $$

Question Number 54767 Answers: 1 Comments: 1

A stone is thrown vertically upwards from a cliff 20m high. After a time of 3 s it passes the edge of the cliff on its way down. Calculate a) the speed of projection b) the speed when it hits the ground c) the times when it is 10m above the top of the cliff d) the time it is 15m above the ground (take g=10ms^(−2) ).

Question Number 54765 Answers: 1 Comments: 0

Question Number 54756 Answers: 1 Comments: 0

solve ∫tan(x−θ)tan(x+θ)tan 2x dx

$${solve} \\ $$$$\int{tan}\left({x}−\theta\right){tan}\left({x}+\theta\right){tan}\:\mathrm{2}{x}\:{dx} \\ $$

Question Number 54755 Answers: 1 Comments: 0

solve for x sin[2cos^(−1) {cot(2tan^(−1) x)}]=0 (x = 1∓(√(2 )) , ∓1 ,−1∓(√2) )

$${solve}\:{for}\:{x} \\ $$$${sin}\left[\mathrm{2}{cos}^{−\mathrm{1}} \left\{{cot}\left(\mathrm{2}{tan}^{−\mathrm{1}} {x}\right)\right\}\right]=\mathrm{0} \\ $$$$ \\ $$$$\left({x}\:=\:\mathrm{1}\mp\sqrt{\mathrm{2}\:}\:,\:\mp\mathrm{1}\:,−\mathrm{1}\mp\sqrt{\mathrm{2}}\:\:\right) \\ $$

Question Number 54745 Answers: 0 Comments: 0

(√(1−x^2 +))(√(1−y^2 ))=a(x−y) (dy/dx)=(√((1−y^2 )/(1−x^2 ))) without solve put x=sinθand y=sinφ direct solve

$$\sqrt{\mathrm{1}−{x}^{\mathrm{2}} +}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }={a}\left({x}−{y}\right) \\ $$$$\frac{{dy}}{{dx}}=\sqrt{\frac{\mathrm{1}−{y}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${without}\:{solve}\:{put}\:{x}={sin}\theta{and} \\ $$$${y}={sin}\phi \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${direct}\:{solve} \\ $$

Question Number 54741 Answers: 1 Comments: 1

such that 1. ((n),(0) )^2 + ((n),(1) )^2 +...+ ((n),(n) )^2 =(((2n)!)/((n!)^2 )) 2. ((n),(0) )+(1/2) ((n),(1) )+...+(1/(n+1)) ((n),(n) )^2 =((2^(n+1) −1)/(n+1))

$$\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{1}.\begin{pmatrix}{\mathrm{n}}\\{\mathrm{0}}\end{pmatrix}^{\mathrm{2}} +\begin{pmatrix}{\mathrm{n}}\\{\mathrm{1}}\end{pmatrix}^{\mathrm{2}} +...+\begin{pmatrix}{\mathrm{n}}\\{\mathrm{n}}\end{pmatrix}^{\mathrm{2}} =\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}.\:\begin{pmatrix}{\mathrm{n}}\\{\mathrm{0}}\end{pmatrix}+\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{1}}\end{pmatrix}+...+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{n}}\end{pmatrix}^{\mathrm{2}} =\frac{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}} \\ $$

Question Number 54740 Answers: 3 Comments: 0

Prove that 1. ((n),(r) ) = (((n−1)),(( r)) ) + (((n−1)),(( r−1)) ) 2. ((n),(r) ) + ((( n)),((r−1)) ) = (((n+1)),(( r)) ) 3. ((n),(0) )+ ((n),(1) )+ ((n),(2) )+..+ ((n),(n) )=2^n

$${Prove}\:{that} \\ $$$$\mathrm{1}.\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\:=\:\begin{pmatrix}{{n}−\mathrm{1}}\\{\:\:\:\:{r}}\end{pmatrix}\:+\:\begin{pmatrix}{{n}−\mathrm{1}}\\{\:{r}−\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{2}.\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\:+\begin{pmatrix}{\:\:\:{n}}\\{{r}−\mathrm{1}}\end{pmatrix}\:=\:\begin{pmatrix}{{n}+\mathrm{1}}\\{\:\:\:\:{r}}\end{pmatrix}\: \\ $$$$\mathrm{3}.\:\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}+..+\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix}=\mathrm{2}^{{n}} \\ $$

Question Number 54737 Answers: 0 Comments: 0

Question Number 54730 Answers: 0 Comments: 3

prove lnx≤x

$${prove}\:{lnx}\leqslant{x} \\ $$

Question Number 54716 Answers: 0 Comments: 6

A light horizontal meter rule PQR has the end P fixed to vertical wall, while a weight of 5N is suspended from the end R. A light string QS of length 50cm fixed to the wall at S is used to maintain the meter rule in equilibrium. If PQ = 40 cm, the tension in the string is ?

$$\mathrm{A}\:\mathrm{light}\:\mathrm{horizontal}\:\mathrm{meter}\:\mathrm{rule}\:\:\mathrm{PQR}\:\:\mathrm{has}\:\mathrm{the}\:\mathrm{end}\:\:\mathrm{P}\:\mathrm{fixed}\:\mathrm{to}\: \\ $$$$\mathrm{vertical}\:\mathrm{wall},\:\mathrm{while}\:\mathrm{a}\:\mathrm{weight}\:\mathrm{of}\:\:\mathrm{5N}\:\mathrm{is}\:\mathrm{suspended}\:\mathrm{from}\:\mathrm{the}\:\mathrm{end} \\ $$$$\mathrm{R}.\:\mathrm{A}\:\mathrm{light}\:\mathrm{string}\:\mathrm{QS}\:\:\mathrm{of}\:\mathrm{length}\:\:\mathrm{50cm}\:\mathrm{fixed}\:\mathrm{to}\:\mathrm{the}\:\mathrm{wall}\:\mathrm{at}\:\mathrm{S}\:\mathrm{is}\: \\ $$$$\mathrm{used}\:\mathrm{to}\:\mathrm{maintain}\:\mathrm{the}\:\mathrm{meter}\:\mathrm{rule}\:\mathrm{in}\:\mathrm{equilibrium}. \\ $$$$\mathrm{If}\:\:\mathrm{PQ}\:=\:\mathrm{40}\:\mathrm{cm},\:\:\:\mathrm{the}\:\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{string}\:\mathrm{is}\:? \\ $$

Question Number 54721 Answers: 0 Comments: 0

Show that ((Σ{sin 2xtan (y−z)cos^2 (y+z)})/(Σ{cos 2xtan (y−z)sin^2 (y+z)} )) = tan 2xtan 2ytan 2z

$$\mathrm{Show}\:\mathrm{that} \\ $$$$\frac{\Sigma\left\{\mathrm{sin}\:\mathrm{2xtan}\:\left(\mathrm{y}−\mathrm{z}\right)\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{y}+\mathrm{z}\right)\right\}}{\Sigma\left\{\mathrm{cos}\:\mathrm{2xtan}\:\left(\mathrm{y}−\mathrm{z}\right)\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{y}+\mathrm{z}\right)\right\}\:\:\:}\:\:=\:\mathrm{tan}\:\mathrm{2xtan}\:\mathrm{2ytan}\:\mathrm{2z} \\ $$

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