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u=ex(x cosy-ysiny),prove that uxx+vyy=0

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Prove that in an MV-algebra, x ∩ y = x if and only if x , ⊕ y ...


Prove that in an MV-algebra, x ∩ y = x if and only if x , ⊕ y = unit , where x ∩ y = def x ⊗ (x , ⊕ y). Prove that the... - 1254648. Study ...
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Prove that it follows from the MV-algebra definition x ∪ y ...


Prove that it follows from the MV-algebra definition x ∪ y = def (x , ∩ y , ) , that x &#8746 ; y = x ... Prove that it follows from the MV-algebra definition ...
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University of Michigan Historical Math ... - quod.lib.umich.edu


... AZ On behalf of Preservation Division The University of Michigan Libraries ... Prove that x'=x cos 0 +y ... Only in Ex. X. 4 did we meet for a moment a ...
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Advanced Mathematical Methods for Scientists and Engineers ...


Advanced Mathematical Methods for Scientists and Engineers. Scribd. ... (x) = ex , are examples of ... First prove that θ →0 ∆x→0 f (x + ∆x) − f (x) g ...
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Introduction to Complex Anlaysis.pdf | Complex Number | Circle


Introduction to Complex Anlaysis.pdf. ... . and (b) x. with 0 < r < R.8 to prove that.12 Use ... only if y — 2kir for some k G Z. ex — 1 — 0 only if x — 0 and ...
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Applied Math Letter – Page 3 – Follow Science


Applied Math Letter. ... 3.1 Transformation to the form u” + a(x) u = 0 ... near the point x = 0. Since the derivative of ex ...
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Introduction to Methods of Applied Mathematics - Follow Science


Introduction to Methods of Applied Mathematics or Advanced ... 1.4 Prove that if limx! u(x) ... function near the point x = 0. Since the derivative of ex is ...
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Document - SEC.gov


... 8-k public document ... 62khk97i m2vsy!&p^e?[6bvq+7z-+pb(pw'0]?x 53>!n,he_8mbn)ex[d!= m*%neq4d7 ... g2"0#y "]7gu*x/9;y;k>y/ [email protected] =rqeivu6#*i*)$q5 ...
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Partial Differential Equations - Handbook of Mathematics - XI


This chapter discusses the quasi-linear partial differential equations of the ... equations A&x+B^ + C&x+D&y = Ex A2ux ... Rx-^R<\ 2U(xp,tp) = U(0,tp-x ...
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Suggested Questions And Answer :


prove that cosx/(sinx+cosy)+cosy/(siny-cosx)=cosx/(sinx-cosy)+cosy/(siny+cosy)

Question: prove that cosx/(sinx+cosy)+cosy/(siny-cosx)=cosx/(sinx-cosy)+cosy/(siny+cosy) There's a typo in your question. The final denominator should be (siny+cosx) not (siny+cosy). Assumption: That the equation is true for all values of x and y, i.e. that it is an identity cosx/(sinx+cosy) + cosy/(siny-cosx) = cosx/(sinx-cosy) + cosy/(siny+cosx)   -- rearrange cosx{1/(sinx+cosy) - 1/(sinx-cosy)} = cosy{1/(siny+cosy)  - 1/(siny-cosx)}   -- rationalise the denominators cosx{[(sinx - cosy) - (sinx + cosy)]/[(sinx+cosy)(sinx-cosy)]} = cosy{[(siny - cosx) - (siny + cosx)]/[(siny+cosy)(siny-cosx)]} cosx{(-2cosy)/(sin^2x - cos^2y)} = cosy{(-2cosx)/(sin^2y - cos^2y)} cosx{(-2cosy)/[(1 - cos^2x) - (1 - sin^2y)]} = cosy{(-2cosx)/(sin^2y - cos^2y)} (-2.cosx.cosy)/[sin^2y - cos^2x] = (-2.cosx.cosy)/(sin^2y - cos^2y) Since lhs is identical with rhs, then the above equation is true for all values of x and y. It is an identity. Hence the Assumption is true Q.E.D.
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prove that cosx\(sinx+cosy)+cosy\(siny-cosx)=cosx\(sinx-cosy)+cosy\(siny+cosy)

Question: prove that cosx/(sinx+cosy)+cosy/(siny-cosx)=cosx/(sinx-cosy)+cosy/(siny+cosy) There's a typo in your question. The final denominator should be (siny+cosx) not (siny+cosy). Assumption: That the equation is true for all values of x and y, i.e. that it is an identity cosx/(sinx+cosy) + cosy/(siny-cosx) = cosx/(sinx-cosy) + cosy/(siny+cosx)   -- rearrange cosx{1/(sinx+cosy) - 1/(sinx-cosy)} = cosy{1/(siny+cosy)  - 1/(siny-cosx)}   -- rationalise the denominators cosx{[(sinx - cosy) - (sinx + cosy)]/[(sinx+cosy)(sinx-cosy)]} = cosy{[(siny - cosx) - (siny + cosx)]/[(siny+cosy)(siny-cosx)]} cosx{(-2cosy)/(sin^2x - cos^2y)} = cosy{(-2cosx)/(sin^2y - cos^2y)} cosx{(-2cosy)/[(1 - cos^2x) - (1 - sin^2y)]} = cosy{(-2cosx)/(sin^2y - cos^2y)} (-2.cosx.cosy)/[sin^2y - cos^2x] = (-2.cosx.cosy)/(sin^2y - cos^2y) Since lhs is identical with rhs, then the above equation is true for all values of x and y. It is an identity. Hence the Assumption is true Q.E.D.
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How do I differentiate x with respect to y when x=√(siny+1)

How do I differentiate x with respect to y when x=√(siny+1) There are two ways tthat you can do this.   1) let u = siny + 1 then du/dy = cosy x = u^(1/2) dx/du = (1/2)u^(-1/2) and dx/dy = (dx/du)*(du/dy) dx/dy = (1/2)(siny + 1)^(-1/2)*(cosy) dx/dy = (1/2)cosy/√(siny+1)   2) Let x^2 = siny + 1 differentiating both sides wrt x, 2x = cosy(dy/dx) dx/dy = cosy/2x dx/dy = cosy/(2√(siny+1))
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(sinx)^cosy=(siny)^cosx

Let x=0 and y=(pi)/2 (90 degrees). Sinx=0, cosy=0 so (sinx)^cosy=0 and (siny)^cosx=1. Therefore the two expressions are not equal.
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if [(cosx)^4/(cosy)^2]+[(sinx)^4/(siny)^2=1 then prove that [(cosy)^4/(cosx)^2]+(siny)^4/(sinx)^2

Take, (cosx)^4/(cosy)^2=m (sinx)^4/(siny)^2=1-m   Now solve, m and you got the answer that [(cosy)^4/(cosx)^2]+(siny)^4/(sinx)^2]=1              Class 10 Safal
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if sinx+cosy=1 then how will i solve its second order derivative?

sin x + cos y =1, so cos y = 1-sin x, so sin y = sqrt(1-(1-sin x)^2)=sqrt(2sin x - sin^2x) If we differentiate with respect to x we get cos x - sin y.dy/dx=0. So we can write dy/dx=(cos x)/(sin y). Differentiate again and we get -sin x - cosy(dy/dx)(dy/dx) - sin y.d2y/dx2=0 [d2y/dx2 represents the second derivative] -sin x - (1-sin x)cos^2x/sin^2y - sin y.d2y/dx2=0 when we make some substitutions. From this we get the second derivative: sin y.d2y/dx2=-(sin x + (1-sin x)cos^2x/(2sin x-sin^2x)=-(2sin^2x-sin^3x + (1-sin x)cos^2x)/(2sin x-sin^2x) The final substitution gives d2y/dx2=-(2sin^2x-sin^3x + (1-sin x)cos^2x)/(2sin x-sin^2x).sqrt(2sin x - sin^2x) =-(2sin^2x-sin^3x+(1-sin x)(1-sin^2x))/(sin x(2-sin x))^(3/2) =-(2sin^2x-sin^3x+1-sin^2x-sin x+sin^3x)/(sin x(2-sin x))^(3/2) =-(sin^2x-sin x+1)/(sin x(2-sin x))^(3/2) or (sin x-sin^2x-1)/(sin x(2-sin x))^(3/2)  
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given : sin (x+y) = sin x cos y + cos x sin y and cos (x+y) = cos x cos y-sin x sin y derive a formula for cot (x+y) and derive a formula for cos(2x)

cot(x+y)=cos(x+y)/sin(x+y)=(cosxcosy-sinxsiny)/(sinxcosy+cosxsiny). There are variations of this: divide top and bottom by cosx: (cosy-tanxsiny)/(tanxcosy+siny), then by cosy: (1-tanxtany)/(tanx+tany). Dividing by sinx and siny instead gives: (cotxcosy-siny)/(cosy+cotxsiny), then: (cotxcoty-1)/(coty+cotx). (Using these variations and putting y=x, you can work out cot(2x).) cos(2x)=cos^2(x)-sin^2(x), putting y=x. This can also be written: 1-2sin^2(x) or 2cos^2(x)-1 because sin^2(x)+cos^2(x).
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chapter complex number....

Question: Find  log(1+i) Suppose z = x + i.y Then, e^z = e^x.e^(iy) = e^x(cosy - i.siny) e^z = e^x(cosy - i.siny) Now, let w = e^z, then log(w) = z setting w = 1 + i, then 1 + i = e^z = e^x.cosy - i.e^x.siny, i.e. 1 = e^x.cosy 1 = -e^x.siny Division of the above eqns gives tan(y) = -1 => y = -π/4. y = -π/4 gives cosy = -siny = 1/√2, therefore e^x = √2 and x = log(√2) log(1 + i) = log(w) = z = x + iy = log(√2) - i.π/4 Answer: log(1 + i) = log(√2) - i.π/4
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sinx+siny=a, cosx+cosy=b, then find yhe value of tan(a-b)/2

Shouldn't this be tan(x-y)/2? The quantities a and b represent values not variables. This question was answered earlier (12 March).
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Cosx÷(cosx+1)-cos÷(cosx-1)

(1+cosx)(1+cosy)(1+cosz)-(1-cosx)(1-cosy)(1-cosz)=k then k= ?
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