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uranmaximum [27]
2 years ago
5

(2x + y)2 – (3x – 2y)2 + (x – 4y)(x + 4y11​

Mathematics
1 answer:
Vinvika [58]2 years ago
8 0

Answer:

Step-by-step explanation:

(a+ b)² = a² +  b² + 2ab

(2x + y)² = (2x)² + y² + 2*2x *y

             = 4x²  + y² + 4xy

(a- b)² = a² + b² - 2ab

(3x - 2y)² = (3x)² + (2y)² - 2*3x *2y

              = 9x² + 4y² - 12xy

(a - b)(a +b) = a² - b²

(x - 4y(x + 4y) = x² - (4y)²

                     = x² - 16y²

(2x + y)² - (3x - 2y)² + (x - 4y)(x +4y)

  = 4x²  + y² + 4xy - (9x² + 4y² - 12xy) + x² - 16y²

  = 4x²  + y² + 4xy - 9x² - 4y² + 12xy + x² - 16y²

  = 4x² - 9x² + x²  + y² - 4y² - 16y²+ 4xy + 12xy

 =    -4x² - 19y² + 16xy

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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.
sweet-ann [11.9K]

Answer:

2.794

Step-by-step explanation:

Recall that if G(x,y) is a parametrization of the surface S and F and G are smooth enough then  

\bf \displaystyle\iint_{S}FdS=\displaystyle\iint_{R}F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})dxdy

F can be written as

F(x,y,z) = (xy, yz, zx)

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\bf G(x,y) = (x, y, 3-x^2-y^2)

with 0≤ x≤1 and  0≤ y≤1

So

\bf \displaystyle\frac{\partial G}{\partial x}= (1,0,-2x)\\\\\displaystyle\frac{\partial G}{\partial y}= (0,1,-2y)\\\\\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y}=(2x,2y,1)

we also have

\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)

and so

\bf F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and  0≤ y≤1

\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794

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Translate the phrase into a variable expression: twice the sum of a number and 5
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Answer:

'Twice the sum of a number and 5' can be translated into variable expression such as:

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Step-by-step explanation:

Given the phrase

<em>Twice the sum of a number and 5</em>

First, let us breakdown the English Phrase

Let the number be = n

The sum of a number 'n' and 5 = n + 5

now, twice the sum of a number and 5 can be determined by multiplying (n+5) with 5.

Thus,

'<em>Twice the sum of a number and 5</em>' can be translated into variable expression such as:

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