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hjlf
3 years ago
15

Write the expression as a polynomial (x+1)(x+2)(y–3)

Mathematics
1 answer:
o-na [289]3 years ago
6 0

\text{Use FOIL method:}\ (a+b)(c+d)=ac+ad+bc+bd\\\\(x+1)(x+2)(y-3)=[(x)(x)+(x)(2)+(1)(x)+(1)(2)](y-3)\\\\=(x^2+2x+x+2)(y-3)=(x^2+3x+2)(y-3)\\\\=(x^2)(y)+(x^2)(-3)+(3x)(y)+(3x)(-3)+(2)(y)+(2)(-3)\\\\=\boxed{x^2y-3x^2+3xy-9x+2y-6}

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Sadie computes the perimeter of a rectangle by adding the length, l , and width, w , and doubling this sum. Eric computes the pe
larisa [96]

a) P = 2(l+w)  Sadie

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2 years ago
Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1
Varvara68 [4.7K]
I assume there are some plus signs that aren't rendering for some reason, so that the plane should be x+y+z=1.

You're minimizing d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2} subject to the constraint f(x,y,z)=x+y+z=1. Note that d(x,y,z) and d(x,y,z)^2 attain their extrema at the same values of x,y,z, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

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L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)

Take your partial derivatives and set them equal to 0:

\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}

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2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43

and plugging this into the first three equations, you find a critical point at (x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right).

The squared distance is then d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43, which means the shortest distance must be \sqrt{\dfrac43}=\dfrac2{\sqrt3}.
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