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Sergeeva-Olga [200]

3 years ago

11

Answer this: x=6 y=-2 -8x-4y+x

2 answers:

olga55 [171]3 years ago

7 0

Answer:

-34

Step-by-step explanation:

first do substitution: -8 · 6 + -4 · -2 + 6

next do the multiplication: -48 + 8 + 6

add: -34

hope this helps!!!

Natali [406]3 years ago

4 0

Answer:

-34

Steps:

-8x - 4y + x

-8x + x = -7x <em>(combine like terms)</em>

-7x - 4y

-7(6) - 4(-2) <em>(plug in the numbers)</em>

-42 +8=

-34

:")

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Let φ(x, y) = arctan (y/x) .

Alexandra [31]

Answer:

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b) $\large \mathbb{R}^2-\{(0,0)\}$

c) the points of the form (x, -x) for x≠0

Step-by-step explanation:

a)

If φ(x, y) = arctan (y/x), the vector field F = ∇φ would be

$\large F(x,y)=(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y})$

On one hand we have,

$\large \frac{\partial \phi}{\partial x}=\frac{\partial arctan(y/x)}{\partial x}=\frac{-y/x^2}{1+(y/x)^2}=-\frac{y/x^2}{1+y^2/x^2}=\\\\=-\frac{y/x^2}{(x^2+y^2)/x^2}=-\frac{y}{x^2+y^2}$

On the other hand,

$\large \frac{\partial \phi}{\partial y}=\frac{\partial arctan(y/x)}{\partial y}=\frac{1/x}{1+(y/x)^2}=\frac{1/x}{1+y^2/x^2}=\\\\=\frac{1/x}{(x^2+y^2)/x^2}=\frac{x}{x^2+y^2}$

So

$\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

b)

The domain of definition of F is

$\large \mathbb{R}^2-\{(0,0)\}$

i.e., all the plane X-Y except the (0,0)

c)

Here we want to find all the points such that

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where k is a real number other than 0.

But this means

$\large -\frac{y}{x^2+y^2}=\frac{x}{x^2+y^2}\Rightarrow y=-x$

So, all the points in the line y = -x except (0,0) are parallel to the vector field F, that is, the points (x, -x) with x≠ 0

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