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MAVERICK [17]
2 years ago
8

|4n - 1 | = - 6" align="absmiddle" class="latex-formula">
what's the answer
Mathematics
1 answer:
Advocard [28]2 years ago
5 0

Answer:

<h3>No solution</h3>

Step-by-step explanation:

Because the absolute value of each number is non negative

|a| ≥ 0 for all real numbers

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\large\boxed{C.\ (-4x+3)(x-2)}

Step-by-step explanation:

-4x^2+11x-6=-(4x^2-11x+6)=-(4x^2-8x-3x+6)\\\\=-\left[4x(x-2)-3(x-2)]=-(x-2)(4x-3)=(-4x+3)(x-2)

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Let φ(x, y) = arctan (y/x) .
Alexandra [31]

Answer:

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

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

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

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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