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Mademuasel [1]
2 years ago
13

Why does it make sense in this situation that the function has this horizontal asymptote?

Mathematics
2 answers:
asambeis [7]2 years ago
5 0

Answer: Option 1

Step-by-step explanation: The average cost per person is always greater than $630 because of the free for the tour guide.

Eva8 [605]2 years ago
4 0

Answer:

the answer is a. The average cost per person is always greater than $630 because of the fee for the tour guide.

Step-by-step explanation:

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Step-by-step explanation:

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marysya [2.9K]
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x = 2
y = 4x +1

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Test the given points to see whether they fall within the solution region.

Point   Pass/Fail
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8 0
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Read 2 more answers
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

8 0
2 years ago
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