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tiny-mole [99]
2 years ago
9

if twice the opposite of a number us increased by 5, the result is the opposite of the number. What is the number?

Mathematics
1 answer:
Likurg_2 [28]2 years ago
6 0

Answer:

If twice the opposite of a number increased by five, then the result is the opposite of a number.    

The asiest way to solve this is the guess and check method. It is asking you what a number doubled then by adding five is the opposite of that number. Now go through a few SINGLE digit numbers to try and solve. I hope this helps.

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If (x^6)^y = x^3, then y = ___.<br><br> Show your work!
musickatia [10]
Y=x^-3...................
8 0
3 years ago
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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larisa [96]

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3 years ago
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omeli [17]

Answer:

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I have no idea how ti so this one can sombody help do this one please it is hard
Vinil7 [7]
Not as hard as you think.
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