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Tcecarenko [31]
3 years ago
7

Required informationSkip to questionNOTE: This is a multi-part question. Once an answer is submitted, you will be unable to retu

rn to this part.Find the sum and product of each of these pairs of numbers. Express your answers as base 3 expansions.The sum of the numbers (20001)3 and (1112)3 is ( )3 and their product is ( )3.
Mathematics
1 answer:
Paladinen [302]3 years ago
6 0

Answer:

The answer is " (21120)_3  and (100011112)_3"

Step-by-step explanation:

Calculating the sum:

a= (20001)_3 \\\\b= (1112)_3

\to a+b = 20001+1112                     // 163+41=204

             =(21120)_3

Explanation:

In the addition, first convert the given value (a,b) into decimal point that is (163, 41) and by adding its value is equal to (204) and after convert its value is equal to (21120)_3.

Calculating the product:

a= (20001)_3 \\\\b= (1112)_3

\to a*b = 20001*1112                //  163 *41=6,683

             =(100011112)_3

Explanation:

In the product calculation we convert the given value (a,b) into decimal point that is (163, 41) and multiplting its value which is equal to (6,683) and after convert its value is equal to (100011112)_3.

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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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2 years ago
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Answer:

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Multiply 3 to both the numerator & denominator of 8/20

(8/20)(3/3) = 24/60

24/60 ≠ 25/60 ∴ these 2 fractions are not proportional.

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Answer:

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Step-by-step explanation:

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