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Simora [160]
3 years ago
11

Part A

Mathematics
1 answer:
valina [46]3 years ago
7 0

*diagram of the dot plot is attached below

Answer:

$2.15

Step-by-step Explanation:

Mean daily pocket money or average daily pocket if the group if students shown in the for plot that is attached below = the sum of the pocket money received by the students ÷ the total number of students.

From the diagram of the dot box attached below, the total number if students = 13 (each dot represents a student)

Sum of the pocket money received by the students = 4(1) + 5(2) + 3(2) + 4(2)

= 4 + 10 + 6 + 8

Sum = 28

Therefore, mean daily pocket money = 28 ÷ 13 = $2.15

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Derek has a bank account that pays 2.1% simple interest. The balance is $910. When will the account grow to $1,000?
swat32

Answer:

The account will grow to 1,000 in 5 years.

Step-by-step explanation:

Since the account will become 1000, the interest is 90 dollars. I=p*r*t.

90=910*.021*t

90=19.11*t.

t=4.71, so it will take 5 whole years.

3 0
2 years ago
What is the image point of (6,1) after a translation left 5 units and down 1 unit?
11Alexandr11 [23.1K]

Answer:

(1, 0 )

Step-by-step explanation:

A translation of 5 units left means subtract 5 from the x- coordinate

A translation of 1 unit down means subtract 1 from the y- coordinate, thus

(6, 1 ) → (6 - 5, 1 - 1 ) → (1, 0 )

6 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
3 years ago
Find the length of the missing side х<br> 10.6<br> 8.6
Yanka [14]

Answer:

13.65

Step-by-step explanation:

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What is the prime factorization of 33
schepotkina [342]
The answer is 11 and 3
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