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3 years ago

The stock of Company A gained 4% today to $4.16. What was the opening price of the stock in the beginning of the day?

Mathematics

1 answer:

Oksana_A [137]3 years ago

6 0

Answer:

The opening price of the stock in the beginning of the day was $4.

Step-by-step explanation:

To be able to calculate the price of the stock at the beginning of the day, you have to consider that the equation to calculate the final price is equal to the opening price multiply for the result of one plus the increase or decrease percentage, which is:

final price=opening price*(1+0.04)

Now, you know that the final price is $4.16, so you can replace this value in the formula and solve for the opening price:

4.16=opening price*1.04

opening price=4.16/1.04

opening price=4

According to this, the answer is that the opening price of the stock in the beginning of the day was $4.

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What is the meaning of the ∝ sign?

ExtremeBDS [4]

Answer:

"proportional to"

Step-by-step explanation:

The sign ∝ means 'is proportional to'. This is used to show that one variable of value is proportional to another value or variable.

Example:

x ∝ y ('x' is proportional to 'y')

Hope this helps.

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3 years ago

Determine if this is a linearly dependent or independent set: (V, V, V ) where syon V = (1,2,2,-1), V = (4,9,9,-4), v = (5,8,9,-

marishachu [46]

Answer:

This is a linearly independent set.

Step-by-step explanation:

We have these following vectors:

$V_{1} = (1,2,2,-1)$

$V_{2} = (4,9,9,-4)$

$V_{3} = (5,8,9,-5)$

In a set of 3 vectors, if one of these vectors can be written as a linear combination of the 2 other vectors, they are linearly dependent. Otherwise, they are linearly independent.

We can verify this by solving the following system:

$xV_{1} + yV_{2} + zV_{3} = (0,0,0,0)$

If the only solution is $(x,y,z) = (0,0,0)$ , they are L.I. Otherwise, they are L.D.

Solution:

$xV_{1} + yV_{2} + zV_{3} = (0,0,0,0)$

$x(1,2,2,-1) + y(4,9,9,-4) + z (5,8,9,-5) = (0,0,0,0)$

We have the following system of equations:

$x + 4y + 5z = 0$

$2x + 9y + 8z = 0$

$2x + 9y + 9z = 0$

$-x -4y - 5z = 0$

I am going to solve this by the row-reduction of the augmented matrix.

This system has the following augmented matrix:

$\left[\begin{array}{cccc}1&4&5&0\\2&9&8&0\\2&9&9&0\\-1&-4&-5&0\end{array}\right]$

To reduce the first row, i am going to make these following operations:

$L_{2} = L_{2} - 2L_{1}$

$L_{3} = L_{3} - 3L_{1}$

$L_{4} = L_{4} + L_{1}$

So the augmented matrix now is:

$\left[\begin{array}{cccc}1&4&5&0\\0&1&-2&0\\0&1&-1&0\\0&0&0&0\end{array}\right]$

Now I reduce the second row, doing:

$L_{3} = L_{3} - L_{2}$

So the matrix is:

$\left[\begin{array}{cccc}1&4&5&0\\0&1&-2&0\\0&0&1&0\\0&0&0&0\end{array}\right]$

Now we can solve the system:

From the third line, we have that

$z = 0$

From the second line:

$y - 2z = 0$

$y - 2(0) = 0$

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From the first line

$x + 4y + 5z = 0$

$x + 4(0) + 5(0) = 0$

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The only solution for this system is $(x,y,z) = (0,0,0)$ . This means that we have a linearly independent set.

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3 years ago

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Answer:

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So, not a right angle triangle

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garik1379 [7]

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