Answer:
The vectors does not span R3 and only span a subspace of R3 which satisfies x+13y-3z=0
Explanation:
The vectors are given as
![v_1=\left[\begin{array}{c}-4&1&3\end{array}\right] \\v_2=\left[\begin{array}{c}-5&1&6\end{array}\right] \\v_3=\left[\begin{array}{c}6&0&2\end{array}\right]](https://tex.z-dn.net/?f=v_1%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-4%261%263%5Cend%7Barray%7D%5Cright%5D%20%5C%5Cv_2%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-5%261%266%5Cend%7Barray%7D%5Cright%5D%20%5C%5Cv_3%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D6%260%262%5Cend%7Barray%7D%5Cright%5D)
Now if the vectors would span the
, the rank of the consolidated matrix will be 3 if it is not 3 this indicates that the vectors does not span the
.
So the matrix is given as
![M=\left[\begin{array}{ccc}v_1&v_2&v_3\end{array}\right] \\M=\left[\begin{array}{ccc}-4&5&6\\1&1&0\\3&6&2\end{array}\right]\\](https://tex.z-dn.net/?f=M%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dv_1%26v_2%26v_3%5Cend%7Barray%7D%5Cright%5D%20%5C%5CM%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%265%266%5C%5C1%261%260%5C%5C3%266%262%5Cend%7Barray%7D%5Cright%5D%5C%5C)
In order to calculate the rank, the matrix is reduced to the Row Echelon form as
![\approx \left[\begin{array}{ccc}-4&5&6\\ 0&\frac{9}{4}&\frac{3}{2}\\ 3&6&2\end{array}\right] R_2 \rightarrow R_2+\frac{R_1}{4}](https://tex.z-dn.net/?f=%5Capprox%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%265%266%5C%5C%200%26%5Cfrac%7B9%7D%7B4%7D%26%5Cfrac%7B3%7D%7B2%7D%5C%5C%203%266%262%5Cend%7Barray%7D%5Cright%5D%20R_2%20%5Crightarrow%20R_2%2B%5Cfrac%7BR_1%7D%7B4%7D)
![\approx \left[\begin{array}{ccc}-4&5&6\\ 0&\frac{9}{4}&\frac{3}{2}\\ 0&\frac{39}{4}&\frac{13}{2}\end{array}\right] R_3 \rightarrow R_3+\frac{3R_1}{4}\\](https://tex.z-dn.net/?f=%5Capprox%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%265%266%5C%5C%200%26%5Cfrac%7B9%7D%7B4%7D%26%5Cfrac%7B3%7D%7B2%7D%5C%5C%200%26%5Cfrac%7B39%7D%7B4%7D%26%5Cfrac%7B13%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20R_3%20%5Crightarrow%20R_3%2B%5Cfrac%7B3R_1%7D%7B4%7D%5C%5C)
![\approx \left[\begin{array}{ccc}-4&5&6\\ 0&\frac{39}{4}&\frac{13}{2\\ 0&\frac{9}{4}&\frac{3}{2}}\end{array}\right] R_2\:\leftrightarrow \:R_3](https://tex.z-dn.net/?f=%5Capprox%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%265%266%5C%5C%200%26%5Cfrac%7B39%7D%7B4%7D%26%5Cfrac%7B13%7D%7B2%5C%5C%200%26%5Cfrac%7B9%7D%7B4%7D%26%5Cfrac%7B3%7D%7B2%7D%7D%5Cend%7Barray%7D%5Cright%5D%20R_2%5C%3A%5Cleftrightarrow%20%5C%3AR_3)
![\approx \left[\begin{array}{ccc}-4&5&6\\ 0&\frac{39}{4}&\frac{13}{2}\\ 0&0&0\end{array}\right] R_3 \rightarrow R_3-\frac{3R_2}{13}\\](https://tex.z-dn.net/?f=%5Capprox%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%265%266%5C%5C%200%26%5Cfrac%7B39%7D%7B4%7D%26%5Cfrac%7B13%7D%7B2%7D%5C%5C%200%260%260%5Cend%7Barray%7D%5Cright%5D%20R_3%20%5Crightarrow%20R_3-%5Cfrac%7B3R_2%7D%7B13%7D%5C%5C)
As the Rank is given as number of non-zero rows in the Row echelon form which are 2 so the rank is 2.
Thus this indicates that the vectors does not span 
<em>Now for any vector the corresponding equation is formulated by using the combined matrix which is given as for any arbitrary vector and the coordinate as </em>
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Now converting the combined matrix as
![\approx \left[\begin{array}{ccccc}-4&5&6&|&x\\ 0&\frac{9}{4}&\frac{3}{2}&|&\frac{4y+x}{4}\\ 3&6&2&|&z\end{array}\right] R_2 \rightarrow R_2+\frac{R_1}{4}\\](https://tex.z-dn.net/?f=%5Capprox%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D-4%265%266%26%7C%26x%5C%5C%200%26%5Cfrac%7B9%7D%7B4%7D%26%5Cfrac%7B3%7D%7B2%7D%26%7C%26%5Cfrac%7B4y%2Bx%7D%7B4%7D%5C%5C%203%266%262%26%7C%26z%5Cend%7Barray%7D%5Cright%5D%20R_2%20%5Crightarrow%20R_2%2B%5Cfrac%7BR_1%7D%7B4%7D%5C%5C)
![\approx \left[\begin{array}{ccccc}-4&5&6&|&x\\ 0&\frac{9}{4}&\frac{3}{2}&|&\frac{4y+x}{4}\\ 0&\frac{39}{4}&\frac{13}{2}&|&\frac{4z+3x}{4}\end{array}\right] R_3 \rightarrow R_3+\frac{3R_1}{4}\\](https://tex.z-dn.net/?f=%5Capprox%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D-4%265%266%26%7C%26x%5C%5C%200%26%5Cfrac%7B9%7D%7B4%7D%26%5Cfrac%7B3%7D%7B2%7D%26%7C%26%5Cfrac%7B4y%2Bx%7D%7B4%7D%5C%5C%200%26%5Cfrac%7B39%7D%7B4%7D%26%5Cfrac%7B13%7D%7B2%7D%26%7C%26%5Cfrac%7B4z%2B3x%7D%7B4%7D%5Cend%7Barray%7D%5Cright%5D%20R_3%20%5Crightarrow%20R_3%2B%5Cfrac%7B3R_1%7D%7B4%7D%5C%5C)
![\approx \left[\begin{array}{ccccc}-4&5&6&|&x\\ 0&\frac{39}{4}&\frac{13}{2}&|&\frac{4z+3x}{4}\\ 0&\frac{9}{4}&\frac{3}{2}&|&\frac{4y+x}{4}\end{array}\right] R_3 \leftrightarrow R_2\\](https://tex.z-dn.net/?f=%5Capprox%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D-4%265%266%26%7C%26x%5C%5C%200%26%5Cfrac%7B39%7D%7B4%7D%26%5Cfrac%7B13%7D%7B2%7D%26%7C%26%5Cfrac%7B4z%2B3x%7D%7B4%7D%5C%5C%200%26%5Cfrac%7B9%7D%7B4%7D%26%5Cfrac%7B3%7D%7B2%7D%26%7C%26%5Cfrac%7B4y%2Bx%7D%7B4%7D%5Cend%7Barray%7D%5Cright%5D%20R_3%20%5Cleftrightarrow%20R_2%5C%5C)
![\approx \left[\begin{array}{ccccc}-4&5&6&|&x\\ 0&\frac{39}{4}&\frac{13}{2}&|&\frac{4z+3x}{4}\\ 0&0&0&|&\frac{13y+x-3z}{13}\end{array}\right] R_3 \rightarrow R_3-\frac{3R_2}{13}\\](https://tex.z-dn.net/?f=%5Capprox%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D-4%265%266%26%7C%26x%5C%5C%200%26%5Cfrac%7B39%7D%7B4%7D%26%5Cfrac%7B13%7D%7B2%7D%26%7C%26%5Cfrac%7B4z%2B3x%7D%7B4%7D%5C%5C%200%260%260%26%7C%26%5Cfrac%7B13y%2Bx-3z%7D%7B13%7D%5Cend%7Barray%7D%5Cright%5D%20R_3%20%5Crightarrow%20R_3-%5Cfrac%7B3R_2%7D%7B13%7D%5C%5C)
From this it is seen that whatever the values of the coordinates does not effect the value of the plane with equation as

So it is verified that the subspace of R3 such that it satisfies x+13y-3z=0 consists of all vectors.
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Answer:
7 workers will be required
Explanation:
We will define actual time required to do a job
= Standard time required for producing each unit x Worker efficiency (%)/100
= 11.65 x 115/100
= 13.3975 minutes
Total available minutes in 4 days = 4 days x 8 hours/ day x 60 minutes / day = 1920 minutes
Therefore ,
Number of units which can be produced by 1 worker in 4 days = 1920 /13.3975
Number of units to be produced = 1000 units
Therefore,
Number of workers required
= Number of units to be produced / Number of units which can be produced by 1 worker in 4 days
= 1000 x ( 13.3975 /1920)
= 6.977 ( 7 rounded to nearest whole number )
The historical cost principle requires that when assets are acquired, they be recorded at cost.
The historical cost principle is an accounting principle under the US GAAP. It entails recording the cost of an asset on the balance sheet at the cost with which the asset was purchased regardless of the changes in the value of the asset.
For example, if a machine was purchased at a cost of £2000. If the historical cost principle is used, the machine would be recorded at £2000 on the balance sheet.
A similar question was answered here: brainly.com/question/14417628
Answer:
The correct answer is 25%
Explanation:
To calculate the value of the tax rate to decide on the municipal bond, we must take the information of the annual yield minus the expenses associated with this product, on the interest of the corporate bond:
Tax Rate = 1 - (0.0525 / 0.0700) = 25%
In this way, 25% or more, is a percentage of the tax rate that can make them decide on the municipal bond option.
Answer:
a. Revenue - Income Statement
b. Common Stock - Balance Sheet
c. Current liabilities - Balance Sheet
d. Long-term Debt - Balance Sheet
e. Dividends - Statement of Shareholder Equity / Statement of Retained Earnings
f. Ending Cash Balance - Balance Sheet
g. Adjustment to reconcile net income to net cash provided by operations -Statement of Cash Flows
h. Cash spent to acquire the Buildings - Statement of Cash Flows
i. Income tax expense - Income Statement
j. Ending Balance of retained earnings - Statement of Shareholder Equity / Statement of Retained Earnings / Balance Sheet
k. Selling general and administrative expenses - Income Statement
l. Total Assets - Balance Sheet
m. Net Income - Income Statement / Statement of Shareholder Equity / Statement of Retained Earnings
n. Income tax payable - Balance Sheet