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Irina18 [472]
3 years ago
5

Kobe is a single dad with two dependent children, Lizzie, age 7 and Leslie, age 3. He has AGI of $51,000 and paid $3,700 to a qu

alified day care center. What amount of credit can Kobe receive for the child and dependent care credit?
Business
1 answer:
strojnjashka [21]3 years ago
3 0

Answer:

amount of credit Kobe receive for child & dependent care credit is  $740

Explanation:

given data

Lizzie age = 7

Leslie age = 3

AGI = $51,000

paid care center =  $3,700

to find out

amount of credit can Kobe receive

solution

we know that here when AGI is more than $43,000

than there child and dependent care credit =  20% of the expenses   ........1

so here amount of credit receive will be as

amount of credit Kobe receive  = 20% ×  paid care center   .............2

amount of credit Kobe receive  = 20% × $3700

amount of credit Kobe receive  = $740

so amount of credit Kobe receive for child & dependent care credit is  $740

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The vectors does not span R3 and only span a subspace of R3 which satisfies x+13y-3z=0

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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]

Now if the vectors  would span the R^3, the rank of  the consolidated matrix will be 3 if it is not 3 this indicates that the vectors does not span the R^3.

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]\\

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}

\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}\\

\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

\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}\\

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 R^3

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<em />v=\left[\begin{array}{c}x&y&z\end{array}\right]<em />

<em />c=\left[\begin{array}{c}c_1&c_2&c_3\end{array}\right]<em />

<em />Mc=v<em />

<em />\left[\begin{array}{ccc}-4&5&6\\1&1&0\\3&6&2\end{array}\right]\left[\begin{array}{c}c_1&c_2&c_3\end{array}\right]=\left[\begin{array}{c}x&y&z\end{array}\right]<em />

<em />M=\left[\begin{array}{ccccc}v_1&v_2&v_3& | &v\end{array}\right] \\M=\left[\begin{array}{ccccc}-4&5&6&|&x\\1&1&0&|&y\\3&6&2&|&z\end{array}\right]\\<em />

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\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}\\

\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}\\

\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\\

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