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2 years ago

don walked 3 3/5 miles on Friday, 3.7 miles on saturday, and 3 5/8 miles on sunday. list the distances from least to greatest

Mathematics

1 answer:

astraxan [27]2 years ago

8 0

Solution:

we are given that

Don walked 3 3/5 miles on Friday

It can be re-written as

$3\frac{3}{5}miles=\frac{18}{5}= 3.6$ miles

3 5/8 miles on sunday.

It can be re-written as

$3\frac{5}{8}miles=\frac{29}{8}= 3.625$ miles

an d He walk 3.7 miles on satarday.

The distances from least to greatest is 3.6, 3.625, 3.7

The distances from least to greatest is $3\frac{3}{5}$ miles, $3\frac{5}{8}$ miles, 3.7 miles.

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ANSWER

EXPLANATION

For a matrix A of order n×n, the cofactor $C_{ij}$ of element $a_{ij}$ is defined to be

$C_{ij} = (-1)^{i+j} M_{ij}$

$M_{ij}$ is the minor of element $a_{ij}$ equal to the determinant of the matrix we get by taking matrix A and deleting row i and column j.

Here, we have

$C_{11} = (-1)^{1+1} M_{11} = M_{11}$

M₁₁ is the determinant of the matrix that is matrix A with row 1 and column 1 removed. The bold entries are the row and the column we delete.

$\begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \end{aligned}$

Since the determinant of a 2×2 matrix is

$\det\left( \begin{bmatrix} a & b \\ c& d \end{bmatrix} \right) = ad-bc$

it follows that

so $C_{11} = M_{11} = 24$

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