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3 years ago

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If the numbers -2.5, -1.6, -3 1/10, - 1/10, and -0.5 were graphed on a number line, which would appear farthest to the left?

1 answer:

Leona [35]3 years ago

7 0

Turn the question around.
If all the numbers were positive: 2.5, 1.6, 3 1/10, 1/10, and 0.5, which would be farthest to the right?

The fact is that the negative numbers are a reflection of the positive numbers.
Since 3 1/10 would be farthest to the right with positive numbers, -3 1/10 will be farthest to the left with the negative numbers.

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Step-by-step explanation:

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devlian [24]

The inverse of matrix A = $\left[\begin{array}{cc}2&2&1&3\\\end{array}\right]$ is $A^{-1} = \frac{1}{4}$ $\left|\begin{array}{cc}3&-2&-1&2\\\end{array}\right|$

The given matrix is A = $\left[\begin{array}{cc}2&2&1&3\\\end{array}\right]$

The inverse of a matrix is given by

$A^{-1} = \frac{1}{|A|} adjA$ ...(1)

Now, determinant of matrix A is

|A| = $\left|\begin{array}{cc}2&2&1&3\\\end{array}\right|$

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Now, adjoint A will be

$A_{11}$ = 3

$A_{12}$ = -1

$A_{21}$ = -2

$A_{22}$ = 2

AdjA = $\left|\begin{array}{cc}3&-2&-1&2\\\end{array}\right|$

Now, substituting the values of |A| and AdjA in equation (1), we get

$A^{-1} = \frac{1}{4}$ $\left|\begin{array}{cc}3&-2&-1&2\\\end{array}\right|$

Therefore, $A^{-1} = \frac{1}{4}$ $\left|\begin{array}{cc}3&-2&-1&2\\\end{array}\right|$

Know more about Inverse of Matrix: -brainly.com/question/12442362

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