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Mrrafil [7]
3 years ago
13

6.258 rounded to the nearest hundredth

Mathematics
2 answers:
Tresset [83]3 years ago
5 0
<h2>Answer:</h2><h2>6.26</h2><h2></h2><h2>Hope this helps!!</h2><h2 />
ElenaW [278]3 years ago
4 0

Answer:

6.26

Step-by-step explanation:

Integer Part: 6

Fractional Part: 258

If the last digit in the fractional part of 6.258 is less than 5, then simply remove the last the digit of the fractional part.

If the last digit in the fractional part of 6.258 is 5 or more and the second digit in the fractional part is less than 9, then add 1 to the second digit of the fractional part and remove the third digit.

If the last digit in the fractional part of 6.258 is 5 or more and the second digit in the fractional part is 9, and the first digit in the fractional part is less than 9, then add 1 to the first digit in fractional part and make the second digit in fractional part 0. Then remove the third digit.

If the last digit in the fractional part of 6.258 is 5 or more and the second digit in the fractional part is 9, and the first digit in the fractional part is 9, then add 1 to the integer part and make the fractional part 00.

With 6.258, rule B applies and 6.258 rounded to the nearest hundredth is 2.26

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Step-by-step explanation:

quotient means division

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In the box below, write the following statement using symbolic language.
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The temperature of the water in a coffeepot does not exceed 210 ° Fahrenheit. Let t be the temperature of the water in the coffe
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krok68 [10]

Since the multiplication between two matrices is not <em>commutative</em>, then \vec A\, \times\,\vec I \ne \vec I \,\times \,\vec A, regardless of the dimensions of \vec A.

<h3>Is the product of two matrices commutative?</h3>

In linear algebra, we define the product of two matrices as follows:

\vec C = \vec A \,\times \vec B, where \vec A \in \mathbb{R}_{m\times p}, \vec B \in \mathbb{R}_{p\times n} and \vec C \in \mathbb{R}_{m \times n}     (1)

Where each element of the matrix is equal to the following dot product:

c_{ij} = \left[\begin{array}{cccc}a_{i1}&a_{i2}&\ldots&a_{ip}\end{array}\right]\,\bullet\,\left[\begin{array}{ccc}b_{1j}\\b_{2j}\\\vdots\\b_{pj}\end{array}\right], where 1 ≤ i ≤ m and 1 ≤ j ≤ n.     (2)

Because of (2), we can infer that the product of two matrices, no matter what dimensions each matrix may have, is not <em>commutative</em> because of the nature and characteristics of the definition itself, which implies operating on a row of the <em>former</em> matrix and a column of the <em>latter</em> matrix.

Such <em>"arbitrariness"</em> means that <em>resulting</em> value for c_{ij} will be different if the order between \vec A and \vec B is changed and even the dimensions of \vec C may be different. Therefore, the proposition is false.

To learn more on matrices: brainly.com/question/9967572

#SPJ1

3 0
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