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poizon [28]
2 years ago
8

Somebody pleaseee help :(

Mathematics
1 answer:
suter [353]2 years ago
3 0

Answer:

answer is A fosha

Step-by-step explanation:

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Square root of 172 ?
Veronika [31]

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13.1148770486

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3 0
2 years ago
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PLS HELP ASAP, WILL MARK YOU BRAINLIEST!!!!
GuDViN [60]

Answer:

Try looking at each angle and add the sides

Step-by-step explanation:

4 0
2 years ago
Two less than three times a number is 28
Alenkinab [10]

Answer:

x = 10

Step-by-step explanation:

Two less than three times a number is 28

Let's break it up.

2 less than: -2

3 times a number: 3x (x is representing a number")

is 28: = 28

So, let's put that together.

Two less than three times a number is 28

3x - 2 = 28

Now, let's solve this.

First, let's isolate the x.

To do this, we must add 2 to each side.

3x - 2 = 28

+2           +2

3x - 2 + 2 = 3x

28 + 2 = 30

We have 3x = 30

Now, we must simplify.

To do this, we must divide each side by 3.

This, in term, will tell us what x equals.

3x = 30

---    ----

3      3

3x/3 = x

30/3 = 10

x = 10

4 0
2 years ago
Find the value of n that makes each equation true. Then, arrange the equations IN increasing ORDER of the values of n found.
andreyandreev [35.5K]

n+5=13 n=8

n+5=10 n=5

2n=3 n=1.5

5n=20 n=4

6 0
3 years ago
If A is a 2 × 2 matrix, then A × I = <br> and I × A =
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
2 years ago
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