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3 years ago

I need the answer along with the procedure

Mathematics

1 answer:

Lera25 [3.4K]3 years ago

7 0

Answer:

Step-by-step explanation:

First, find the slope of the given line by finding two points. (0, 3) and (1, 1)

Use the equation for slope rise/run to find the slope.

From the left point to the right point, it goes down 2 units and to the right by 1 unit

-2/1 = -2 = slope

We already know the y-intercept which is y = 3, when x=0 and when the line crosses the y-axis

y=-2x+3 is the equation of the line.

Perpendicular line's slope is the opposite reciprocal of the original line's slope. Opposite reciprocal of -2 would be 1/2

y = 1/2(x)

y=1/2x -5/2

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Which expression is equivalent to 5(x + 7)?

larisa86 [58]

You need to multiply both numbers inside the parentheses by 5:

5 times x = 5x
5 times 7 = 35

So your answer is B. 5x + 35

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4 years ago

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Diano4ka-milaya [45]

Answer:

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4 years ago

What are the domain and range of the function below?

irina1246 [14]

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3 years ago

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Solve for w. -3(-8w+1) -w=5(w-1)-1 Simplify your answer as much as possible. W =

oksian1 [2.3K]

Answer:

W = - 6

Step-by-step explanation:

-3(-8w+1) - w = 5(w-1) - 1 -> (Original equation)

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3 years ago

If A is a 2 × 2 matrix, then A × I = and I × A =

krok68 [10]

Since the multiplication between two matrices is not commutative, 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 commutative because of the nature and characteristics of the definition itself, which implies operating on a row of the former matrix and a column of the latter matrix.

Such "arbitrariness" means that resulting 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

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