7.) Plumbers earn $160 a day and plumbers' helpers earn $100 a day. Twenty

Mathematics

1 answer:

9966 [12]2 years ago

6 0

Answer:

Step-by-step explanation:

p and h are number of plumbers and helpers, respectively.

Twenty workers earn a total of $2600

p+h = 20

h = 20-p

160p + 100h = 2600

160p + 100(20-p) = 2600

60p = 600

p = 10

h = 20-p = 10

Ten plumbers and ten helpers.

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Please help find the percent of increase?!

Contact [7]

Subtract to find the difference:

2000 - 500 = 1500

Divide the difference by the original population:

1500 / 500 = 3

Multiply that by 100 to get the percentage:

3 x 100 = 300%

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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.