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3 years ago

4 questions ! will mark and rate!

Mathematics

1 answer:

Dennis_Churaev [7]3 years ago

7 0

Answer:

1. 617.5 cm³

2. 335.6 in³

3. 151.6 m³

4. 3479.1 ft³

Step-by-step explanation:

B1: 19.5 × 9 = 175.5

B2: 6 × 13 = 78

sqrt(B1 ×B2) = 117

h = 5

Volume = ⅓ × 5 × [175.5 + 78 + 117]

617.5

B1: 3.14 × 4.5² = 63.585

B2: 3.14 × 3² = 28.26

sqrt(B1 ×B2) = 42.39

h = 7.5

Volume = ⅓ × 7.5 × [63.585 + 28.26 + 42.39]

335.5875

B1: ½(8+12) × 6 = 60

B2: ½(3+4.5) × 2.25 = 8.4375

sqrt(B1 ×B2) = 22.5

h = 5

Volume = ⅓ × 5 × [60 + 8.4364 + 22.5]

151.5625

B1: 3.14 × 12² = 452.16

B2: 3.14 × 7² = 153.86

sqrt(B1 ×B2) = 263.76

h² = 13² - (12-7)²

h² = 13² - 5²

h² = 144

h = 12

Volume = ⅓ × 12 × [452.16 + 153.86 + 263.76]

3479.12

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<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.