All categories

2 years ago

Hi !! can someone help prove these please

Mathematics

1 answer:

beks73 [17]2 years ago

5 0

Step-by-step explanation:

Here's the first one for ya. I can't tell what the first denominator of the second one is [CosB, CscB, or SecB?], so unless you get back to me on that, then I can't help you with it.

You might be interested in

Write the sentence as a equation b is 380 more than the quality y times 268

Svet_ta [14]

A: b = 268y + 380
is : =

5 0

3 years ago

Kenya plans to make a down payment plus monthly payments in order to buy a motorcycle. At one dealer she would pay $2,350 down a

tatuchka [14]

I think the amount would be $4,100

4 0

3 years ago

Find the equation of the line shown below...

vovangra [49]

Answer:

y = -1/4x+2

Step-by-step explanation:

We can find the slope by finding two points on the line, and doing (y1-y2) ÷ (x1-x2). The two points I chose were (0,2), and (4,1). We can plug this into the equation to get (2-1) ÷ (0-4), to get -1/4, which is our slope. We can see on the graph, that the line intersects with the y axis when the y coordinate is 2, making the y-int 2.

4 0

2 years ago

Read 2 more answers

ANSWER ASAP!! WILL GIVE OR MORE POINTS TO THE FIRST ONE WHO ANSWERS!

Sergio [31]

Answer: See the picture attached.

Step-by-step explanation:

Given the following mixed numbers:

$-2\frac{3}{4}\\\\-4\frac{1}{2}$

You need to convert them to decimal numbers. The steps to do this are shown below:

1. Divide the numerator of the fraction by the denominator.

2. Add the quotient obtained to the whole number part.

Then:

$-2\frac{3}{4}=-(2+0.75)=-(2.75)=-2.75\\\\-4\frac{1}{2}=-(4+0.5)=-(4.5)=-4.5$

Therefore, the point is:

$(-2.75,-4.5)$

Where $-2.75$ is the x-coordinate and $-4.5$ is the y-coordinate.

Knowing this, you can plot it on the coordinate plane, as you can see in the picture attached.

8 0

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

3 0

2 years ago