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3 years ago

Can some one help me pleases

Mathematics

1 answer:

vaieri [72.5K]3 years ago

6 0

Answer:

It's A. 28 degrees

Step-by-step explanation:

First, you determine the angle measure directly below x. That is a straight line, so it measures 180 degrees. You can add 38 and 90 degrees, and then subtract from 180 to get the last angle measure. This gives you 52. The right 3 angles measure 90 degrees, because they are on the opposite side of a right angle marker. You can add 10 and 52 and then subtract from 90 to get the missing x value, 28.

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Please help me. I suck at math

serg [7]

Answer:

Answer is ...Obtuse

Step-by-step explanation:

$\frac{m$

$\frac{5x + 4}{2} = \frac{3}{2} x + 21$

$\frac{5x + 4}{2} = \frac{3x + 42}{2}$

$5x + 4 = 3x + 42$

$5x - 3x = 42 - 4$

$2x = 38$

$x = 19$

$substituting \: x = 19 \: \:in \: \: \: \: \: \: \: \\ m$

We get,

m<ABC =

$(5 \times 19) + 4 = 99$

Hence <ABC is obtuse

4 0

3 years ago

What is the answer to this too?

Yuri [45]

Answer:

Step-by-step explanation:

This one sounds the most confident, being a call to action. It is good to end a thought with motion.

7 0

3 years ago

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Please please help please

nikklg [1K]

Answer: It will still measure the same at 30 degrees.

Step-by-step explanation:

A rotation just changes orientation, not size or measure

5 0

3 years ago

Read 2 more answers

Square ABCD has side length a. Half circles with

Inga [223]

The area of the shaded region is $0.215a^2$

<h2>Area of composite objects</h2>

The area of the shaded region is expressed according to the formula:

$A = A_s - A_c$

Get the area of the square As

As = a²

$A_c = \pi (0.5a)^2\\
A_c = 0.785a^2$

Take the difference in the areas

Area of the shaded part = $a^2-0.785a^2\\
$

Area of the shaded part = $0.215a^2$

Hence the area of the shaded region is $0.215a^2$

Learn more on area of composite objects here: brainly.com/question/22716761

7 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