All categories

2 years ago

What is the next step in constructing an angle (MNO ) congruent to the given angle (ABC ) with a straightedge and a compass?

Mathematics

1 answer:

nata0808 [166]2 years ago

5 0

The next step in constructing an angle (MNO ) congruent to the given angle (ABC ) is, using N as the center and the same radius as before, draw an arc that intersects |NO|. label the arc PQ where Q intersects |NO|.

To construct angle MNO congruent to the given angle ABC, the following conditions must be satisfied:

The two angles must be equal (∠ABC = ∠MNO).
The radius of the two triangles must be equal.
The radius from center N must be equal to radius BA and BC.
The radius of angle MNO is NO and it must equal BA and BC

Thus, we can conclude that, the next step in constructing an angle (MNO ) congruent to the given angle (ABC ) is, using N as the center and the same radius as before, draw an arc that intersects |NO|. label the arc PQ where Q intersects |NO|.

Learn more about construction of congruent angles here: brainly.com/question/20097419

You might be interested in

Pls helppp <br> 1/2•7•24=

Kazeer [188]

Answer:

Step-by-step explanation:

0.5 · 7 · 24

0.5 · 168

3 0

3 years ago

Read 2 more answers

Brian tied a 136-foot rope around the perimeter of his rectangular house. He knows that the length of the house is 10 feet less

STALIN [3.7K]

Answer:

length 42 cm and width 26 cm

Step-by-step explanation:

Given

Since Brian tied 136 foot rope around the perimeter , hence perimeter of house is 136 foot.

Given that shape of house is rectangular.

formula to calculate perimeter of rectangle can be used which is 2*(L+B)

where L is length of rectangle and B width of rectangle

Let the L and B be length of rectangular side of house

According to question

length of house is 10 feet less than twice the width of house

mathematically it can be represented as

L = 2*B - 10 --->eq. 1

Perimeter of house = 136 feet

using formula of Perimeter of rectangle

136 = 2*(L+B)

substituting the value of length in terms of width from eq. 1 in formula of perimeter

=> 136/2 = 2*B - 10 + B (2*B+ B = 3B)

=> 68 = 3*B - 10

=> 78 = 3*B

B = 78/3 = 26

Therefore B which is width = 26 CM

length = 2*B - 10= 26*2 - 10 = 52-10=42

length of house = 42 cm

Hence dimension of Brian's house is length 42 cm and width 26 cm

5 0

3 years ago

A) y B) y>x<br> C) y+x = 1<br> D) y/x = 1

Bond [772]

Answer:

The answer should be C

6 0

2 years ago

Solve the following matrix equations: (matrices)

Masja [62]

Step-by-step explanation:

$3X + \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix} - 1 & 6 \\ 10 & 14 \end{pmatrix} \\ \\ 3X = \begin{pmatrix} - 1 & 6 \\ 10 & 14 \end{pmatrix} - \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} \\ \\ 3X = \begin{pmatrix} - 1 - 2 & 6 - 3 \\ 10 - 4 & 14 - 5 \end{pmatrix}\\ \\ 3X = \begin{pmatrix} - 3 & 3 \\ 6 & 9\end{pmatrix}\\ \\ X = \frac{1}{3} \begin{pmatrix} - 3 & 3 \\ 6 & 9\end{pmatrix}\\ \\ X = \begin{pmatrix} \frac{ - 3}{3} & \frac{3}{3} \\ \\ \frac{6}{3} & \frac{9}{3} \end{pmatrix}\\ \\ \huge \red{ X} = \purple{ \begin{pmatrix} - 1 &1 \\ 2 & 3 \end{pmatrix}}$

$3X + 2I_3=\begin{pmatrix} 5 & 0 & -3 \\6 & 5 & 0\\ 9 & 6 & 5\end{pmatrix} \\\\3X + 2\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix} =\begin{pmatrix} 5 & 0 & - 3\\6 & 5 & 0\\ 9 & 6 & 5 \end{pmatrix} \\\\3X + \begin{pmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2\end{pmatrix} =\begin{pmatrix} 5 & 0 & - 3\\ 6 & 5 & 0 \\ 9 & 6 & 5 \end{pmatrix} \\\\3X =\begin{pmatrix} 5 & 0 & -3 \\ 6 & 5 & 0 \\ 9 & 6 & 5 \end{pmatrix} - \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \\\\3X =\begin{pmatrix} 5-2 & 0-0 & -3-0 \\ 6-0 & 5-2 & 0-0 \\ 9-0 & 6-0 & 5-2 \end{pmatrix} \\\\3X =\begin{pmatrix} 3 & 0 & - 3 \\ 6 & 3 & 0 \\ 9 & 6 & 3 \end{pmatrix} \\\\X =\frac{1}{3} \begin{pmatrix} 3 & 0 & - 3 \\ 6 & 3 & 0 \\ 9 & 6 & 3 \end{pmatrix} \\\\X =\begin{pmatrix} \frac{3}{3} & \frac{0}{3} & \frac{-3}{3} \\\\ \frac{6}{3} & \frac{3}{3} & \frac{0}{3} \\\\ \frac{9}{3} & \frac{6}{3} & \frac{3}{3} \end{pmatrix} \\\\\huge\purple {X} =\orange{\begin{pmatrix} 1 & 0 & - 1\\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}}\\$

8 0

3 years ago

In a for loop, is it possible to use indexing?

kenny6666 [7]

Answer:

Yes

Step-by-step explanation:

A for loop basically relies on repeating the same code for a pre-set number of times. During which you can make any code repeat inside of it, including indexing through a type of list. Many times a for-loop will use the indexes in a list to calculate the number of times that the loop has to repeat. This is usually done in order to search or apply changes in an array of other types of data structures that have countless values stored in it.

4 0

2 years ago