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liq [111]
3 years ago
10

What affects the size of an angle? A. The length of its sides B. The direction of one side C. Its orientation in space D. The am

ount of rotation of one side
Mathematics
2 answers:
AlexFokin [52]3 years ago
5 0

Answer:

The Correct answer is D

The size of an angle depends only on the

<u>amount of rotation necessary</u> to turn one side to a specific point.

Took the PF test and it was Correct!!

Step-by-step explanation:

MrMuchimi3 years ago
4 0

Answer:

<h3>D. The amount of rotation of one side.</h3>

Step-by-step explanation:

We need to explain the statement " What affects the size of an angle? ".

Given options are

A. The length of its sides

B. The direction of one side

C. Its orientation in space

D. The amount of rotation of one side.

The correct option would be :

<h3>D. The amount of rotation of one side.</h3>

<em>Because as the one side rotated, it makes a greater or smaller arc between the two lines. That would affects the size of an angle. </em>

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nlexa [21]

Answer:

(x,y) = (5,-6)

Step-by-step explanation:

\underline{\textbf{Determinant of a matrix.}}\\\\\text{For a}~ 2 \times 2 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2\\b_1&b_2 \end{vmatrix} = a_1b_2 - a_2b_1\\\\\\\text{For a}~ 3 \times 3 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix} = a_1\begin{vmatrix} b_2&b_3\\c_2&c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1&b_3\\c_1&c_3 \end{vmatrix}+ a_3 \begin{vmatrix} b_1&b_2\\c_1&c_2 \end{vmatrix}\\\\\\

                     ~~~~~~~~~~~~~~~~~~=a_1(b_2c_3-b_3c_2) -a_2(b_1c_3-b_3c_1) +a_3(b_1c_2-b_2c_1)

\underline{\textbf{Cramer's Rule to solve a system of two equations.}}\\\\\text{Consider the system of two equations:}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_1x + b_1 y= c_1\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_2x +b_2 y = c_2\\\\\text{Here,}\\\\x = \dfrac{D_x}{D}= \dfrac{\begin{vmatrix} c_1&b_1\\c_2&b_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\\\ y= \dfrac{D_y}{D}= \dfrac{\begin{vmatrix} a_1&c_1\\a_2&c_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\

\underline{\textbf{Solution:}}\\\\~~~~~~~~~~~~~~~~~~~~~~~8x-5y = 70~~~~~~...(i)\\\\~~~~~~~~~~~~~~~~~~~~~~~9x +7y = 3~~~~~~~...(ii)\\\\\text{Applying Cramer's rule:}\\\\x = \dfrac{D_x}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 70& -5 \\3&7 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{70(7) -(-5)(3)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{490+15}{56+45}\\\\\\~~=\dfrac{505}{101}\\\\\\~~=5

y = \dfrac{D_y}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 8& 70 \\9&3 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{(8)(3) -(70)(9)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{24-630}{56+45}\\\\\\~~=-\dfrac{606}{101}\\\\\\~~=-6

\textbf{Hence, the solution to the system of equation is}~ (x,y) = (5,-6)

7 0
1 year ago
What is the midpoint of the segment whose endpoints are 17,1 and -9,3
Veseljchak [2.6K]
Easy ill tell u and then do a equation first sense it has - it means u have to subtract alright it goes like this. 17,1
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                                                   --------
                                                    7,8
and this is ur answer as u see in my equation ur welcome only give me brainiest only if im right 
5 0
3 years ago
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