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Ludmilka [50]
3 years ago
7

Can some one help me with this

Mathematics
2 answers:
Fittoniya [83]3 years ago
8 0
The answer is <u>A, parallel</u>. The slopes are the same (<em>m=-4/5</em>)

I hope this helped :D
anyanavicka [17]3 years ago
3 0
Its parallel for this math problem
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4y + 4 = 2y - 10 <br> can somebody break this down to solve it ? please
vredina [299]

Answer:

y=-7

Step-by-step explanation:

first, you want variables on one side, numbers on the other side of the equation.

4y+4=2y-10

*plus ten on both side to get rid of -10*

4y+14=2y

*minus 4y on each side to get rid of 4y*

14=-2y

*divide by -2 on each side to get the value of just one y itself*

y=-7

8 0
3 years ago
What is 411, 600, 000 in scientific notation
Morgarella [4.7K]

Answer:

4.116 × 108

Step-by-step explanation:

I simply used a Scientific Notation Converter

5 0
3 years ago
Read 2 more answers
What is the relationship between 0.04 and 0.004
olga_2 [115]
0.04 is ten times more than 0.004
4 0
3 years ago
determine if the ordered pair is a solution of the equation is (2,4) a solution of y = 10 -3×? true or false​
anzhelika [568]

Answer:

False

Step-by-step explanation:

4 = 10 - 3²

4 = 10 - 9

4 = 1

7 0
3 years ago
Find the length of each side of the triangle determined by the three points P1,P2, and P3. State whether the triangle is an isos
Tanya [424]

Answer:

The triangle is both an Isosceles triangle and a right triangle.

Step-by-step explanation:

Given the vertices of a triangle.

$ P_{1} = (- 1, 4) $

$ P_{2} = (6, 2) $    and

$ P_{3} = (4, - 5) $

We find the distance between all the points to determine the length of each side of the triangle.

Distance between any two points, say, $ (x_1, y_1) $ and $ (x_2, y_2) $ is:

                                 $ \sqrt{\bigg ( \textbf{x}_{\textbf{2}} \hspace{1mm} \textbf{- x}_{\textbf{1}} \bigg )^{\textbf{2}} \textbf{+}   \bigg( \textbf{y}_{\textbf{2}} \hspace{1mm} \textbf{- y}_{\textbf{1}} \bigg)^ {\textbf{2}} $

Length between $ P_1 $ and $ P_{2} $ , (Side 1) :

$ (x_1, y_1) = (- 1, 4) $     and

$ (x_2, y_2) = (6, 2) $

Distance = $ \sqrt{\bigg(6 - (-1) \bigg)^{2} \hspace{1mm} + \hspace{1mm} \bigg( 2 - 4 \bigg )^2 $

$ = \sqrt{7^2 + 2^2} $

$ = \sqrt{49 + 4} $

$ = \sqrt{\textbf{53}} $

Length of Side 1 = $  \sqrt{\textbf{53}} $ units.

Distance between $ P_1 $ and P_2 , (Side 2):

$ (x_1, y_1) = (-1, 4) $

$ (x_2, y_2) = (4, - 5) $

Distance = $ \sqrt{ \bigg( 4 + 1 \bigg)^2 \hspace{1mm} + \bigg( - 5 - 4 \bigg ) ^2 $

$ = \sqrt{ 25 + 81 } $

$ = \sqrt{\textbf{106}} $

Length of Side 2 = $ \sqrt{\textbf{106}} $ units.

Distance between $ P_2 $ and $ P_3 $ , Side 3 :

$ (x_1, y_1) = (4, 5) $

$ (x_2, y_2) = (6, 2) $

Distance = $ \sqrt{ 2^2 \hspace{1mm} + \hspace{1mm} 7^2} $

$ = \sqrt{49 + 4} $

$ = \sqrt{\textbf{53} $

Length of Side 3  = $ \sqrt{\textbf{53}} $ units.

Note that the length of Side 1 = Length of Side 3.

That means the triangle is Isosceles.

Also, For a triangle to be right angle triangle, using Pythagoras theorem we have:

(Side 1)² + (Side 3)² = (Side 2)²

$ \bigg( \sqrt{53} \bigg )^2 \hspace{1mm} + \hspace{1mm} \bigg( \sqrt{53} \bigg)^2 \hspace{1mm} = \hspace{1mm} \bigg ( \sqrt{106} \bigg ) ^2 $

i.e, 53 + 53  = 106

Hence, the triangle is a right - angled triangle as well.

7 0
3 years ago
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