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olasank [31]
3 years ago
8

The perpendicular bisector or to the line segment with an points(-4,9) and (8,5). The equation of the line is

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
8 0
The equation of the line is x=5/8y
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Answers for the 2 boxes please :)​
scoray [572]

Answer:

(8, - 5 )

Step-by-step explanation:

Given the 2 equations

3x + 4y = 4 → (1)

- x - 3y = 7 → (2)

Multiplying (2) by 3 and adding to (1) will eliminate the x- term.

- 3x - 9y = 21 → (3)

Add (1) and (3) term by term to eliminate x, that is

0 - 5y = 25

- 5y = 25 ( divide both sides by - 5 )

y = - 5

Substitute y = - 5 into either of the 2 equations and solve for x

Substituting into (1)

3x + 4(- 5) = 4

3x - 20 = 4 ( add 20 to both sides )

3x = 24 ( divide both sides by 3 )

x = 8

solution is (8, - 5 )

7 0
3 years ago
Read 2 more answers
Use the coordinate grid to determine the coordinates of point A:
Simora [160]

Based on the given coordinate grid, and the increment for each grid line, the coordinates of point A is ( fraction negative 1 and 1 over 4, fraction 3 over 4) or (-1¹/₄, ³/₄).

<h3>What are point A's coordinates?</h3>

The x-value of point A is said to be 5 grid lines to the left of y-axis. This means that the x value will be negative.

The increments are 1/4 so the x-value is:
= -1/4 x 5

= -1¹/₄

The y-value is given by the 3 grid lines above the x-axis which means that this value is positive:

= 1/4 x 3

= 3 /4

The coordinates of A are:

(-1¹/₄, ³/₄)

Find out more on coordinates at brainly.com/question/14355852

#SPJ1

4 0
1 year ago
What is the answer to this: 12= 4+x/2
Fittoniya [83]

Answer:

16

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
A triangle has vertices at (-2,-3),(4, -3), and (3,5). What is the area of the triangle?
Bezzdna [24]

Answer:

this triangle is ABC with A(-2 ; -3), B(4; - 3) and C(3;5)

=> AB=\sqrt{(4-(-2))^{2}+\sqrt{(-3 -(-3))^{2} }  }=\sqrt{6^{2} }=6\\\\AC=\sqrt{(3-(-2))^{2}+(5-(-3))^{2}  }=\sqrt{5^{2}+8^{2}  }=\sqrt{89}\\\\BC =\sqrt{1^{2}+8^{2}  }=\sqrt{65}

using Heron theorem, we have:

S=\sqrt{p(p-AB)(p-AC)(p-BC)}\\\\S=\sqrt{(\frac{6+\sqrt{89}+\sqrt{65} }{2})(\frac{\sqrt{89}+\sqrt{65}-6  }{2} )(\frac{6+\sqrt{65}-\sqrt{89}  }{2})(\frac{6+\sqrt{89}-\sqrt{65}  }{2})   }\\\\S=24  \\\\

with S is the area of the triangle

       p=\frac{AB+AC+BC}{2}=\frac{6+\sqrt{89}+\sqrt{65}  }{2}

Step-by-step explanation:

5 0
3 years ago
What's the answer for this 3,4,5,6?
blondinia [14]
Every 2 minute..............

8 0
2 years ago
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