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Ivan
3 years ago
6

Triangle HIJ has coordinates H(-3,5), (8,2), and J(-3,-5). Find

Mathematics
1 answer:
Vlad1618 [11]3 years ago
5 0

Answer:

H (x= -3 , y=5)

I (x=8 , y=2)

J (x= -3 , y= -5)

D is the distance between the 2 points so just plug in the x and y cords of each of the corresponding points

so for (HI)

you plug in 2 on the red x

and 8 on the red y

then -3 in the blue x

and 5 in the blue y

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1. Subtract 15 to the other side (5x^2 = 36)

2.Square root both sides to remove the squared of 5x^2 (sqrt(5x^2) = sqrt(36) simplified to 5x = 6)

3. divide by 5 to get x = 6/5

Final result is 6/5
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A statement we accept without proof is called a:<br> theorem<br> definition<br> postulate<br> term
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Step-by-step explanation:

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Solve 3(2 x-1)= 4 x +2
slega [8]
First, r<span>earrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :  </span><span>3*(2*x-1)-(4*x+2)=0 
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6 0
4 years ago
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Determine la ecuación general de la recta que pasa por los puntos (1,4); (- 2, - 5) y grafíquela.
Rudiy27

The equation of the line that passes through the points (1 , 4) and (-2, -5) is y = 3x + 1

<h3>Further explanation</h3>

Solving linear equation mean calculating the unknown variable from the equation.

Let the linear equation : y = mx + c

If we draw the above equation on Cartesian Coordinates , it will be a straight line with :

<em>m → gradient of the line</em>

<em>( 0 , c ) → y - intercept</em>

Gradient of the line could also be calculated from two arbitrary points on line ( x₁ , y₁ ) and ( x₂ , y₂ ) with the formula :

\large {\boxed {m = \frac{y_2 - y_1}{x_2 - x_1}} }

If point ( x₁ , y₁ ) is on the line with gradient m , the equation of the line will be :

\large {\boxed{y - y_1 = m ( x - x_1 )} }

Let us tackle the problem.

Let :

(1 , 4) → (x₁ , y₁)

(-2, -5) → (x₂ , y₂)

To find the straight line equation, the following formula can be used :

\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}

\frac{y - 4}{-5 - 4} = \frac{x - 1}{-2 - 1}

\frac{y - 4}{-9} = \frac{x - 1}{-3}

\frac{y - 4}{3} = \frac{x - 1}{1}

y - 4 = 3 ( x - 1 )

y = 3x - 3 + 4

\large {\boxed {y = 3x + 1} }

<h3>Learn more</h3>
  • Infinite Number of Solutions : brainly.com/question/5450548
  • System of Equations : brainly.com/question/1995493
  • System of Linear equations : brainly.com/question/3291576

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Linear Equations

Keywords: Linear , Equations , 1 , Variable , Line , Gradient , Point

4 0
4 years ago
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