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2 years ago

What is the midpoint of C(5,3 and D(-3,-6)

Mathematics

1 answer:

Nat2105 [25]2 years ago

6 0

Answer:

Step-by-step explanation:

Midpoints of two coordinates is expressed using the formula;

M(X, Y) = (x2+x1/2, y2+y1/2)

Given the coordinates c(5,3) and d(-3,-6)

x1 = 5, y1 = 3, x2 = -3 and y2 = -6

X = x1+x2/2

X = 5+(-3)/3

X = 5-3/2

X = 2/2

X = 1

Also;

Y = y1+y2/2

Y = 3+(-6)/2

Y = 3-6/2

Y = -3/2

Y = -1.5

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Two boats depart from a port located at (–8, 1) in a coordinate system measured in kilometers and travel in a positive x-directi

miss Akunina [59]

Answer:

$\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.$

Step-by-step explanation:

1st boat:

Parabola equation:

$y=ax^2 +bx+c$

The x-coordinate of the vertex:

$x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=1\\ \\b=-2a$

Equation:

$y=ax^2 -2ax+c$

The y-coordinate of the vertex:

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Parabola passes through the point (-8,1), so

$1=a\cdot (-8)^2-2a\cdot (-8)+c\\ \\80a+c=1$

Solve:

$c=10+a\\ \\80a+10+a=1\\ \\81a=-9\\ \\a=-\dfrac{1}{9}\\ \\b=-2a=\dfrac{2}{9}\\ \\c=10-\dfrac{1}{9}=\dfrac{89}{9}$

Parabola equation:

$y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}$

2nd boat:

Parabola equation:

$y=ax^2 +bx+c$

The x-coordinate of the vertex:

$x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=0\\ \\b=0$

Equation:

$y=ax^2+c$

The y-coordinate of the vertex:

$y_v=a\cdot 0^2+c\Rightarrow c=-7$

Parabola passes through the point (-8,1), so

$1=a\cdot (-8)^2-7\\ \\64a-7=1$

Solve:

$a=-\dfrac{1}{8}\\ \\b=0\\ \\c=-7$

Parabola equation:

$y=\dfrac{1}{8}x^2 -7$

System of two equations:

$\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.$

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