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lana [24]
2 years ago
11

Find the distance between the two points 5,6 and 0,-6. Show your work​

Mathematics
1 answer:
valkas [14]2 years ago
3 0

Answer:

The distance between two points is √169 or 13.

Step-by-step explanation:

In the question we need to find the distance between two points 5,6 and 0,-6 given.

Distance between two points can be found using the formula:

d(P1,P2) = \sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2}

Putting the given points in the formula we get,

x₂ = 0, x₁ = 5, y₂= -6, y₁= 6

=\sqrt{(0-(5))^2 + (-6-(6))^2} \\=\sqrt{(0-5)^2 + (-6-6)^2}\\=\sqrt{(-5)^2 + (-12)^2}\\=\sqrt{25 + 144}\\=\sqrt{169}\\=13

So, the distance between two points is √169 or 13.

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you bought a 7 notebooks at the school supply store cost $4.76 what is the unit rate explain how you determined your answer use
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Simplify<br><br>Square root of 50 over 2 times -4 + square root of 121
kobusy [5.1K]

Answer:

  -9

Step-by-step explanation:

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Your calculator can help with this.

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Find an equation of the line passing through the points (2, 2) and (-1,-6)?
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3 years ago
Is x+y+1=0 a tangent of both y^2=4x and x^2=4y parabolas?
Lubov Fominskaja [6]

Answer:

  yes

Step-by-step explanation:

The line intersects each parabola in one point, so is tangent to both.

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For the first parabola, the point of intersection is ...

  y^2 = 4(-y-1)

  y^2 +4y +4 = 0

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  y = -2 . . . . . . . . one solution only

  x = -(-2)-1 = 1

The point of intersection is (1, -2).

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For the second parabola, the equation is the same, but with x and y interchanged:

  x^2 = 4(-x-1)

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  x = -2, y = 1 . . . . . one point of intersection only

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If the line is not parallel to the axis of symmetry, it is tangent if there is only one point of intersection. Here the line x+y+1=0 is tangent to both y^2=4x and x^2=4y.

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Another way to consider this is to look at the two parabolas as mirror images of each other across the line y=x. The given line is perpendicular to that line of reflection, so if it is tangent to one parabola, it is tangent to both.

7 0
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