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3 years ago

The midpoint of is point P at (−6,−2). If point E is at (2,−4), what are the coordinates of point F?

Mathematics

2 answers:

Shtirlitz [24]3 years ago

8 0

F would be (-14, 0)

In order to find this, we first have to note that the coordinates of P would simply be the average of it's E and F counterparts. We'll first start by finding the x value.

(Ex + Fx)/2 = Px

(2 + Fx)/2 = -6

2 + Fx = -12

Fx = -14

Now we'll do the same for the y coordinate.

(Ey + Fy)/2 = Py

(-4 + Fy)/2 = -2

-4 + Fy = -4

Fy = 0

mr_godi [17]3 years ago

5 0

Answer:

A). (-14, 0)

Step-by-step explanation:

I just took the quiz on Edge... Hope this helps!!

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Answer:

Step-by-step explanation:

first we will break the parts of the question down:

d= distance

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the independent variable is t, time, because the (distance) that the car travels depends on how long the car has been traveling for (time) meaning that the dependent value is distance.

A table we can create with 5 coordinate pairs is:

(time) (distance)

1 hr 60miles

2 hr 120 miles

3 hr 180 miles

4 hr 240 miles

5 hr 300 miles

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4 years ago

9. Given the point (6,-8) values of the six trig function.

lozanna [386]

Answer:

Here's what I get.

Step-by-step explanation:

9. (6, -8)

The reference angle θ is in the fourth quadrant.

∆AOB is a right triangle.

OB² = OA² + AB² = 6² + (-8)² = 36 + 64 = 100

OB = √100 = 10

$\sin \theta = \dfrac{-8}{10} = -\dfrac{4}{5}\\\\\cos \theta =\dfrac{6}{10} = \dfrac{3}{5}\\\\\tan \theta = \dfrac{-8}{6} = -\dfrac{4}{3}\\\\\csc \theta = \dfrac{10}{-8} = -\dfrac{5}{4}\\\\\sec \theta = \dfrac{10}{6} = \dfrac{5}{3}\\\\\cot \theta = \dfrac{6}{-8} = -\dfrac{3}{4}$

10. cot θ = -(√3)/2

The reference angle θ is in the second quadrant.

∆AOB is a right triangle.

OB² = OA² + AB² = (-√3)² + (2)² = 3 + 4 = 7

OB = √7

$\sin \theta = \dfrac{2}{\sqrt{7}} = \dfrac{2\sqrt{7}}{7}\\\\\cos \theta = \dfrac{-\sqrt{3}}{\sqrt{7}} = -\dfrac{\sqrt{21}}{7}\\\\\tan \theta = \dfrac{2}{-\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}\\\\\csc \theta = \dfrac{\sqrt{7}}{2} \\\\\sec \theta = \dfrac{\sqrt{7}}{-\sqrt{3}} = -\dfrac{\sqrt{21}}{3}\\\\\cot \theta = -\dfrac{\sqrt{3}}{2}$

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The midpoint of ￼ is point P at (−6,−2). If point E is at (2,−4), what are the coordinates of point F?

The midpoint of is point P at (−6,−2). If point E is at (2,−4), what are the coordinates of point F?