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alex41 [277]
3 years ago
14

What are the coordinates of point J'?(-2,-4)(-2,-6)

Mathematics
1 answer:
Alla [95]3 years ago
4 0

(-2,-4) is the right answer because it makes sentence

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Vanessa draws one side of equilateral ΔABC on the coordinate plane at points A(–2, 1) and B(4,1). What are the possible coordina
grandymaker [24]

Important question. No menu of choices given so we'll just have to figure it out.

It's not possible in two dimensions for all three vertices of an equilateral triangle to be lattice points, i.e. points with integer coordinates.  

Since A(-2,1) and B(4,1) have integer coordinates, our other point won't. There will be a √3 involved, the algebraic tell that there's a 30/60/90 triangle lurking somewhere in the problem.

Here the given side is parallel to the x axis which makes things easier.  

|AB| = 4 - -2 = 6

The x coordinate of the third vertex will be the same as that of the midpoint of AB, let's call it M,

M = ( (-2 + 4) / 2, (1 + 1)/2 ) = (1, 1)

We're making some progress.  Whatever our height h is our two candidates for the third vertex C are (1, 1 ± h).

Since |AB|=6, we get |AB|=|BC|=|AC|=6 because it's an equilateral triangle.

The altitude CM bisects the triangle into 2 30/60/90 triangles.  Let's take one of them, AMC.  Angle AMC is a right angle, so we have a right triangle with legs |AM|=3 and |CM|=h, and hypotenuse |AC|=6. The Pythagorean Theorem tells us:

3² + h² = 6²

9 + h² = 36

h² = 27

h = 3√3

Answer: C is (1, 1 + 3√3) or (1, 1 - 3√3)

4 0
3 years ago
Hey can I have some help with this I am very confused
aniked [119]

#a

m<5=40°

  • Reason-Its Given

#b

m<2=140

  • Reason-Given

#c

<5 and <2 are linear pair(looked properly in diagram) Hence they are supplementary.

#d

<5 and 2 are same side interior angles (Clearly shown on diagram .

  • We can call them co responding interior angles and their sum will be 180 if lines are parallel

#e

a||b

Let's verify

Let the angle below <1 be <3

Now

<2+<3=180

<5+<4=180

  • <5=<2

\\ \sf\longmapsto

Hence verified

4 0
3 years ago
Help me please..........
aleksklad [387]

Answer:

66.4°

Step-by-step explanation:

→ Find the side which is not given

Hypotenuse = 15 cm

Opposite = ?

Adjacent = 6 cm

→ Utilise the formula triangle with out opposite in it

Cos⁻¹ = Adjacent ÷ Hypotenuse

→ Substitute in the terms

Cos⁻¹ = 6 ÷ 15 ⇔ 66.4°

3 0
3 years ago
What are the coordinates of vertex J of the pre-image?(14 POINTS. PLS HELP, GEOMETRY)
DaniilM [7]
If we look at where it is now, we would have multiplied the x and y values by 1/2. since we are working backwards, lets multiply by 2 instead...giving us the vertex for the pre-image....

J is now (0, -2)      multiply by 2
J was     (0, -4)
5 0
3 years ago
Read 2 more answers
Math homework help needed.
Hatshy [7]
Answer:

The input of -1 is not allowed for the function.
7 0
3 years ago
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