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Westkost [7]
3 years ago
10

Pentagon ABCDE is shown on the coordinate plane below:

Mathematics
2 answers:
goblinko [34]3 years ago
4 0
It's (5,2) , I took it already so here you go
nikitadnepr [17]3 years ago
3 0

Answer:  The correct option is (B) (5, 2).

Step-by-step explanation:  Given that the co-ordinates of the vertices of pentagon ABCDE are A(-2, 4), B(-6, 2), C(-5, -2), D(1, -2) and E(2, 2).

We are to find the co-ordinates of the point C', if the pentagon ABCDE is  rotated 180° around the origin to create pentagon A'B'C'D'E'.

We know that a rotation of 180° changes the co-ordinates of a point (x, y) according to the following rule:

(x, y)  ⇒  (-x, -y).

Therefore, the vertices of pentagon ABCD will transform to the vertices of pentagon A'B'C'D'E' as follows:

A(-2, 4)  ⇒ A'(2, -4),

B(-6, 2)   ⇒ B'(6, -2),

C(-5, -2)  ⇒  C'(5, 2),

D(1, -2)    ⇒  D'(-1, 2),

E(2, 2)    ⇒   E'(-2, -2).

Thus, the co-ordinates of the point C' are (5, 2).

Option (B) is correct.

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Ahat [919]

Answer:

Cos θ  = -4/5

Tan θ = -3/4

Step-by-step explanation:

The question is as following:

Select the correct answer from each drop-down menu.

Angle θ lies in the second quadrant, and sin θ =3/5.

cos θ = -4/5, -3/5, 3/5, 4/5

tan θ = -4/3, -3/4, 3/4, 4/3

==================================================

Since, the angle lies in second quadrant (negative x axis and positive y axis) we can deduce that cos θ is negative and tan θ is also negative.

If sin θ =3/5

cos \  \theta =\sqrt{1-sin^2 \theta} = \sqrt{1-(\frac{3}{5} )^2} =\sqrt{1-\frac{9}{25} }=\sqrt{\frac{16}{25} } =\frac{4}{5}

∴ Cos θ = -4/5 ⇒ because θ lies in the second quadrant.

And

Tan θ = (sin θ)/(cos θ) = (3/5) / (-4/5) = -3/4.

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3 years ago
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