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s344n2d4d5 [400]
4 years ago
14

Hello! Please help with this fill in the blank, thanks!

Mathematics
1 answer:
Ivenika [448]4 years ago
7 0
<h3>Answer:</h3>

Triangle ABC is rotated <u>180 degrees around point E</u>, therefore, A maps to <u>D</u>, B maps to <u>C</u>, and C maps to <u>B</u>.

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

Explanation:

If you were to rotate triangle ABC 180 degrees around point E, then it would line up perfectly with triangle DCB

  • Point A lands on point D
  • Point B lands on point C
  • Point C lands on point B

Note how in the sequence ABC we have A first, B second, then C third

Then for the sequence DCB we have D first, C second, then B third

This order matters so we can pair up the points properly

  • A goes to D (A is first of ABC; D is first of DCB)
  • B goes to C (B is second in ABC; C is second in DCB)
  • C goes to B (C is third in ABC; B is third in DCB)
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If you do the math you'll get 3/4 of the bag is dirt.
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X = -135°, cos 2x = _______
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Answer:

cos 2 x=1-sin^2 x

Step-by-step explanation:

x=-135°

cos 2x=cos (-270)=cos (-270+360)=cos 90=0

or cos 2x=1-2sin ²x=1-2(sin (-135))²

=1-2(-sin 135)²

=1-2(sin 135)²

=1-2(sin (180-45))²

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3 years ago
Help meh plz! i will give brainleist and new short sound track!
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Answer:

I believe The first one is 1.5 and the second one is 12

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7 0
3 years ago
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Which of the following is a solution to 2cos2x − cos x − 1 = 0?
Pavel [41]

Answer:

Option A is correct.

Solution for the given equation is, x = 0^{\circ}

Step-by-step explanation:

Given that : 2\cos^2x -\cos x -1 =0

Let \cos x =y

then our equation become;

2y^2-y-1= 0           .....[1]

A quadratic equation is of the form:

ax^2+bx+c =0.....[2] where a, b and c are coefficient and the solution is given by;

x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}

Comparing equation [1] and [2] we get;

a = 2 b = -1 and c =-1

then;

y = \frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}

Simplify:

y = \frac{ 1 \pm \sqrt{1+8}}{4}

or

y = \frac{ 1 \pm \sqrt{9}}{4}

y = \frac{ 1 \pm 3}{4}

or

y = \frac{1+3}{4} and y = \frac{1 -3}{4}

Simplify:

y = 1 and y = -\frac{1}{2}

Substitute y = cos x we have;

\cos x = 1

⇒x = 0^{\circ}

and

\cos x = -\frac{1}{2}

⇒ x = 120^{\circ} \text{and} x = 240^{\circ}

The solution set:  \{0^{\circ}, 120^{\circ} , 240^{\circ}\}

Therefore, the solution for the given equation  2\cos^2x -\cos x -1 =0 is, 0^{\circ}





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