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Georgia [21]
3 years ago
12

Write and solve the equation to find the measure of

Mathematics
1 answer:
kvv77 [185]3 years ago
4 0
I'm not sure the equation is written correctly but the measure of the given angle is 126 degrees. In this instance, the variable in 3x (x) would be equal to 42. Hope this helped.

180-54= 126
126 / 3 = x
x = 42
:)
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What is the ratio of 30 divided by 4:2
klasskru [66]
4:2 can represented as 4 / 2. Or also 2.

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7 0
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Please show work and thank youuu
Assoli18 [71]

Answer:  6\sqrt{3}

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

Explanation:

Method 1

We can use the pythagorean theorem to find x.

a^2+b^2 = c^2\\\\6^2+x^2 = 12^2\\\\36+x^2 = 144\\\\x^2 = 144-36\\\\x^2 = 108\\\\x = \sqrt{108}\\\\x = \sqrt{36*3}\\\\x = \sqrt{36}*\sqrt{3}\\\\x = 6\sqrt{3}\\\\

-----------------------------------

Method 2

Use the sine ratio to find x. You'll need a reference sheet or the unit circle, or simply memorize that sin(60) = sqrt(3)/2

\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(60^{\circ}) = \frac{x}{12}\\\\\frac{\sqrt{3}}{2} = \frac{x}{12}\\\\x = 12*\frac{\sqrt{3}}{2}\\\\x = 6\sqrt{3}\\\\

-----------------------------------

Method 3

Similar to the previous method, but we'll use tangent this time.

Use a reference sheet, unit circle, or memorize that tan(60) = sqrt(3)

\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(60^{\circ}) = \frac{x}{6}\\\\\sqrt{3} = \frac{x}{6}\\\\x = 6\sqrt{3}\\\\

-----------------------------------

Method 4

This is a 30-60-90 triangle. In other words, the angles are 30 degrees, 60 degrees, and 90 degrees.

Because of this special type of triangle, we know that the long leg is exactly sqrt(3) times that of the short leg.

\text{long leg} = (\text{short leg})*\sqrt{3}\\\\x = 6\sqrt{3}\\\\

The short leg is always opposite the smallest angle (30 degrees).

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Answer:

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Step-by-step explanation:

B. The quadrilaterals cannot be placed such that each occupies one-quarter of the circle because the vertices of parallelogram 1 do not form right angles.

E. The quadrilaterals cannot be placed such that each occupies one-quarter of the circle because the vertices of parallelogram 4 do not form right angles.

Step-by-step explanation:

hope it helps:3

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