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DIA [1.3K]
3 years ago
5

Check the apply π\6 is the reference angle for

Mathematics
2 answers:
belka [17]3 years ago
5 0
\pi = 180\\
 \frac{\pi}{6} = x\\ \pi x =  \frac{180\pi}{6}\\ \pi x = 30 \pi \\ x = 30 
Tanzania [10]3 years ago
5 0
X=30 hope it helps........
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Simplify −0.6(−6.3) answer??
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Consider the equation and the relation “(x, y) R (0, 2)”, where R is read as “has distance 1 of”. For example, “(0, 3) R (0, 2)”
Leviafan [203]

Answer:

The equation determine a relation between x and y

x = ± \sqrt{1-(y-2)^{2}}

y = ± \sqrt{1-x^{2}}+2

The domain is 1 ≤ y ≤ 3

The domain is -1 ≤ x ≤ 1

The graphs of these two function are half circle with center (0 , 2)

All of the points on the circle that have distance 1 from point (0 , 2)

Step-by-step explanation:

* Lets explain how to solve the problem

- The equation x² + (y - 2)² and the relation "(x , y) R (0, 2)", where

 R is read as "has distance 1 of"

- This relation can also be read as “the point (x, y) is on the circle

 of radius 1 with center (0, 2)”

- “(x, y) satisfies this equation , if and only if, (x, y) R (0, 2)”

* <em>Lets solve the problem</em>

- The equation of a circle of center (h , k) and radius r is

  (x - h)² + (y - k)² = r²

∵ The center of the circle is (0 , 2)

∴ h = 0 and k = 2

∵ The radius is 1

∴ r = 1

∴ The equation is ⇒  (x - 0)² + (y - 2)² = 1²

∴ The equation is ⇒ x² + (y - 2)² = 1

∵ A circle represents the graph of a relation

∴ The equation determine a relation between x and y

* Lets prove that x=g(y)

- To do that find x in terms of y by separate x in side and all other

  in the other side

∵ x² + (y - 2)² = 1

- Subtract (y - 2)² from both sides

∴ x² = 1 - (y - 2)²

- Take square root for both sides

∴ x = ± \sqrt{1-(y-2)^{2}}

∴ x = g(y)

* Lets prove that y=h(x)

- To do that find y in terms of x by separate y in side and all other

  in the other side

∵ x² + (y - 2)² = 1

- Subtract x² from both sides

∴ (y - 2)² = 1 - x²

- Take square root for both sides

∴ y - 2 = ± \sqrt{1-x^{2}}

- Add 2 for both sides

∴ y = ± \sqrt{1-x^{2}}+2

∴ y = h(x)

- In the function x = ± \sqrt{1-(y-2)^{2}}

∵ \sqrt{1-(y-2)^{2}} ≥ 0

∴ 1 - (y - 2)² ≥ 0

- Add (y - 2)² to both sides

∴ 1 ≥ (y - 2)²

- Take √ for both sides

∴ 1 ≥ y - 2 ≥ -1

- Add 2 for both sides

∴ 3 ≥ y ≥ 1

∴ The domain is 1 ≤ y ≤ 3

- In the function y = ± \sqrt{1-x^{2}}+2

∵ \sqrt{1-x^{2}} ≥ 0

∴ 1 - x² ≥ 0

- Add x² for both sides

∴ 1 ≥ x²

- Take √ for both sides

∴ 1 ≥ x ≥ -1

∴ The domain is -1 ≤ x ≤ 1

* The graphs of these two function are half circle with center (0 , 2)

* All of the points on the circle that have distance 1 from point (0 , 2)

8 0
3 years ago
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