Triangle fgh is inscribed in circle o the length of radius oh is 6 and fh is congruent to og. what is the area of the sector for
1 answer:
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Lester and Carey are 14 feet apart
Step-by-step explanation:
Mrs. Kim is teaching a 5th grade class
- She is standing 48 feet in front of Lester
- Carey is sitting to Lester's right
- Carey and Mrs. Kim are 50 feet apart
We need to find how far apart Lester and Carey
Look to the attached figure
The positions of Mrs. Kim, Lester, and Carey formed a right triangle
∵ The distance between Mrs. Kim and Carey is 50 feet
∴ The length of the hypotenuse of the Δ = 50 feet
∵ The distance between Mrs. Kim and Lester is 48 feet
∴ The length of one leg of the Δ = 48 feet
The formula of length of the sides of the right Δ is:
Substitute the length of the hypotenuse and the length of the given leg in the formula above
∵ (50)² = (48)² + (leg)²
∴ 2500 = 2304 + (leg)²
- Subtract 2304 from both sides
∴ 196 = (leg)²
- Take square for both sides
∴ 14 = leg
∵ This leg represents the distance between Lester and Carey
∴ The distance between Lester and Carey = 14 feet
Lester and Carey are 14 feet apart
Learn more:
You can learn more about the right triangle in brainly.com/question/1238144
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Answer:
The equation determine a relation between x and y
x = ±
y = ±
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 = ±
∴ 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 = ±
- Add 2 for both sides
∴ y = ±
∴ y = h(x)
- In the function x = ±
∵ ≥ 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 = ±
∵ ≥ 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)