Answer:
c
Step-by-step explanation:
Given that:
since cos (kπ) =
Then, the series can be expressed as:
In the sum of an alternating series, the best bound on the remainder for the approximation is related to its term.
∴
Find the perimeter of the polygon. Assume that lines which appear to be tangent are tangent. Round to the nearest tenth if neces
sary.
1 answer:
Answer:
Step-by-step explanation:
we know that
The <u><em>Two-Tangent Theore</em></u>m states that if two tangent segments are drawn to one circle from the same external point, then they are congruent
so
Applying the Two-Tangent Theorem at each vertex of the triangle
The perimeter of the triangle is equal to
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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)