Solutions for Two distinct parabolas
1 answer:
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9514 1404 393
Answer:
(0, π/4] ∪ (π, 5π/4]
Step-by-step explanation:
Multiplying by 2 gives ...
cot(x) ≥ 1
The cotangent function decreases from ∞ to 1 in the domain (0, π/4], and again in the domain (π, 5π/4]. The solution is the union of these two intervals.
x ∈ (0, π/4] ∪ (π, 5π/4]
_____
(a, b] is interval notation for a < x ≤ b
<h2>Answer:</h2>
The ratio of the area of region R to the area of region S is:
The sides of R are in the ratio : 2:3
Let the length of R be: 2x
and the width of R be: 3x
i.e. The perimeter of R is given by:
( Since, the perimeter of a rectangle with length L and breadth or width B is given by:
)
Hence, we get:
i.e.
Also, let " s " denote the side of the square region.
We know that the perimeter of a square with side " s " is given by:
Now, it is given that:
The perimeters of square region S and rectangular region R are equal.
i.e.
Now, we know that the area of a square is given by:
and
Hence, we get:
and
i.e.
Hence,
Ratio of the area of region R to the area of region S is: