a circle with radius 3 has a sector with a central angle of 17/9 pi radians. what is the area of the sector?
1 answer:
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Answer:
Since a/2⁽ⁿ ⁺ ¹⁾b < a/2ⁿb, we cannot find a smallest positive rational number because there would always be a number smaller than that number if it were divided by half.
Step-by-step explanation:
Let a/b be the rational number in its simplest form. If we divide a/b by 2, we get another rational number a/2b. a/2b < a/b. If we divide a/2b we have a/2b ÷ 2 = a/4b = a/2²b. So, for a given rational number a/b divided by 2, n times, we have our new number c = a/2ⁿb where n ≥ 1
Since = a/(2^∞)b = a/b × 1/∞ = a/b × 0 = 0, the sequence converges.
Now for each successive division by 2, a/2⁽ⁿ ⁺ ¹⁾b < a/2ⁿb and
a/2⁽ⁿ ⁺ ¹⁾b/a/2ⁿb = 1/2, so the next number is always half the previous number.
So, we cannot find a smallest positive rational number because there would always be a number smaller than that number if it were divided by half.
Answer:
Step-by-step explanation:
So, on substituting all these values, we get
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<h2>ADDITIONAL INFORMATION :-</h2>Sign of Trigonometric ratios in Quadrants
- sin (90°-θ) = cos θ
- cos (90°-θ) = sin θ
- tan (90°-θ) = cot θ
- csc (90°-θ) = sec θ
- sec (90°-θ) = csc θ
- cot (90°-θ) = tan θ
- sin (90°+θ) = cos θ
- cos (90°+θ) = -sin θ
- tan (90°+θ) = -cot θ
- csc (90°+θ) = sec θ
- sec (90°+θ) = -csc θ
- cot (90°+θ) = -tan θ
- sin (180°-θ) = sin θ
- cos (180°-θ) = -cos θ
- tan (180°-θ) = -tan θ
- csc (180°-θ) = csc θ
- sec (180°-θ) = -sec θ
- cot (180°-θ) = -cot θ
- sin (180°+θ) = -sin θ
- cos (180°+θ) = -cos θ
- tan (180°+θ) = tan θ
- csc (180°+θ) = -csc θ
- sec (180°+θ) = -sec θ
- cot (180°+θ) = cot θ
- sin (270°-θ) = -cos θ
- cos (270°-θ) = -sin θ
- tan (270°-θ) = cot θ
- csc (270°-θ) = -sec θ
- sec (270°-θ) = -csc θ
- cot (270°-θ) = tan θ
- sin (270°+θ) = -cos θ
- cos (270°+θ) = sin θ
- tan (270°+θ) = -cot θ
- csc (270°+θ) = -sec θ
- sec (270°+θ) = cos θ
- cot (270°+θ) = -tan θ