Option B is the answer because angle A is greater than C and A
The area bounded by the 2 parabolas is A(θ) = 1/2∫(r₂²- r₁²).dθ between limits θ = a,b...
<span>the limits are solution to 3cosθ = 1+cosθ the points of intersection of curves. </span>
<span>2cosθ = 1 => θ = ±π/3 </span>
<span>A(θ) = 1/2∫(r₂²- r₁²).dθ = 1/2∫(3cosθ)² - (1+cosθ)².dθ </span>
<span>= 1/2∫(3cosθ)².dθ - 1/2∫(1+cosθ)².dθ </span>
<span>= 9/8[2θ + sin(2θ)] - 1/8[6θ + 8sinθ +sin(2θ)] .. </span>
<span>.............where I have used ∫(cosθ)².dθ=1/4[2θ + sin(2θ)] </span>
<span>= 3θ/2 +sin(2θ) - sin(θ) </span>
<span>Area = A(π/3) - A(-π/3) </span>
<span>= 3π/6 + sin(2π/3) -sin(π/3) - (-3π/6) - sin(-2π/3) + sin(-π/3) </span>
<span>= π.</span>
Answer:
12, 8, and 4
Step-by-step explanation:
a + d + a + a - d = 24
3a = 24 ⇒ [ a = 8 ]
and (a + d)² + a² + (a - d)² = 224
3a² + 2d² = 224
2d² = 224 - 192 = 32
d² = 16
[ d = ±4 ]
∴ The first three terms to A.P are
a + d = 8 + 4 = 12
a - = 8
a - d = 8 - 4 = 4
∴ Three terms are 12 , 8, and 4
Answer: the square root of 3
Step-by-step explanation:
