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>
The complete proof statement and reason for the required proof is as follows:
Statement Reason
m<PNO = 45 Given
MO Given
<MNP and <PNO are a
linear pair of angles Definition of linear pairs of angles
<MNP and <PNO are
supplementary angles Linear Pair Postulate
m<MNP + m<PNO = 180° Definition of supplementary angles
m<MNP + 45° = 180° Substitution property of equality
m<MNP = 135° Subtraction property of equality
Step-by-step explanation:
3/8 and 11/15
LCM = 30
3/8x4 and 11/15x2
12/30 and 22/30
22/30-12/30
=10/30
The difference is 10/30
The larger sum is 11/15
vous êtes les bienvenus, baisers!
Answer:
D) 3 × (40 + 7) = 141
Step-by-step explanation:
(3 × 40 = 120) + (3 × 7 = 21) = 141
