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
Step-by-step explanation:
5 + b ÷ (11 - 9)
Substitute b = 14
5 + 14 ÷ (11 - 9)
Work the order of operations from left to right
Since there are no exponents, parentheses first
5 + 14 ÷ (11 - 9)
5+14 ÷ 2
Then division
5+7
Then addition
12
I believe the answer to your question is b.
Imagine that the class only had 4 students (unrealistic most likely, but small numbers help much better I think)
If we had 4 students and 3 were boys, then 4-3 = 1 girl is in the class. This makes the ratio of boys to girls be 3 to 1. In other words, there are 3 times as many boys compared to girls.
Divide the number of girls (1) over the number of students total (4) to get 1/4 = 0.25 = 25%
<h3>Answer: 25%</h3>
