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:
approx 7.81 Exactly: √61
Step-by-step explanation:
Use Pythag theorem: 5^2 + 6^2 = hyp^2 (line AS)
25 + 36 = 61
hyp = √61 or about 7.81
(8.6 x 108) x (3.2 x103
place them in order
(8.6 x 3.2) x (108 x 103)
27.52 x (108 x 103)
27.52 x 11124
That is your answer to this question.
Answer:
IM pretty sure b) and c)
Step-by-step explanation:
Answer:
(a)18
(b)1089
(c)Sunday
Step-by-step explanation:
The problem presented is an arithmetic sequence where:
- First Sunday, a=1
- Common Difference (Every subsequent Sunday), d=7
We want to determine the number of Sundays in the 120 days before Christmas.
(a)In an arithmetic sequence:
Since the result is a whole number, there are 18 Sundays in which Aldsworth advertises.
Therefore, Aldsworth advertised 18 times.
(b)Next, we want to determine the sum of the first 18 terms of the sequence
1,8,15,...
The sum of the numbers of days published in all the advertisements is 1089.
(c)SInce the 120th day is the 18th Sunday, Christmas is on Sunday.
