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:
The surface area is 
Step-by-step explanation:
we know that
The surface area of a square pyramid is equal to the area of the square base plus the area of its four lateral triangular faces.
so
![SA=b^{2}+4[\frac{1}{2}(b)(l)]](https://tex.z-dn.net/?f=SA%3Db%5E%7B2%7D%2B4%5B%5Cfrac%7B1%7D%7B2%7D%28b%29%28l%29%5D)
we have
substitute the values
![SA=0.4^{2}+4[\frac{1}{2}(0.4)(0.6)]=0.64\ m^{2}](https://tex.z-dn.net/?f=SA%3D0.4%5E%7B2%7D%2B4%5B%5Cfrac%7B1%7D%7B2%7D%280.4%29%280.6%29%5D%3D0.64%5C%20m%5E%7B2%7D)
Answer:
Step-by-step explanation:
we know that
The phrase " the sum of x and 8" is equivalent to adds the number x and the number 8 or sum the number x plus the number 8
so
16=2/3 times number of students
times both sides by 3
48=2 times number of students
divide both sides by 2
24=number of students
there aer 24 students i the club
Answer:
26.28 m
Step-by-step explanation:
the whole explanation along with solution has been found in attachment. Please go through it.
