Answer:no the -2/3 should be before -1/5
Step-by-step explanation:
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:
Step-by-step explanation:
= - 4 - 2( - 3) = 2
= - 4 - 2( - 2) = 0
= - 4 - 2( - 1) = - 2
= - 4 - 2( 0 ) = - 4
= - 4 - 2( 1 ) = - 6
Range: { 2, 0, - 2, - 4, - 6 }
(x/100-1)(x/100+1)=kx^2-1
(x/100-1)(x/100+1)=(x/100)^2 - 1^2 because it's a difference of squares ( (a-b)(a+b) = a^2-ab+ab-b^2 = a^2 - b^2):
(x/100)^2 - 1^2=kx^2 - 1
(x/100)^2 - 1=kx^2 - 1
(x/100)^2=kx^2
(x^2)/(100^2)=kx^2
1/(100^2)=k
1/10000=k
k=0.0001
Subtract the answer of the division problem from the sum of 33 and 7. Your answer would be:
40 - 5
Simplified:
35
