Answer:
90
Step-by-step explanation:
The first term = 6* 2^0 = 6
The second term = 6 * 2^ (2-1) = 6*2 = 12
The third term = 6* 2^(3-1) = 6*2^2 = 6*4 = 24
The fourth term = 6* 2^(4-1) = 6* 2^3 = 6*8 = 48
S4 is the sum of the 1st four terms
S4 = 6+ 12+24+ 48 = 90
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:
-4
Step-by-step explanation:
distributing gets you to : 8h + 38 = -18 - 6h
then add 6h to 8h and you have 14h + 38 = -18
solve normally & thats how you get your answer
