Answer:-288
Step-by-step explanation:
Given
First term
is common for both AP and GP
For AP
---1
----2
From 1 & 2 we get
d=4

for AP

For GP
----3
----4
From 3 & 4 we get



For 

Answer:
the answer is c h2=k2=120%
Step-by-step explanation:

Here, in the given figure. Angle y° forms vertical opposite angle pair with Angle 163°, therefore they have equal values.
that means ~

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:
27
Step-by-step explanation: