19/50
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Answer:
x = 62
Step-by-step explanation:
x and 118 form a linear pair (added up, they both equal 180 degrees)
So, 180-118 = 62 which is the measurement of x
−2x + 3y + 5z = −21
−4z = 20
6x − 3y = 0
do -4z=20 first
divide both sides by -4 to get z by itself
-4z/-4=20/-4
z=-5
Use z=-5 into −2x + 3y + 5z = −21
-2x+3y+5(-5)=-21
-2x+3y-25=-21
move -25 to the other side
sign changes from -25 to +25
-2x+3y-25+25=-21+25
-2x+3y=4
6x-3y=0
find x by eliminating y
Add the equations together
-2x+6x+3y+(-3y)=4+0
-2x+6x+3y-3y=4
4x=4
Divide by 4 for both sides
4x/4=4/4
x=1
Use x=1 into 6x − 3y = 0
6(1)-3y=0
6-3y=0
Move 6 to the other side
6-6-3y=0-6
-3y=-6
Divide both sides by -3
-3y/-3=-6/-3
y=2
Answer:
(1, 2, -5)
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>
