Answer:
C
Step-by-step explanation:
(x+2)² + (y-3)² = 3²
x² + y² + 4x - 6y + 4 + 9 - 9 = 0
x² + y² + 4x - 6y + 4 = 0
Answer:
17°
Step-by-step explanation:
∠AOC is a right angle, so it measures 90°
∠AOB and ∠BOC are in ∠AOC, so m∠AOB + m∠BOC = 90
m∠AOB = 73°
m∠BOC = x (shown in the diagram)
73 + x = 90
x = 17
Centimeter
Milimeter
Micrometer
Nanometer
For the answer to the question above,
<span>r = 1 + cos θ
x = r cos θ
x = ( 1 + cos θ) cos θ
x = cos θ + cos^2 θ
dx/dθ = -sin θ + 2 cos θ (-sin θ)
dx/dθ = -sin θ - 2 cos θ sin θ
y = r sin θ
y = (1 + cos θ) sin θ
y = sin θ + cos θ sin θ
dy/dθ = cos θ - sin^2 θ + cos^2 θ
dy/dx = (dy/dθ) / (dx/dθ)
dy/dx = (cos θ - sin^2 θ + cos^2 θ)/ (-sin θ - 2 cos θ sin θ)
For horizontal tangent line, dy/dθ = 0
cos θ - sin^2 θ + cos^2 θ = 0
cos θ - (1-cos^2 θ) + cos^2 θ = 0
cos θ -1 + 2 cos^2 θ = 0
2 cos^2 θ + cos θ -1 = 0
Let y = cos θ
2y^2+y-1=0
2y^2+2y-y-1=0
2y(y+1)-1(y+1)=0
(y+1)(2y-1)=0
y=-1
y=1/2
cos θ =-1
θ = π
cos θ =1/2
θ = π/3 , 5π/3
θ = π/3 , π, 5π/3
when θ = π/3, r = 3/2
when θ = π, r = 0
when θ = 5π/3 , r = 3/2
(3/2, π/3) and (3/2, 5π/3) give horizontal tangent lines
</span>---------------------------------------------------------------------------------
For horizontal tangent line, dx/dθ = 0
<span>-sin θ - 2 cos θ sin θ = 0 </span>
<span>-sin θ (1+ 2 cos θ ) = 0 </span>
<span>sin θ = 0 </span>
<span>θ = 0, π </span>
<span>(1+ 2 cos θ ) =0 </span>
<span>cos θ =-1/2 </span>
<span>θ = 2π/3 </span>
<span>θ = 4π/3 </span>
<span>θ = 0, 2π/3 ,π, 4π/3 </span>
<span>when θ = 0, r=2 </span>
<span>when θ = 2π/3, r=1/2 </span>
<span>when θ = π, r=0 </span>
<span>when θ = 4π/3 , r=1/2 </span>
<span>(2,0) , (1/2, 2π/3) , (0, π), (1/2, 4π/3) </span>
<span>At (2,0) there is a vertical tangent line</span>
