A square is always symmetrical because if you cut it in half or draw a line in the middle the 2 pieces will be the exact same, also because the square has all sides and angles that is congruent (the same). Hoped my answer helped Cx.
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
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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>
Answer:
1
Step-by-step explanation:
5x1+3=8
Answer:
40%
Step-by-step explanation:
11+24+16+9=60
24/60=2/5
.40
40%