Answer:
qgg
Step-by-step explanation:
grrqg
Answer:
11in
Step-by-step explanation:
a^2+b^2=c^2 when c is the hypotenuse
10^2+sqrt21^2=c^2
100+21=c^2
121=c^2
11=c
X = 4(cosθ - cos²θ) = 4cosθ - 4cos²<span>θ
</span>dx/dθ = -4sinθ + 8sinθcosθ = -4sinθ(1 - 2cos<span>θ)
dy/d</span>θ = 4(cosθ - cos²θ + sin²θ) = 4(cos<span>θ - 1)
</span>∴dy/dx = 4(cosθ - 1)/-4sinθ(1 - 2cosθ) = 1-cosθ/(sinθ - 2sinθcosθ<span>)
Now, we're finding a horizontal tangent, which is when dy/dx = 0 (horizontal tangent).
1-cos</span>θ/(sinθ - 2sinθcosθ) = 0
1-cosθ = 0
cosθ = 1
θ = 0, 360 in the interval [0, 360].
At θ = 0, x = 0, y = 0
Only points in the open interval [-2π, 2π] are at (0, 0), unless I made a mistake somewhere.
I think it would be 2 the positive and negative
