Answer:
r = -12cos(θ)
Step-by-step explanation:
The usual translation can be used:
Putting these relationships into the formula, we have ...
(r·cos(θ) +6)² +(r·sin(θ))² = 36
r²·cos(θ)² +12r·cos(θ) +36 +r²·sin(θ)² = 36
r² +12r·cos(θ) = 0 . . . . subtract 36, use the trig identity cos²+sin²=1
r(r +12cos(θ)) = 0
This has two solutions for r:
r = 0 . . . . . . . . a point at the origin
r = -12cos(θ) . . . the circle of interest
I hope this helps you
2 1/2=2.2+1/2=5/2
5/2-1/2
4/2
2 used paint
2 1/2=5/2
5/2-2
2,5-2
0,5
1/2 left
KE=1/2(m)(v)^2
KE=1/2(80)(1.5)^2
KE=90 J
Step-by-step explanation:
it looks like this is a regular (right) triangular prism with 3 equal side areas and one bottom area.
the surface area of the pyramid is the sum of all 4 of these areas.
all these areas are triangles.
the area of a triangle is
baseline × height / 2
in our case
1 side area is 6×8.2/2 = 3×8.2 = 24.6 yd²
all 3 side areas together are 24.6×3 = 73.8 yd²
the bottom triangle is
6×5.2/2 = 3×5.2 = 15.6 yd²
so, the total surface area is
73.8 + 15.6 = 89.4 yd²
