Answer:
The correct option is;
Step-by-step explanation:
The given parameters are
The equation of motion of one (the first) object is r = 4·cos(θ)
The equation of motion of the other (the second) object is r = -1 + 2·cos(θ)
Equating both equations gives;
4·cos(θ) = -1 + 2·cos(θ)
4·cos(θ) - 2·cos(θ) = -1
2·cos(θ) = -1
cos(θ) = -1/2
θ = Arccos(-1/2) = 120° = 2·π/3
Therefore, the two equations are equal when θ = 2·π/3 for which we have;
r = 4·cos(2π/3) = -2 and r = -1 + 2·cos(2π/3) = -2
∴ r = 4·cos(2π/3) = -1 + 2·cos(2π/3) = -2
The coordinate that represents a possible collision point of the objects in the form (r, θ) is therefore
Answer:
A. without changing the setting of the compass, place the point of the compass on point D and drawn out the intersects line L and extends below line L.
Step-by-step explanation:
In the figure attached, the procedure is shown.
In this case, we are at step 2, the first arc was made, and you have to make the second arc (without changing the setting of the compass) placing the point of the compass on point D.
Step-by-step explanation:
Answer:
9. 
8. 
7. 
Step-by-step explanation:
![\displaystyle [x - h]^2 - [y - k]^2 = r^2 → Hyperbola\:Equation \\ [x - h]^2 + [y - k]^2 = r^2 → Circle\:Equation \\ [h, k] → Centre](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Bx%20-%20h%5D%5E2%20-%20%5By%20-%20k%5D%5E2%20%3D%20r%5E2%20%E2%86%92%20Hyperbola%5C%3AEquation%20%5C%5C%20%5Bx%20-%20h%5D%5E2%20%2B%20%5By%20-%20k%5D%5E2%20%3D%20r%5E2%20%E2%86%92%20Circle%5C%3AEquation%20%5C%5C%20%5Bh%2C%20k%5D%20%E2%86%92%20Centre)
According to the equations in the exercises, in the parentheses, ALL NEGATIVE SIGNS give the OPPOSITE terms of what they REALLY are, so be EXTREMELY CAREFUL with your translations:
9. [4, 11]
+ 3 + 4
_____
[7, 15] → (x - 7)² and (y - 15)²
8. [−2, 8]
+ 3 + 2
_____
[1, 10] → (x - 1)² and (y - 10)²
7. [−8, −14]
+ 3 - 5
______
[−5, −19] → (x + 5)² and (y + 19)²
** NOTISE THAT THE RADII <em>NEVER</em><em> </em>ALTER.
I am joyous to assist you anytime.
<h3>
Answer: Choice B) 468 cm^2</h3>
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Explanation:
As the diagram shows, we have a triangular face with a base of b = 18 and height of h = 13. The area of this triangular face is:
A = 0.5*b*h
A = 0.5*18*13
A = 117 square cm or cm^2
This represents the area of one face, but there are 4 identical copies of this triangular face assuming this tetrahedron is a regular polyhedron. So we have a total surface area of 4*117 = 468 cm^2 which points us to choice B
This is the amount of paint you'll need to cover the outer surface of the pyramid, or you can think of it as the amount of wrapping paper needed. This assumes there are no gaps or overlaps.
