Radius, r = 3
The equation of a sphere entered at the origin in cartesian coordinates is
x^2 + y^2 + z^2 = r^2
That in spherical coordinates is:
x = rcos(theta)*sin(phi)
y= r sin(theta)*sin(phi)
z = rcos(phi)
where you can make u = r cos(phi) to obtain the parametrical equations
x = √[r^2 - u^2] cos(theta)
y = √[r^2 - u^2] sin (theta)
z = u
where theta goes from 0 to 2π and u goes from -r to r.
In our case r = 3, so the parametrical equations are:
Answer:
x = √[9 - u^2] cos(theta)
y = √[9 - u^2] sin (theta)
z = u
Answer:
C
Step-by-step explanation:
The equation of a parabola in vertex form is
y = a(x - h)² + k
where (h, k) are the coordinates of the vertex and a is a multiplier
Given
y = 2x² + 12x + 1
To express in vertex form use the method of completing the square.
The coefficient of the x² term must be 1 , thus factor out 2 from 2x² + 12x
y = 2(x² + 6x) + 1
add/ subtract ( half the coefficient of the x- term)² to x² + 6x
y = 2(x² + 2(3)x + 9 - 9) + 1
= 2(x + 3)² - 18 + 1
= 2(x + 3)² - 17 → C
Step-by-step explanation:
With reference to the regular hexagon, from the image above we can see that it is formed by six triangles whose sides are two circle's radii and the hexagon's side. The angle of each of these triangles' vertex that is in the circle center is equal to 360∘6=60∘ and so must be the two other angles formed with the triangle's base to each one of the radii: so these triangles are equilateral.
The apothem divides equally each one of the equilateral triangles in two right triangles whose sides are circle's radius, apothem and half of the hexagon's side. Since the apothem forms a right angle with the hexagon's side and since the hexagon's side forms 60∘ with a circle's radius with an endpoint in common with the hexagon's side, we can determine the side in this fashion:
tan60∘=opposed cathetusadjacent cathetus => √3=Apothemside2 => side=(2√3)Apothem
As already mentioned the area of the regular hexagon is formed by the area of 6 equilateral triangles (for each of these triangle's the base is a hexagon's side and the apothem functions as height) or:
Shexagon=6⋅S△=6(base)(height)2=3(2√3)Apothem⋅Apothem=(6√3)(Apothem)2
=> Shexagon=6×62√3=216
3.482×10^9
Imagine the decimal at the end of the last 0 and move it 9 places to the left
I think that would be 40 correct me if I’m wrong
