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: y = 5d + 7 / c - 3
Step-by-step explanation:
Step 1: Add -3y to both sides
cy - 7 + -3y = 5d + 3y + -3y
= cy - 3y - 7 = 5d
Step 2: Now, add 7 on both sides
cy - 3y - 7 +7 = 5d +7
= cy - 3y = 5d +7
Step 3: Factor out y
y(c - 3) = 5d + 7
Step 4: Divide both sides by c-3
y(c - 3) / c - 3 = 5d+7 / c-3
Your answer for this should be
y = 5d + 7 / c - 3
Hope this helped!
Answer:
can you send it clearly it is blur see
Answer:
10. it is correct it is 24
11. 21
Step-by-step explanation:
I could not see the answer but it should look like this:
11.6-3(3-[2-3]-10)-3
6-3(3-[-1]-10)-3
6-3(3-(-1)-10)-3
6-3(4-10)-3
6-3(-6)-3
6-(-18)-3
(6-(-18))-3
24-3
21
A perpendicular line has a slope of infinity. Therefore it is not possible to write the equation in point-slope form.
The equation is x = -4.
