Answer:
(-3, 3)
Step-by-step explanation:
Substitute for y:
x+6 = -2x -3
3x = -9 . . . . . . . add 2x -6 to both sides
x = -3 . . . . . . . . divide by 3
y = (-3) +6 = 3 . . substitute for x in the first equation
The solution is ...
(x, y) = (-3, 3)
Answer:
I think it could be 8+6+y
Step-by-step explanation: I dont really know but dont delete this. Can I be Brainlyest please?
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
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