How to calculate moments with 3 separate weights of different amounts at different points?
200n 100n 300n

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what would the moments be?

I don't completely understand your drawing, although I can see that you certainly
did put a lot of effort into making it.  But calculating the moment is easy, and we
can get along without the drawing.

Each separate weight has a 'moment'.
The moment of each weight is: 

             (the weight of it) x (its distance from the pivot/fulcrum) .

That's all there is to a 'moment'.

The lever (or the see-saw) is balanced when (the sum of all the moments
on one side) is equal to (the sum of the moments on the other side).

That's why when you're on the see-saw with a little kid, the little kid has to sit
farther away from the pivot than you do.  The kid has less weight than you do,
so he needs more distance in order for his moment to be equal to yours.

Global warming was a major concern some people do not care global warming is fake it is not real the percentage of it happening  is slight 

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s