Let the point be P, and the masses be located at P1 and P2.
The centre of mass is such that the moments of masses m1 and m2 exert an equal moment at the point P,
and the distance P1P2=d
namely,
m1(mPP1)=m2(mPP2)
The distance of the centre of mass from m1 is therefore
d1=d*m2/(m1+m2)
Similarly, the distance of the centre of mass from m2 is
d2=d*m1/(m1+m2)
Hope this help! I’m sure this is right!
Answer:
Step 4
part a: no answer yet still having trouble showing my work
part b:
-0.125(x-24)^2+50
-0.125(x-24)(x-24)+50
-0.125(x^2-24x-24x+576)+50
-0.125(x^2-48x+576)+50
-0.125x^2+6x-75+50
-0.125x^2+6x-22
a=-0.125
b=6
c=-22
part c: yes he went through
part d:
plug in roots (4,0) and (44,0)
if x is 4 and y is 0
0=-0.125(4)^2+6(4)-22
0=-2+24-22
0=-2+2
0=0
if x=44 and y=0
0=-0.125(44)^2+6(44)-22
0=-242+264-22
0=-242+242
0=0
Answer:
(f - g)(x) = (3x + 1) - (x^2 - 6) = 3x + 1 - x^2 + 6 = -x^2 + 3x + 7
The answer to your question is 1
