The center of mass of a set of points p1, . . , pn is defined as the point mn = (p1 + · · · + pn)/n. (a) for a tetrahedron p1p2p3p4, consider the point dividing the segment m3p4 in the ratio 1: 3, where m3 is the center of mass of the vertices p1p2p3. show that this point is the center of mass of the tetrahedron vertices, and that its location is independent of the order in which the vertices are taken. (b) two edges of the tetrahed

solution:

a)

M₃= P₁+P₂+P₃/3

Now the let point A be center of mass tetrahedran then,

Acm = P₁+P₂+P₃+P₄/4

4Acm =P₁+P₂+P₃+P₄

P₁+P₂+P₃= 3M₃

4Acm = 3M₃ + p₄   1

Let a point o be the point that devide line  

3M₃p₄ into radio 1:3

i.e M₃o/o P₄  = 1/3

so, center of mass of this point will be,

0 = xcm = 3M₃o + o P₄ /3+1

40 = 3M₃o + o P₄   2

From 1 and 2 ,

We can see that o = Acm

Hence, we have considered arbitrary points.

xcm = P₁+P₂+P₃+P₄/4

now, the mid point of P₁P₂ = P₁+P₂/2

         x P₁P₂ = P₁+P₂/2

         x P₃P₄ = x P₃ + P₄/2

now center of mass of  x P₁P₂ & x P₃P₄  

  = XP₁P₂P₃P₄ = x P₁P₂ + x P₃P₄/2

XP₁P₂P₃P₄  =  x P₁+P₂+ P₃+P₄ = Xcm

Hence it is proved that line joining midpoint passes through c.m

answer: $34.31

step-by-step explanation:

$11.99 x 3 = $35.97

$35.97 - 10% = $32.37

$32.37 + 6% = $34.31