Let vk be vectors, and suppose that a point mass of mk is located at the tip of each vector. the center of mass for this set of point masses is equal to: v = [(m1v1 ++ mkvk)/m] where m = m1 ++ mk. determine how to divide a total mass of 11 kg among the vectors u1 = (−1, 3), u2 = (3, −2), and u3 = (5, 2) so that the center of mass is (21/11, 6/11).

m1 = 4

m2 = 5

m3 = 2

Step-by-step explanation:

given (21/11, 6/11) = m1 (-1/3) + m2 (3, -2) + m3 (5, 2)

= (-m1 + 3m2 + 5m3) / 11 = 21/11

= (3m1 + (-2)m2 + 2m3) / 11 = 6/11

so that m1 + m2 +m3 = 11

-m1 + 3m2 + 5m3 = 21

3m1 - 2m2 + 2m3 = 6

from this, we get the augmented matrix as

\left[\begin{array}{cccc}-1&1&1&11\\-1&3&5&21\\3&-2&2&6\end{array}\right]

= \left[\begin{array{cccc}-1&1&1&11\\0&4&6&32\\0&-5&-1&-27\end{array}\right]  \left \{ {{R2=R2 + R1} \atop {R3=R3 -3R1 }]} \right.

= \left[\begin{array}{cccc}-1&1&1&11\\0&1&3/2&8\\0&-5&-1&-27\end{array}\right]

= \left[\begin{array}{cccc}-1&1&1&11\\0&1&3/2&8\\0&0&13/2&13\end{array}\right]

(R3 = R3 + 5R2)

this gives m1 + m2 + m3 = 11

m2 + 3/2 m3 = 8

13/2 m3 = 8

13/2 m3 = 13

m3 = 2

m2 = 8 -3/2 (2) = 5

= m1 = 11- 5 - 2 = 4

this gives

m1 = 4

m2 = 5

m3 = 2

1/120

Step-by-step explanation:

There are 5! = 120 ways to place these disks, but only one of them will have them alphabetical by artist. So, if you randomly choose one of these ways, there will be a 1/120 chance you placed them alphabetically by artist.