Prove the part of the theorem which lets w be any solution of Ax = b, and defines vh = w - p. Show that v_h is a solution of Ax = 0. This shows that every solution of Ax = b has the form w = p + v_h, with p a particular solution of Ax = b and v_h a solution of Ax = 0. Let w and p be solutions of Ax = b. Substitute for v_h from the equation w = p+ v_h.

It is proved that is a solution of Ax=b.

Step-by-step explanation:

As given let w and p be a solution of Ax=b. Then,

and

Defining,

To show is a solution of Ax=b, we have to show . Then,

       (By putting value of )

      (By distributive law)

                    (Since w and p are solution of Ax=b)

Hence become a solution of the homogenous system Ax=b.

Since w and p are solution of Ax=b, and any linear combination of the solutions again a solution of Ax=b, this shows that every solution of Ax=b is of the form  where is a solution of Ax=0.

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts

just take the area of the square and then the area of the circle then subtract the circle area from the squares area

step-by-step explanation: