CPhill

Midpoint  of   (0,0)  and (2,3)   =   (  [ 2 + 0 ] / 2 , [ 0 + 3 ]  / 2 )  = ( 1, 3/2)

Midpoint of (5,0) and ( 6,3)  =    (   [ 5 + 6 ] / 2 , [0 + 3 ] / 2 )   =  (11/2, 3/2)

The slope of the line containing these two  midpoints =  0   [ since both  y coordinates are the same ]

D(t)   =  14t   if   0 ≤ t ≤ 2 ,

            √ [  (14t )^2  + ( 10t)^2 ]  =  t√ [296 ]  = 2t√74    if  t > 2

Basically...if I read this correctly....we want the surface area of the "front" of this object

The area of the top triangle  is   (1/2) base of triangle * height of triangle =   (1/2) (6)(7)  = 21 m^2

And the area of the rectangle below this  =  base * height =  6 * 2   = 12 mm^2

So...the area specified   is  [ 21  + 12 ]   =  33 mm^2

OK....good deal   !!!!

Yep....this makes it easier when we can reduce the size of the coefficients.....

20 )

x  - 2y   =  18      rearrange as    x  = 18  + 2y    (1)

-6x - 4y  = 4        (2)

Put  (1)  into (2)  for  x

-6(18 + 2y) - 4y   = 4       simplify

-108 - 12y  - 4y  = 4

-108 - 16y   = 4      add 108 to  both sides

-16y  = 112    divide both sides by  -16

y  = -7

And using (1)....   x  = 18  + 2(-7)   =  18  - 14    =  4

19)

21x  - 3y   = -42     divide through by  3  ⇒    7x  - y  = -14   (1)

y  = 7x + 14     (2)

Sub  (2)  into (1)     for "y"  and we have

7x -  [ 7x + 14]  = -14    simplify

-14   = -14       this is always true...whenever this happens....we have infinite solutions...!!!!

5 0 0     =  3 ways   -  any of the 5 candies can  be given to one person

4 1 0  =   6 ways  = one candy can be given to any one person (3 ways) and the other 4 can be given to one of the other two (2 ways)  ...so...3 x 2  =  6

3 2 0  =   6 ways -  any of the three candies can br given to one person (3 ways)  and the other two can be given to one of the other two (2 ways)...so....3 x  2  = 6

3 1 1  =   3 ways    -  any of the 3 candies can be given to any one person (3 ways) ...since the other two candies are indistinguishable...we can distribute them to the other two people in any manner

2 2 1   =   3 ways  -  any 1 of the candies can be given to any of the three people....the other 4 can be split equally between the remaining two people

So...  3 + 6 + 6 + 3 + 3  = 21 ways

Good....!!!!

2)   294n^3  + 343n^2  + 126n  + 147

This one isn't as obvious....

Take out the GCF between 294n^3  and 343n^2   =  49n^2

Take out the GCF between  126  and 147  = 21

So we have

49n^2 [ 6n + 7 ]  + 21 [ 6n + 7 ]      

[6n + 7 ] [ 49n^2 + 21]      note....in the second factor...we can take out a  "7"  and we have

[6n + 7]  [ 7n^2 + 3] *7  

7 [ 6n + 7 ] [7n^2 + 3 ]