CPhill

The answer is 6.....to see this....let C = (0, 0)  and let D  = (9, 0)

[These are arbitrary...we can pick any two points that are 9 units apart ]

Then CD  = 9

The first transformation doesn't do anything to the length....it just shifts  C  to the right 3 units and down 2 units  ...so.... C'  =  (3, - 2)

And it does the same to D.....so we have D' =  (12, - 2)

The second transformation  is applied to both of these points.....so 

C''  = (2/3)(3, - 2)  =   (2, - 4/3)

And D''  = (2/3)(12, - 2)  =  ( 8, - 4/3)

And the distance between these points is

sqrt [  (8 - 2)^2 + ( -4/3  -  - 4/3)^2  ]  =

sqrt [ 6^2  + 0^2 ]  =

sqrt [ 36 ]  =  

6     .....just as we suspected !!!!!!

We can equate the area of the the triangle on both sides of an equation

(1/2)*  (3 + 5) * sqrt [ (x + 2)^2 - 3^2 ]   =  (1/2) (x + 2) * sqrt  [ ( 8^2 - 2^2 ]

8 * sqrt [ x^2 + 4x + 4 - 9]  = (x +2 ) sqrt (60) 

8 * sqrt [ x^2 + 4x - 5]  = (x+ 2) sqrt (60)          square both sides

64 (x^2 + 4x  - 5 ]  =  (x^2 + 4x + 4 ) * 60

16(x^2 + 4x - 5)  = 15(x^2 + 4x + 4)

16x^2 + 64x - 80  =  15x^2 + 60x + 60

x^2 + 4x - 140  = 0       factor

(x - 10) ( x + 14)  = 0

Setting both factors to 0 and solving for x gives that x  = -14 (reject)  or  

x = 10

Here's my best attempt.....

Consider the sets  where one of the Americans wins gold...we have 3 possibilities for the American gold medal

And  of the other 5 sprinters, we want  to choose any of the other 5 non-Americans to receive the silver and any 4 of the non-Americans to recieve the bronze

So...this gives  3 * 5 * 4  = 60 possible ways

And we will have the same number of sets for an American winning silver  and the same number of sets for an American winning bronze

So far... this gives 60 * 3  =  180 ways

To this..we must add the sets where no American wins any medals....we have 5 ways to award the gold medal, 4 ways to award the silver and 3 ways to award the bronze = 5 * 4 * 3 = 60 ways

So  180 + 60  =   240 ways where, at most, one American wins a medal

The focus lies above the directrix....so this parabola turns upward

The vertex  lies midway between these and is given by  ( 4, (2 - 3) / 2 )  = (4, -1/2)

The focal distance,p, is the distance between the vertex and the focus....since the parabola turns upward this is positive and is given by  2 - (-1/2)  =  2 + 1/2  = 5/2

So....we have this form :

4p (y - k)  = (x - h)^2    where  the vertex is (h, k)  = (4, -1/2)  and p = 5/2

Filling in what we know, we have

4 (5/2) (y -  -1/2)  = (x - 4 ) ^2

10 ( y +1/2)  =  ( x - 4 )^2         divide both sides by 10

y +1/2  = (1/10) (x - 4)^2        subtract 1/2 from both sides

y  = (1/10) (x - 4)^2  - 1/2       this is in vertex form

The "npr" button at the bottom left ....

93.

P ( Having a pet  l Having a sibling)  = 

15 students have a sibling...of these 7 have a pet

So....the probability is just 7/15 ≈ 46.7%

94.

Note.......P(A or B)  = P(A) + P(B) - P(A and B)

P(ace  or 2)  = P(ace) + P(2)  - P(ace and 2)   =

                         4/52  +  4/52  -  1/52  =

                                     7 / 52

95.  P (5 or diamond)  = P(5) + P(diamond)  - P(5 and diamond) =

                                        4/52  +  13/52    -         1/52   =

                                              16/52  =

                                               4/13

96.   We have 26 black cards....2 of them are queens....so  2/26  = 1/13

97. We have 4 kings....only one is a diamond....so  1/4

98. Don't know the answer to this one, LF   ...sorry  !!!

Don't worry about it, LF......!!!

90.    We have 40 marbles....and we are drawing with replacement.......

Blue then Red  =

(4/40) * (12/40)  =

(1/10) * (3/ 10)

3 / 100

91.  Two yellows  =  (8/40) * (8/40)  =  (1/5) * (1/5)  =  1/25

92.  Not a green on the first pick...= 24/40  =  3 / 5

3. 

\( \sqrt{5 + x} + \sqrt{20 - x} = 7\)

We probably can  just solve this by inspection....note.....x = 4 seems to work

So

(5 + 4) (20 - 4)   =  

9 * 16 =

144

Don't worry about it, LF....as long as I can see it, the form doesn't matter

We can write the numerator  as  x^(5/2)  y ^(3/2)

And the denominator is x^(2) y ^(3)

So....using   aⁿ / a^m  =  a^n-m   we have

x^(5/2 - 2)  *  y^ (3/2 - 3)   =

x^(5/2- 4/2) * y^(3/2 - 6/2)  =

x^(1/2) * y ( -3/2) ⇒   "B"

Thanks, tertre.....may I add.....the events are dependent......if we were drawing "with replacement," they would be independent....but here, the probability of drawing a diamond on the second draw changes depending upon what card was drawn first