CPhill

8^x  = 69

log₈ 69  =  x

Call the total number in the bag, N

Call the number  of red marbles, R

Call the number white, W

Call the number blue, B

Call the number green, G

4 reds are chosen has  the same probability as  1 blue  and 3 reds

R  *  (R-1) * (R - 2) *   (R - 3)             B      R          R- 1     R - 2

__   ____       _____   ______  =     ___   _____    _____  ____

 N    N - 1      N - 2        N - 3             N     N - 1      N  - 2   N - 3

Note that the denominators in each case will be the same  in all cases...so...we can ignore these

So...this simplifies to  R - 3  =  B     (1)

4 reds   has the same probability  as  1  white , 1  blue  and 2 reds

So....subbing (1)  in for B

R (R-1) (R - 2) (R - 3)   =  W (R - 3)  R (R -1)

This simplifies to

R - 2  =   W      (2)

4 reds  has the same probability  as one of each color

Subbing  (1) and (2)  in for B, W   we have

R (R - 1) (R - 2) (R - 3)   =  R (R - 2) (R -3) G

This simplifies to

R - 1   =  G

Since there has to be at least one of each color....let the number of Red  = 4

Let the number of Green  = R - 1  =  3

Let the number of White  = R  - 2  =  2

Let the number of Blue  = R - 3  =  1

So.....the fewest number possible is 10

Here you go  :

(x + y)^2  =  45    ⇒  x^2 + 2xy + y^2   =  45     (1)

xy  = 10    (2)

Sub (2)  into (1)

x^2 + 2(10) + y^2  = 45

x^2 + y^2  + 20  = 45       subtract 20 from both sides

x^2 + y^2  =  25 

Finally....

(x  - y)^2  = x^2 - 2xy + y^2  = 

(x^2 + y^2) - 2xy  =  

(25)- 2(10)

25 - 20  =

5

Let  a  be the first number

a + d   = the second

a + 2d  = the third

If it's this :

a (a + d)  =  5

(a + d) + (a + 2d)  = 10

WolframAlpha  gives these solutions :

https://www.wolframalpha.com/input/?i=a+(a+%2B+d)%C2%A0+%3D%C2%A0+5,++(a+%2B+d)+%2B+(a+%2B+2d)%C2%A0+%3D+10

If  it's this

a(a + d)  = 10

(a + d) + (a + 2d)  = 5

https://www.wolframalpha.com/input/?i=a+(a+%2B+d)%C2%A0+%3D%C2%A0+10,++(a+%2B+d)+%2B+(a+%2B+2d)%C2%A0+%3D+5

Either way....it's U-G-L-Y  !!!!

x^2  =  16    take both roots

x  = ±√16

x  =  -4, 4

So...  the sum  = 0

x^2 - 5x + 4  =  0     factor as

(x - 4) ( x - 1)  = 0

Set  both factors to 0 and solve for x

x  - 4  = 0          x  - 1  = 0

x  = 4                   x  = 1

So

x  = 1, 4

The  unknown distance, D, at the top of the triangle on the right can be found as follows

D/ 14.4  =  14.4 /10.8      cross-multiply

10.8 * D   =  (14.4)^2 

10.8 * D  =  207.36    divide both sides by 10.8

D  = 19.2

So....the total distance is  10.8  + 19.2   =   30 m

[ In a right triangle like this, the  "14.8"  is a geometric mean to the bases  ]

Note that  we added   -5  to the x coordinate  of  P  to get the x coordinate of P'

And we added 10  to the y coordinate of P  to get the y coordinate of P'

So  translating (7,2)   using the same rules  becomes  :

( 7 + -5 , 2 + 10)  =  ( 2, 12)

We have the following relationship

 ΔPQR  similar to  ΔQTR  similar to  ΔPTQ

Note  that

PT            QT

___   =     ___

TQ            TR

So... "A"  is correct

We  have  that

16             x

___   =    ___      cross-multiply

  x             9

16 * 9  =  x^2

144 = x^2     take the positive root of both sides

12  =  x