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OK....good deal   !!!

5x - 17  = -1x  + 7       add 17, 1x to both sides

6x  = 24   divide both sides by 6

x  = 4

QR  splits the sides PS and PT proportionally.....

So

PQ / PR  =  PS / PT

PQ / PR  = [ PQ + QS] / PT

8/ PR  =  [ 8+ 6 ]  / 10.5

8 /PR  = 14/10.5

PR / 8  =  10.5 / 14

PR  =  8 (10.5) /14  =  6

Thank you so much! I checked my work and corrected it. 

f(0)  =  0^(0+1) * (0 + 2)^(0 + 3)  =  0^1  * 2^3  =   0

f(-1) =  (-1)^(-1 + 1) * (-1 + 2)^(-1 + 3)  = (-1)^0 * (1)^(2)  = 1 * 1  = 1

f(-2)  =  (-2)^(-2 + 1) * (-2 +2)^(-2 + 3)  =  (-2)^(-1)* (0)^(1)  =  0

f(-3)  =  (-3)^(-3 + 1) * (-3 + 2)^(-3 + 3)  = (-3)^(-2) * (-1)^0  =  (1/9)*(1)  = 1/9

The sum of these  =   1 1/9    =   10/9

a *  b   =  a  + a/b   =   12 + 12/3   =  12 + 4   =   16

f(x)  =  g (g(x))

g (g (x))    means that we sub g(x) into itself....so we have that

g(g(x) ) = 2 (2x - 1) - 1  =

4x - 2 - 1  =

4x - 3

So....since this is  f(x)

Then f(3)  =

4(3)  - 3  =

12 - 3  =

9

3.  

t/ 4  = [ t + 1 ]  / 6      cross-multiply

6t  =  4 [ t + 1 ]

6t  =  4t + 4       subtract  4t from both sides

2t  =  4   divide both sides by 2

t  = 2

t + 1  =  3

To find the remaining missing side in the first figure we have that

Missing side / t  =   10/ [ t + 1 ]

Missing side / 2  =  10/ 3    

Missing side  =   2 * 10  / 3  =   20/3

2. 

6/8  =  [ w - 5 ] / w     cross-multiply

6w  =  8 [ w - 5 ]

6w  =  8w - 40       subtract 8w  from both sides

-2w  = -40        divide both sides by  -2

w  =  20

w - 5  =  15

To find one missing side  in the second figure that would be the couter-part to the side of 9 in the first figure we have that

Missing side/ w  =   9/8

Missing side 20  =  9/8      multiply both sides by 20

Missing side  =  20 * 9 / 8   =  180 / 8  = 22.5

And to find the last missing side of the second figure that is a counter-part to the side of 14 in the first figure we have that

Missing side / 22.5  = 14/ 9

Missing side  =  22.5 * 14 / 9   =    35

1.   We have that

6/9  =  t / [ t + 2 ]    cross-multiply

6 [ t + 2]  = 9t

6t + 12  =  9t    subtract 6t from both sides

12  =  3t          divide both sides by 3

4  =  t

t + 2  =  6

We can find the missing side S  of the figure on the right, thusly

11/9  = S /[ t + 2]

11/9  =  S  / 6      cross-multiply

11* 6  =  9S

66 =  9S       divide both sides by   9

66/9  =  S   =  22/3

Since MN is parallel to PQ, then triangle MNO is similar to triangle QPO

Then 

OM / OQ  =  ON/ OP

9/6  =  ON/ OP

3/2  = ON/OP

Then  PN  is divided into  5 equal parts  and OP  is  2 of them

So

OP  = (2/5)PN   = (2/5) 25    =   10

I fixed the link so it must work now. Can you help me?

The image(s) doesn't/don't display for me when I pull up the link.....

Suppose y varies directly with​ x, and y equals=20 when x equals = -5. Find x when y equals = 24. 

We  have

y  =  k x       

And we know that

20  = k (-5)      divide both sides by -5

-4  = k

So.......when y = 24, we have

24  =  - 4 x       divide both sides by -4

-6   =  x

And Nani means What?

CBF and DEF are similar.

DEF's side lengths are half the length of CBF's. Therefore, if EF's length is 9, then BF's length is double that. Can you solve it?

Please don't post AoPS questions (or at least spam them).

Use the Message Board, or ask a friend! Only if you really need the answer, ask for hints.

PQR and PST are similar. PQ + QS = 14, and PQ = 8, so PQ is 8/14, or 4/7, the length of PS.

Same proportion for PT. PT is 10.5, and PR is 4/7 of that length. Can you find out from here?

If triangle OPQ's side lengths are 2/3 of OMN, then PN can be split so that OP is 2/3 of ON.

Basically, OP is 2/3 of ON. Together, they make up 25 (PN).

Can you solve from here?

We can multiply each by 5, like saying how many 0.4*5 = 2 inches in 8*5 = 40 inches?

That's easier to comprehend.

Can you solve now? How many 2 inches are in 40 inches?

\(-5 \cdot -2 = 10 \text{ is the largest value for $b$}\)

Here is basically Melody's solution but rearranged a bit to make more intutive sense.

\(\text{First off there are $N=\dbinom{16}{4}\dbinom{12}{4}\dbinom{8}{4} =63063000 $ possible hands}\\ \text{First person gets their cards. There are 4 1's to choose from and $\dbinom{12}{3}$ ways to choose the other 3 cards}\\ \text{Second person has 3 1's and $\dbinom{9}{3}$ ways to pick the other cards}\\ \text{Third person has 2 1's and $\dbinom{6}{3}$ other cards}\\ \text{This gets us the total number of hands $n$ where each player has a 1 in their hand}\\ n = 4\dbinom{12}{3}\cdot 3\dbinom{9}{3}\cdot 2\dbinom{6}{3}\\ p = \dfrac n N = \dfrac{\dbinom{12}{3}\dbinom{9}{3}\dbinom{6}{3} 4!}{\dbinom{16}{4}\dbinom{12}{4}\dbinom{8}{4}} = \dfrac{8870400}{63063000 } = \dfrac{64}{455}\)