CPhill

Let N be the number of junior high students. Then 100 + N = the number of senior high students.

And we are told that :

N / [100 + N ] = 7/ 9    cross multiply

9N = [100 + N] * 7   simplify

9N = 700 + 7N    subtract 7N  from both sides

2N = 700    divide both sides by 2

N = 350   and this is the number of junior high students....the number of senior high students is 450.....and their ratio is 350 / 450  = 7 /9

If Janene runs 1 mile in 9 minutes, each minute, she runs (1/9) of a mile.

And if Emiliy runs 1 mile in 8.25 minutes, then each minute she runs (1/8.25) = (4/33) of a mile

Then, in each minute, the amount of ground that Emily "makes up" on Janene = (4/33) - (1/9) = 1/99 mi

And, if Janene runs for 16 minutes before Emily arrives, she is 16(1/9) = 16/9 miles ahead of Emily when Emily starts.

So.....the time it will take Emily to catch Janene is given by (1/99)M = 16/9  where M is the time in minutes.

Multiply both sides by 99  and we have  M = (99/9)*16 = 11 * 16  = 176 minutes  for Emily to catch Janene

Let N be the number of seventh-graders.....then, 1.5N = the number of eigth--graders....so we have

[.06(1.5)N + .03(N)] / [N + 1.5N]  =

[.09N + .03N] / [ 2.5N]  =

[.12N] / [2.5N]  =     ["cancel" the "N's" ]

.12/2.5 =  .048  =  4.8% 

There are  (2014 - 1) - 1 = 2012 offices between them

Each minute they close the distance by (25 + 36) = 61 offices.  So....the number of minutes required to close the whole distance is just  2012/61 = about 33 minutes

So   1 + 25(33)  = 826    ....and office 826 is approximately where they meet

Proof [ from the other end]..... 2014 - 36(33) =  826

There were $10 in the seventh pile....since each pile had the same amount, there must have been 7 * $10  = $70  in the plies. But.....this is (1/2) of the money he started with.....so....he must have started witrh $140.

Let N be the number...I assume you mean to double the number and take the product with it's triple???  ...If so, we have....... 

(2N)(3N) = (1/2)N        

6N^2 = (1/2)N 

12N^2  = N   .... subtract N from both sides

12N^2 - N = 0     factor

N(12N - 1)  = 0    so...N = 0  or N = 1/12

We can break this down as :

(8.73 / 9)  * (10^8 / 10^3)  =

.97  * 10^5   =

9.7 * 10^4    (in scientific notation)

(1 + 1) - (1 * 1) / 1  =

2 - (1) / 1  =

2 - 1  =   1

Since we have a SSA situation, we could have more than one triangle ...Using the Law of Sines,  we have......

sin B / 2 = sin 51 /  5   ...... so......

sin^-1 ((2/5)* sin 51)  = about 18.11º  = angle B

And......subtracting this from 180 we have (180 - 18.11)° =  161.89º.......but....this angle added to the known angle of 51° is greater than 180°, so only one triangle is possible.......as Melody found......

The angles are

A = 51  B = 18.11  C = 110.89

And side c  =  c / sin 110.89 =  5 /sin 51    ...so....   c = 5(sin110.89) / sin 51   = about 6.01

So the sides are  a = 5,  b = 2 and c = 6.01

1 in = 2.54 cm...so.....

1cm/25km  = 2.54cm / ?? km   cross-multiply

(??) (1)  = (25)(2.54)  = 63.5 km