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f(t) = 200 000 (1 + .02)^t           Where t is in years

put in t = 3    and calculate the value in 3 years

P = 100 *  (1+ .02)^t        millions          Where t = # years after 2014

3 9 27 81    3^5 = 243

3^6 = 243*3 = 729

So the units digiit appears to cycle  in this roup of 4:      3   9   7   1       3  9  7  1  

2086/4 = 2086/4 = 521  R 2          so units digit will be   9

(1 - .12)^5 = .88^5 = 0.5277319168     this is the ending percentage of original 100%

    it goes down to the ending percentage

1 - .5277 = -0.5277319168+1 = 0.4722680832        it is down 47.23 %

town A population  47750 - 125m + 200m

town B population  50000 - 175m

Equate the towns' populations , then solve for 'm'

47750 - 125m+200m = 50000-175m               Answer A

 Now solve for 'm'......

Find slope, m, between the two given points

(y1-y2)/ (x1-x2) = (7-5)/(-2 - 3) = -2/5

Perpindicular slope m of the new line = -1/m = 5/2

Line through 2,28 with this new slope =

(y-28) = 5/2 ( x-2)

(3/5)^3+8(1.15)^2+1.08^4 = 12.15648896     

                                           by pasting into Web2.0 calculator

Area of triangle = 1/2 b*h =   1/2 (11)(5)     cm²

Area of Rectangle below triangle =  5 * 11   cm²

Area of rectangular base = 15 * 3  cm²

Add these toogher.... then multiply by the depth (shown as 7 cm) to get your answer in cm³

The link does not work.

It looks like it is the link to a test, that has protection on it. Good for them.

Unfortunately, this also does not work as your input had equal signs instead of modular congruence signs. I don't believe you can input modular congruence signs into Wolfram Alpha. 

Thanks for trying to help me though!

Look at many, many solutions here:

https://www.wolframalpha.com/input/?i=%282ab%2Bbc%2Bca%29+mod+13+%3D0%2C++%28ab%2B2bc%2Bca%29+mod+13%3D+6abc%2C++%28ab%2Bbc%2B2ca%29++mod+13+%3D+8abc%2C+solve+for+a%2Cb%2Cc

Sorry, but it says that is incorrect. 

Thank you for trying to help out though!

Also, if someone wants to post a solution to this, it'd be best if they provided an explanation, because I'd like to learn from it. Thanks!

????

Look for the same solution as under "Mr & Mrs Johnson" some 10 questions below this one and follow exactly the same method.

Let the price of child's ticket = C

Let the price of an adult's ticket = A

1C + 2A = 16.50.......................(1)

8C + 17A = 138.00..................(2), solve for C, A

a - C = 4.50 dollars - child's ticket   and    A = 6.00 dollars -  adult's ticket

b - [12 * $4.50] + [3 * $6] =$54 + $18 = $72.00 - Total cost for 3 adults and 12 children.

a=1;b=0;c=0;p=0; cycle:d=a*100+b*10+c;if(LCM(8*d, 10^10)==4*LCM(d, 10^10), goto loop, goto next); loop:printd," ",;p=p+1; next:c++;if(c<10, goto cycle, 0);c=0;b++;if(b<10, goto cycle, 0);c=0;b=0;a++;if(a<10, goto cycle,0);print"Total = ",p

OUTPUT: m = 512

[x + 24] + [x + 123] = 141

x = -3

Angle JGF =[-3 + 24] = 21 degrees, or "C".

Angle JGF = 32 degrees.

a + b + c leaves a remainder of 8 when divided by 13.

[BCD] = 100/3.

16/26  /  1/4   =

16/26  *  4/1 = 64/26       simplify if you want 

A-C-D = 7:11 =  7 + 11 = 18 'units'

A-B-D = 1:5   =  3:15    = 18 'units'

AB : BC : CD   = 3    :   ?    :  11          ? = 18-11-3 = 4

3:4:11   

x=0;p=0; y=0;z=0;n=x^2;if(n==y^2 + z^2 and y+z>x and y*z/2==2*(x+y+z), goto loop, goto next);loop:p=p+1;printp," =",x,y,z;next:x++;if(x<200, goto4,0);x=0;y++;if(y<200, goto4, 0);x=0;y=0;z++;if(z<200, goto4,0)

OUTPUT:

1  = 41 40 9
2  = 26 24 10
3  = 20 16 12

Total areas =[40 * 9 / 2] + [24 * 10/2] + [16 * 12/2] =180 + 120 + 96 = 396 units^2.

Surprised that no one else noticed this before me! (Or maybe they did, just didn´t post)

a=0;b=0;c=0;d=1;p=0; cycle:n=a*1000+b*100+c*10+d;if(482 % n==74 and 992 % n==74 and 1094 % n==74, goto loop, goto next); loop:printn," ",;p=p+1; next:d++;if(d<10, goto cycle, 0);d=1;c++;if(c<10, goto cycle, 0);d=1;c=0;b++;if(b<10, goto cycle,0);b=0;c=0;d=1;a++;if(a<10, goto cycle,0);print"Total = ",p

D = 102