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yeah if you have done a slightly more advanced AoPS before, you should be perfectly fine.

I took it today actually

Just like last year, my subconscious mind just decided to check over the work that I did. The questions were fairly easy for me, but because I let myself drift off, I had to guess the last few. Be sure to go as fast AS POSSIBLE without pushing yourself to the point of almost guessing.

idrk what the AMC8 might have on it...since i haven't done the AMC's in like a thousand years :') 
but what i can tell you is that you should look at the past year's problems or do some problems on like AOPS or something just to warm up your brain for tomorrow! :D 

you will do awesome don't worry 

The center of the hyperbola  is  (0,0)  = (h, k)

 c = the dist ance form the center to either focal point  = 100

c^2 =  100^2   =  10000

The differences  from the receiver to the transmitters  =  2a

So     

2a =  180

a   =  90

a^2 = 8100

And 

b^2  = c^2 - a^2

b^2  = 10000 - 8100

b^2  =1900

The form is

(x - h)^2              (y - k)^2

_______   -          ______    =      1

a^2                          b^2

(x - 0)^2               (y - 0)^2

_______   -        _______  =    1

 10000                  1900

   x^2                   y^2

______       -    _____   =    1

 10000               1900

See the graph here : https://www.desmos.com/calculator/uebrg1j0dy

LOL ok i was gonna do it but now im scared

I think this might be

f(x) = 7^x     and    g(x)  =     7^{x+ 2}   - 4

If so....the transformation is

Horizontally 2 units to the left

Vertically 4 units down

SO I didn;t record my answers- just circled the phamplet...

T-T

Last time I didn't get an answer sheet...

T-T

AB =  9√2

BC = 10√2

CA  = 11√2

The  semiperimeter  of  the triangle   =  30√2 / 2 =  15√2

Using Heron's formula.....the area A  is given by

√[ 15√2  (15√2 - 9√2) (15√2 - 10√2) ( 15√2 - 11√2)  ]    =

√[ 15√2  (6√2) ( 5√2) (4√2)  ]   =

√ [1800 * 4]  =

√(3600 * 2 )  =

60√2

Since BC  = 10√2....the altitude from A  can be found as

60√2 =  (1/2) (10√2) (Altitude)

60  = (1/2)(10) (Altitude )

60  = 5 (Altitude)

12  = Altitude from A to BC

Note that angles EBA  and ECD   are  equal  right angles

And angle BEA  = angle CED

So triangle  EBA  is similar to triangle ECD

So

BE / BA  = CE / DC

BA  = 12

Let EB =   20 - x

Let CE  = x

DC  = (12 + 9 + 9)  - 27  = 3

So  we have that

[20 - x] / 12 =  x / 3              cross-multiply

3[20 - x]  =  12x              simplify

60 - 3x  = 12x                rearrange as

60 = 15x       divide both sides by  15

x = 4

She should aim 4 ft to the right of  "C"

EDIT TO CORRECT A PREVIOUS MISTAKE.....THANKS TO EP FOR CATCHING THIS    !!!!

Please do not post soluitons to this problem!  This is a homework problem.

If you need assistance, please use the resources of the online class.

So this is a RIGHT triangle  ('bounded by the coordinte axes")

 y intercept is where x = 0    y = 6

x intercept is where y= 0      x = 3   

so the base is 3 and the height is 6   and it is a right triangle

AREA = 1/2  base * height         Can you take it from here?  

The graph crosses the y = - 1 line      FOUR times

I have reduced the similar triangles it to a simple ratio problem....you should be able to solve it:

1/4 = CE/EB      and   CE + EB = 20

so out of FIVE parts   (1+4)    CE is ONE of them  and EB is FOUR of them

20 divided by 5 parts = 4 per part

   so CE = 4    EB = 16         CE/EB = 4/16 = 1/4       DO you see how to do ratios nw?

Skip the problems you do not have a clue how to solve.....

   Answer all of the easy ones first, then go back to the more difficult ones...

      Double check your work....check to see if your answer is 'reasonable'    ....

          If they do not count off for incorrect answers, make SURE all of the questions have an answer/guess before time is up !

not gunna lie, but why don't you solve your own problems. the questions you post are kind of... skeptical....

do you ACTUALLY have confusion on them?

tips:

take your time

if there is a problem you can't solve, skip it or give up and CHECK your work.

Similar triangles....

3 is to 12   as   CE is to EB       AND    CE+ EB =20

3/12 = CE/EB   

1/4  = CE/ EB              Want me to finish it or can you try from here?  Do you remember how to solve ratios?

Im taking it friday, bro my heart is pounding

Is that right?

Can you keep the radicals and simplify it from there?

Is this supposed to be   - (3^x) - 5  ??     Appears to be.....

Well, obviously, she shoud aim at point E.....   haha   jk

Distance D = (12+9+9)  -  27  = 3

   The two angles at E are equal    (angle of incidence = angle of reflection, generally) 

       the large and the smaller triangle are similar triangles.....

           Does all of that give ya a clue?