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Haha.

tan A  =  y / x  = -5/-12  =  5/12

csc A =   r / y

r =  sqrt (5^2 + 12^2)  =  sqrt (25 + 144)  = sqrt (169)  = 13

Since  y is negative in Q3....then csc A =   13/-5  =  -13/5

YAY!!! 
                                                                      
HAPPY BIRTHDAY HECTICTAR!! 

We have the following  equation

M  =  ae^(kt)

Where M is the mass after t hours

a  = the beginning mass

k  is the growth constant

We need to find k   thusly

5  = 2e^(k 30)       divide both sides by  2

2.5  =  e^(30k)       take the Ln  on both sides

Ln 2.5  =  Ln e^(30k)         and we can write

Ln 2.5  =  30k   * Ln e           [Ln e = 1....so we can ignore this ]

Ln 2.5 =  30 k       divide both sides by  30

Ln (2.5) /30 = k

After 25 hours we have that

M  = 2e^(Ln (2.5)/30* 25)   ≈  4.29  mg

OHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH! I SEEEEEEEEEEEEEE

*mind blown*

I am one year older than you SVS lol

Not too bad.....

We need to figure this

    3x^4  + 2x  + 1

+     h(x)

______________

  5x^2  - 6x - 1

Note that the first term of h(x)  must have been     -3x^4   since there is no fourth power term in the final result

The second term must have been  5x^2

The third term must have been    2x + ax = - 6x  ....so  ax  = -8x

And the last term must have been -1 + k = -1    ⇒  k  = 0

So  h(x)  must have been     -3x^4  + 5x^2  - 8x  

Check :

3x^4  +  0x^2   +  2x   -  1

-3x^4 +  5x^2   -  8x    + 0

_____________________

             5x^2   -  6x   - 1

Note that  (x - 1)   has a degree of 1

So......we must  multiply this by a polynomial of degree 3  to  get a polynomial of degree 4

So.... "f"  has a degree of 3

Note that when x = 0....f(x)  =   1

As x grows larger and larger.....f(x)   tends to approach y =  0   (the x axis)

So....the range is   (0, 1]

See the graph here :   https://www.desmos.com/calculator/5s4fsm8zqo

( a + b + c) (a + b - c)  = 3ab

[ (a + b) + c ]  [(a + b) - c ]  = 3ab

(a + b)^2  - c^2  = 3ab

a^2 + 2ab + b^2 - c^2 = 3ab

a^2 + b^2   =  ab + c^2     (1)

Using the Law of Cosines

c^2  = [a^2 + b^2] - 2(a)(b) cos C      (2)

Sub (1)  into (2)    and we have that

c^2  = [ ab + c^2]  - (2ab) cos C

c^2  = ab + c^2 - (2ab) cos C       subtract c^2 from both sides

0   =  ab - 2ab cos C        subtract ab from both sides

-ab  = - 2ab cos C            divide both sides by -2ab

1/2  =  cos C

arccos (1/2)  = C  =  60°

Thanks to asinus for the hint of applying the Law of Cosines  !!!

Thanks, heureka!!!

An alternative approach :

Note that triangles EFC  and BDE  are similar......which means that

EF/FC  =  BD/DE

5 / 7  =  BD / 5      multiply both sides by  5

25/7  = BD

Happy B-day(I'm old as your digits reversed around lol)

you add a g to it 

to make it gone!

Just multiply each wage rate by  2000....that will give you the annual income for each profession...

HAPPY 21ST BIRTHDAY HECTICTAR !

HERE'S  THE  KEY TO YOUR  FUTURE!      

May it be an extremely happy and sucessful one.      

Yes, that's what I'm talking about. Several users on this website have made it their job to spread hate, and that is not what we want.

Thank you so so much!!!!

Happy b-day Hectictar, what's wrong with Alabama?

Happy 21st birthday, you are in the prime of your life, with so much ahead of you!

Try using fact that there's a right triangle between ABA'. You're hunting for A' coordinates based off A or B coordinates. 

Reflecting across origin is normal, disregard other dudes rants. 

thanks guest and melody! <3

We need to solve this system  .....it's a little tedious, so I used WolframAlpha to solve it....however....I could go through it if you wanted.....

(-28- h)^2 + (784 - k)^2  = r^2

(-2 - h)^2 + ( 4 - k)^2 = r^2

(13 - h)^2 + (169 - k)^2  = r^2

The center of the circle is at  (h, k)  = (-2475, 312)   and r^2  = 6210593

The equation of the circle is  ( x+ 2475)^2 + (y - 312)^2  = 6210593

Here's a graph : https://www.desmos.com/calculator/7uay3uslee

We can find the other  "x" intersection point thusly  ...sub x^2  for y in the equation of the circle and we have that

(x^2 + 2475)^2 + (x^2 - 312)^2  =  6210593

I also used WolramAlpha to solve this....again....I can go through it if you want

The last intersection point x value  = 17

So  y  = x^2  = 17^2  = 289

So the last intersection point  is (17, 289)

The focal distance, a, is given by

4ay = x^2

4(1/4)y = x^2

y = x^2

So  a  = 1/4

The vertex of the parabola = (0, 0)

And the focal point  =  (0, 0 + a)  =  ( 0 , 0 + 1/4)  =   (0, 1/4)  = (0, .25)

So....now all youhave to do is to find each of the distances  from the four intersection points to the poiint (0, .25)

Can you do this????  [ Note.....some technology (like WolframAlpha) might help  with these calculations ]

OK, man....no prob   !!!