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If angle X is 120.... the other two angles sum to only 60 degrees..... using a calculator   tan^-1 (3) =~ 71 degrees

   so this is not possible

Here is my take on this one:

Try again:

Re-take:

How many possible combos are there?

10 c 3 * 10 C1 = 1200 possible combos        you have a 1 in 1200 chance of doing this

Matching two AND Superball

10 c 2 = 45

10 c 1 = 10     (for Superball)           45 * 10 = 450  you have a 450 out of 1200  chance of doing this

Matching two 

45        out of 1200

matching three

120  out of 1200

matching superball

10  out of 1200

total prob of winning is then      (1 + 450 + 45 + 120 + 10)/1200 = 626/1200 = 313/600

Hmmmmm.....getting closer perhaps!

Thanks so much for replying! But this isnt correct

Where the two intersect , y will be equal....so equate the two equations

2x^2 - 7x +1 = 8x^2 - 25x+13   solve for x

6x^2 -18x+12 = 0       Use Quadratic Formula to find x = 1  and 2     

                                          use these values of 'x' in either of the equations to calculate the corresponding 'y' values

6*r *r = b

a*r = b

a*r = 6 * r * r

a = 6 * r

a * r * r = 1296

6*r *r * r = 1296

r = 6

6 * r * r = b

b = 216

then a = 36

Here is my take on this one:

Matching two AND Superball

10 c 2 = 45

10 c 1 = 10     (for Superball)           45 * 10 = 450  possibles for matching TWO numbers correct and Super

Matching THREE and superball

10C3 = 120

10C1 = 10      (for Superball)            120*10 = 1200  possibles for matching THREE numbers and Super

Matching two 

45

matching three

120

matching superball

10

YOU get ONE ticket     1 out of 450   + 1 out of 1200 + 1 out of 45 + 1 out of 120 + 1 out of 10

                                         1/450                1/1200            1/45               1/120              1/10       = 481/3600

Different answer....I do NOT know if it is correct!

YOU are welcome !          Thanx for the 'thanks' !

sixty eight      x = 1  6   11   16   21    26 ........ 336

                     y =  201 ........0

h(2x^2)    means the g(x) function must fit within the -8 , 8 interval

domain of h only goes to 8

2x^2 = 8

x^2 = 4

x = +-2                  

when h goes to -8

2x^2 = -8

x^2 = -4

x= +- 2i    not real

So I think the domain of g(x) is     [-2,2]       I have not seen anyone else answer these type of questions....I need to learn the correct answer too!

12x^2 + 6x + 9 = 0

use quadratic fromula to find the roots are    -1/4 +-  i sqrt (11) / 4

    (not positive of this answer)

a=3 b=2

can u explain?

It has two imaginary roots....no REAL roots.

ok thanks

First, simplify the equations

2.2x - 3y = -3.2

.4x +y = 8.8         Multiply this equation by 3 and add to the top equation

3.4x = 23.2             Now you can solve for x......use this value of x in one of the equations to calculate y

f(3) = 5   and  g(3) = 4      so s(3) = 9

11111111111

1/30, I think

Thanks for all the help!

@ElectricPavlov Thank you for working out this question!

I'd like someone to answer this too.    

I have not answered this but I would start by substituting   

1) sin^2x - 1 = cos^2x

2) then I'd tidy the equation

3) substitute y for sinx

4) solve as a quadratic equation

5) solve sinx= your solutions.

I have not seen notation like this but I am sure it must mean find  f(3)     Just by substituting 3 into the equation.

Then put your answer to the power of 4

\([f(3)]^4\)

I'm not sure what you mean by coordinates...