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thank you so much :)

If we  have a double root , the disctiminant =   0      ....so......

b^2  - 4 * 2 * 2  =   0

b^2  -  16  =   0

b^2  =  16      take both roots

b = -4      and  b =  4

30/50  = 3/5

So

3/5   = sqrt  ( y / 50)         square  both sides

9/25   = y / 50        cross-multiply

50  * 9   =  25  y

450  = 25 y

y  =  450  / 25    =    18

We  just  need  to sum the 8th  row entries of Pascal's  triangle  starting with  the  3rd element

This  is

28  +  56  +  70  +  56  + 28  +  8  +  1  =

247  different  committees

g(0)  =   -4  - 3   =   -7

55₆  =  5(6) + 5   =  35₁₀

So....in  base 10, we are  summing  all the  integers  from 1  to  35

This   is        (35 )  ( 36)    / 2   =  630

Converting this  back to base 6

630  / 6     =   105  R  0

105 / 6  =         17  R  3

17 / 6   =            2  R  5

2 / 6  =               0  R  2

Reading the remainders  from bottom to top

2530₆

ok, thanks!

For each of the 10 toppings, you can choose to have it or don't have it, 2 options. 

2^10 = 1024

=^._.^=

This is not a math or right answer, but if a + b = c, maybe x + y = z?

=^._.^=

11 + 997a = 100x

Since 11 + 997a needs to be divisable by 100, the last 2 digits need to be 00. 

11 + 997a = 0 (mod 100)

11 + 97a = 0 (mod 100)

a = 37

11 + 997(37) = 100x

36900 = 100x

x = 369

=^._.^=

10 choices for the first role, 9 for the second, 8 for the third, and 7 for the last. 

10*9*8*7

=^._.^=

The number of ways is 6!/36 = 20.

I found 120, but it is not correct, please HELP.

If it has a height of AD, then 4/5

I'm not sure why they gave that angle A is 90 degrees, maybe I drew it wrong.

Expand A(y - B)(y - C)

A*(y² - (B+C)y + BC)

Ay² - A(B+C)y + ABC

Compare the coefficients with those of  6y² - y - 15

Can you take it from here?

Don't forget  -8 and  \(\pm 64^{1/4}\) (fourth root of 64 is real, but not an integer). 

Also 2 is in the domain of the function.

WEll thought through MPS       

(spelling leaves a bit to be desired though    )

Just use a Venn diagram of 2 overlapping circles.

One circle is dogs the other is cats the overlapping is both.

so 7 goes in the overlap

so how many go in the other two bits?

You work it out.  It really is not hard.

I drew it.

Obviously, that is not exact but I believe it is close enough to answer the question.

Of course you are not meant to do it this way.  I was just having some fun.

Perhaps you can use my pic to help you with the exact geometrical calculations.  

Keep in mind that  180-135 = 45

This happens when $x = 64, x^2 = 64, x^3 = 64, x^4 = 64$ because then the denominator will be $0.$

Thus the answer is $4.$

If you are looking for the values, they are $64, 8, 4, 2.$

Let $\gamma = \frac{1}{a}$ and $\theta = \frac{1}{b}.$

We have $\gamma x + \theta y = 1$

We have $\gamma \cdot 1 + \theta \cdot 11 = 1$

We also have $\gamma \cdot 4 + \theta \cdot 8 = 1$

Multiplying first equation by $4,$ we have $4 \gamma + 44 \theta = 4$

Subtracting, we have $36 \theta = 3$

$\theta = \frac{1}{12}$

$\gamma = \frac{11}{12}$

Thus, $a = \frac{1}{\gamma} = \frac{1}{\frac{11}{12}} = \boxed{\frac{12}{11}}$