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we k**l and eat Gilligan

I don't see any really clever way of doing this.  You can expand it using the binomial theorem to get

\(\left(2+\sqrt{3}\right)^9 = \sum \limits_{k=0}^9~\dbinom{9}{k}2^k~3^{(9-k)/2}\)

if you then expand all that out and combine things you'll have an expression that looks like

\(a + b \sqrt{3}\)

b above turns out to be 40545 so you'll need to take the square root of 3 out to 5 decimal places

there are 3 rolls whose product totals 180

(5,6,6), (6,5,6), (6,6,5)

There are 216 possible rolls of 3 six sided dice.

\(P[\text{product of 3 dice = 180}] =\dfrac{3}{216}=\dfrac{1}{72}\)

I'd say... uhhh... a low probability!

\(\left(\dfrac{1}{13}\right)\left(\dfrac{4}{51}\right)\left(\dfrac{4}{50}\right)= \dfrac{16}{33150} \approx 4.83\times10^{-4}\)

first off we regroup things so that we have 5 Republicans and 8 non-Republicans

this gets us

\(P[R]=\dfrac{5}{13},~P[!R] = \dfrac{8}{13}\)

now we have a simple Binomial distribution with parameters  \(n=3,~p=\dfrac{5}{13}\)

\(P[3] = \left(\dfrac{5}{13}\right)^3 = \dfrac{125}{2197}\)

@200% 1 becomes 3 every 2 years. And:
@800% 1 becomes 9 every n years
The year to year inctrease =3^1/2 =sqrt(3) - 1 x 100 =73.20% annual comp.rate.
9 = 1 x [1 + .732]^n
n =Log(9) / Log(sqrt(3))
n = 4 Years, or 2003 + 4 = 2007.

Date            Increase          number of visitors            

2003                0                          N                           

2005                2N               N+2N = 3N                    

2007             2*3N = 6N      3N+6N = 9N                   

2009             2*9N = 18N    9N+18N = 27N             

2011             2*27N = 54N   27N+54N = 81N            

800% increase on N = 8N so total visitors would have been  N + 8N = 9N 

Hence year was 2007.

Certainly there is such a value of x!  Melody's original reply is correct.

.

https://web2.0calc.com/questions/from-2003-onward-the-number-of-daily-visitors-to  this was a question from a while back. It might have a better explanation!

Well this question is very odd if you think about it, since most people would say 2011 because it seems obvious. They would be wrong! This isn't the same as doubleing because you are not doing 200% + 200%, but instead it will be 200% * 400% so the answer to this problem is 2007 using our equation. Could I get confirmation?

Simplify the following:
(4 x^2 + 7 x + 9) - (2 x^2 + 10 x + 4)

Factor 2 out of 2 x^2 + 10 x + 4:
4 x^2 + 7 x + 9 - 2 (x^2 + 5 x + 2)

-2 (x^2 + 5 x + 2) = -2 x^2 - 10 x - 4:
-2 x^2 - 10 x - 4 + 4 x^2 + 7 x + 9

Grouping like terms, 4 x^2 - 2 x^2 + 7 x - 10 x - 4 + 9 = (4 x^2 - 2 x^2) + (7 x - 10 x) + (9 - 4):
(4 x^2 - 2 x^2) + (7 x - 10 x) + (9 - 4)

4 x^2 - 2 x^2 = 2 x^2:
2 x^2 + (7 x - 10 x) + (9 - 4)

7 x - 10 x = -3 x:
2 x^2 + -3 x + (9 - 4)

9 - 4 = 5:
=2x^2 - 3x + 5

Day 2 temp - Day 1 temp.

-1 - *(-5) =

-1 + 5      =+4 - increase in temp from Day 1 to Day 2

What number can be added to 9 to get a result of 0?

+9 + (-9) =0

What is the sum of -10 and -25?

(-10) + (-25) =-35

What is the value of -3 - (-8)?

-3 -*(-8) =

-3 + 8     =+5

Thanks Rom,

Would it be fair to say that the fallacy in my answer is that i assumed such an x existed in the first place ?

Thanks

Solve for y:
11 sqrt(2) sqrt(y) = 16 sqrt(2) sqrt(y) - 5

Subtract 16 sqrt(2) sqrt(y) from both sides:
-5 sqrt(2) sqrt(y) = -5

Divide both sides by -5 sqrt(2):
sqrt(y) = 1/sqrt(2)

Raise both sides to the power of two:
y = 1/2

When you do equations like this, reorganize the numbers because you are subtracting. Reorganise them like this: 20-9.50-4x.

It only works with equations like this: 20-17-2 or 19-6x-4-3 because of the communitive properties. Btw, it factors out like this: 11.50-4x.

This was wrong.  Melody's approach is correct.

The 10 tulips, 5 of one color and 5 of another is a bit of a red herring.

You are basically choosing 2 colors out of n.

\(\text{so you need to find }n \ni \dbinom{n}{2}=15 \\ \\ \text{a bit of testing quickly reveals that }n=6\)

Notice that 25 = 5*5,  65 = 13*5,  60 = 12*5.  One simple way is then as follows:

Other choices are possible for a, b and c, but they all lead to the same final result.