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A=2/3=50/75

B=1536

C=8/25=24/75

D=-16

E=\(3\sqrt{2}\)\(\approx4.24264068\)

So the order is

\(C,A,E,D,B\)

Compute \(\dbinom53+\dbinom63+\dbinom73\).

You can use \(e^{i \theta} = \cos \theta + i \sin \theta\)

Here, theta works out to 40 degrees.

I think that might be correct, but I forgot to include a part in my question. Could you make your answer in addition instead of substraction?

The answer is 3x - 2y - 5 > 0

Number of ways of drawing 3 white balls out of 12 white balls are:

12 C 3 ==220 ways

Number of ways of drawing 1 red out of [20 - 13 +1]==8 red balls are:

8 C 1 ==8 ways

Number of ways of drawing all 4 balls are:

220  x  8 ==1,760 ways - or your chances of winning the SuperLottery are: 1 in 1,760 chances.

153  -  5 ==148

Factor(148) ==2^2  x  37................(1)

267  -  8 ==259

Factor(259) ==7  x  37....................(2)

x == GCD(148,  259) ==37

Thanks, that's right!

If they sum to 2, there are 2 stars and 9 bars which makes for \({11 \choose 9} = 55\) solutions. 

If they sum to 1, there is 1 star and 9 bars which makes for \({10 \choose 9} = 10\) solutions. 

If they sum to 0, there are 0 stars and 9 bars which makes for \({9 \choose 9} = 1\) solution. 

So, there are \(55 + 10 + 1 = \color{brown}\boxed{66} \) solutions.

The triangle inequality states that the two legs (smaller numbers) must have a larger sum than the hypotenuse (largest number)

Can you take it from here?

The answer is 7*pi/60.

\((2^{3})^{1\over6}\cdot2^x=32^{1\over2}\)

\(2^{1\over2}\cdot2^x=32^{1\over2}\)

\(2^{{1\over2}+x}=2^{6\cdot{1\over2}} = 2^3\)

\({1\over2}+x=3\)

\(x={5\over2}\)

Children a, b, c, and 2t need 7 identical candies:

Case 1: t = 0

a + b + c = 7, nonnegative a and b and c: stars and bars = 9 choose 2 = 36

Case 2: t = 1

a + b + c = 5, nonnegative a and b and c: stars and bars = 7 choose 2 = 21

Case 3: t = 2

a + b + c = 3, nonnegative a and b and c: stars and bars = 5 choose 2 = 10

Case 4: t = 3

a + b + c = 1, nonnegative a and b and c: stars and bars = 3 choose 2 = 3

36 + 21 + 10 + 3 = 70 total ways.