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(a) There are 285 ways.

(b) There are 66 ways.

a - b = 5*sqrt(3).

There are 570 ways.

The area of the triangle is 5*sqrt(11).

The order is B < E < D < A < C.

1. The equation f(f(x)) = 4 has only one solution.

2. A + B = 8.

The probability is 3/10.

The expected value of the number of sixes is all the number of cases for the number of sixes * probability to get that number of 6s, all added together. 

The probability to get three sixes is (1/6)^3=1/216 

The probability to get two sixes is (1/6)*(1/6)*(5/6)*3 = 15/216

The probability to get one six is (1/6)*(5/6)*(5/6)*3 = 75/216

The expected value of sixes is 3*1/216 + 2*15/216 + 1*75/216 = 3/216 + 30/216 + 75/216 = 108/216 = 1/2

We have:

\(3x^2+9x+15=2x^2+8x-17\)

Start off by simplifying.

\(x^2+x+32=0\)

We want to find (r+1)(s+1) = rs+r+s+1

We apply Vieta's formula:

rs = 32

r+s = -1

32-1+1 = 32

The first, last, and middle numbers must be red. 

This means that the 2nd and 4th squares can be either yellow or blue. There are 2 options for each which makes for \(2 \times 2 = \color{brown}\boxed{4}\) cases

We want to minimize \(\sqrt{x^2+ y^2} \)

Substituting \(y = {1 \over 2}\times(x^2 - 8)\), we get \( \sqrt{x^2 + {x^4 \over 4} - 4x^2 + 16}\), which simplifies to \( \sqrt{{x^4 \over 4} - 3x^2 + 16}\). 

Now, let \(z = x^2\). We have \( \sqrt{{z^2 \over 4} - 3z + 16}\). Because this is a quadratic, the minimum occurs at \(-{b \over 2a} = -{3 \over 2 \times {0.25}} = 6\). 

Substituting this in gives us \( \sqrt{{6^2 \over 4} - 3\times 6^2 + 16} =\color{brown}\boxed{ \sqrt{7}}\)

Find a - b.

Hello Guest!

\(a^3 + 3ab^2 = 679\\ \underline{a^3 - 3ab^2 = 673}\\ 2a^3=1352\\ a^3=676\\ a=\sqrt[3]{676}=8,7638...\)

\(b=\sqrt{\frac{a^3-673}{3a}}\\ b=\sqrt{\frac{676-673}{3\cdot \sqrt[3]{676}}}\\ b=\sqrt{\frac{1}{\sqrt[3]{676}}}\\ \color{blue}a-b=\sqrt[3]{676}-\sqrt{\frac{1}{\sqrt[3]{676}}}=8.4388297647395448\\\)

  !

a=8,776382955329128

b=0,3375531905895829

Let the children be a, b, c, d, and e since they are different.

a + b + c + d + e = 12. By stars and bars if each variable is nonnegative, we have (12 + 5 - 1 choose 5 - 1) = 16 choose 4 = 1820 ways.

However we overcounted the ways for when the variables are more than 5. But these cases are easy to count:

Case 1: There was a 6.

a + b + c + d = 6. Stars and bars = 9 choose 3 = 84 ways, multiply this by 5 to get the cases when d, c, b, and a are 6 = 420 ways.

Case 2: There was a 7. 

... 

Then once you finish up the cases, all you have to do is take 1820, and subtract the overcounts. (The answer is not >1400)

(r + 1)(s + 1) = -16.

There are 14 ways.

(a) The probability is 3/10.

(b) The probability is 17/500.

That answer is wrong.

First we use Stars and bars to find all possible cases:

\({x}_{1}+{x}_{2}+{x}_{3}...+{x}_{7}=9\)

Stars and bars tells us that there are (9+7-1) choose (7-1) = 5005 nonnegative integer solutions to this.

Now we assume Dhruv (assume that he is x₇) gets 2 chocolates. We can find the number of cases which Dhruv gets 2 chocolates. 

\({x}_{1}+{x}_{2}...+{x}_{6} = 7\)

Now we use stars and bars again to find that there are (7+6-1) choose (6-1) = 792 nonnegative integer solutions.

So the probability that Dhruv gets 2 chocolates is (792/5005) = (72/455)

(There might be a calculation error if there is tell me)

Case 1: x = 10

u + v + w + y + z = 8, stars and bars means (8 + 5 - 1 choose 5 - 1) = 495 ways.

Case 2: x = 9

u + v + w + y + z = 9, stars and bars mean (13 choose 4) = 715 ways.

Case 3: x = 8

Stars and bars means (14 choose 4) = 1001 ways.

...

Case 11: x = 0

(22 choose 4) = 7315 ways.

Add (12 choose 4) + (13 choose 4) + (14 choose 4) + ... + (21 choose 4) + (22 choose 4) is your answer = 32857 total solutions.

(b) By stars and bars, instead of (n + k - 1 choose k - 1), we have (n - 1 choose k - 1), so (19 choose 4) = 3876 ways.

(a) Friends, a, b, c, d, and e.

a + b + c + d + e = 20. All friends are nonnegative, so by stars and bars we have (20 + 5 - 1 choose 5 - 1) = (24 choose 4) = 10626 ways to give out 20 identical stickers to his friends.