proyaop

(b) This is a stars and bars question again. However, the formula is not (n + k - 1) choose (k - 1), it is (n - 1) choose (k - 1) where k is the # of variables and the second formula is for positive constraint. Thus the second problem is (20 - 1) choose (5 - 1) = 19 choose 4 = 3876 ways.

(a) This is a stars and bars question, you can look up the why on stars and bars but I will apply it here. a + b + c + d + e = 20. The variables are nonnegative. Total ways is (20 + 5 - 1) choose (5 - 1) = 24 choose 4 = 10626 ways.

3^(n + 3) = 3^n * 3^3

2^(n + 5) = 2^n * 2^5.

So our equation is 3^n * 3^3 + 2^n * 2^5 * 31^n.

27(3^n) + 32(2^n) * 31^n

Note that 59 = 32 + 27

Also, 27(3^n) will always be divisible by 27, and 32(2^n) * 31^n will also always be divisible by 32. 

Not saying that the sum will be divisible by 59, but good luck from here!

For the first letter in the top row, we have 1 option, for the second letter in the second row we have 2 options, for the third letter in the third row we have 2 options, for the last row we have one option. The total permutations/cases are 1 x 2 x 2 x 1 = 4 different ways to spell ARCH (for this problem, not 4!)

Oh, mixed numbers... the math police have arrived. Let's change 15 3/5 to an improper fraction of 78/5, they are still the same. 

After 6 days, 78/5 * 6 = 93.6 bushels of corn sold. 93.6 is not an integer, but that's the change in the amount of bushels of corn from his original inventory from day 1.

\(f(x) = {x+5\over x-7}\)

\(h(x) = (x + 5)(x-7)\)

These two functions are obviously not the same because when you give quantity x + 5 = a, and x - 7 = b, a/b is not equal to ab, unless a is 0 which is conditional and not for all values of x. The two functions only intersect when x is -5 at first glance, and they do not overlap each other consistently so they are not the same function.

Try some cases, x = 10.

U + v + w + y + z = 8. Stars and Bars (8 + 5 - 1) choose (5 - 1) = 12 choose 4 = 495

x = 9.

u + v + w + y + z = 9. Star and bars 9 + 5 - 1 choose 5 - 1 = 13 choose 4 = 715.

Etcetera until you reach case x = 0, u + v + w + y + z = 18, Stars and bars 18 + 4 choose 4 = 7315.

Our lowest case is 12 choose 4, and we add up all the way until 22 choose 4.

12 choose 4 + 13 choose 4 + 14 choose 4 ... + 21 choose 4 + 22 choose 4 = 32857 solutions.

24x^2 + kx - 8.

Factors of 24: 1, 24. 2, 12. 3, 8. 4, 6. 

(x    )(24x    ) or (2x _ _)(12x _ _), etc

You could so something like this while plugging in factors of 8 into the blank spots and adding opposite parity operations. (_x + 4)(_x - 2) etc.

gl

You can do casework: And the other 3 kids are children, a, b, and c. (STARS AND BARS QUESTION)

1. The twins both get no candy.

a + b + c = 17. Total cases = (17 + 3 - 1) choose (3 - 1) because a and b and c all are nonnegative. 171 total cases.

2. The twins both get 1 candy.

a + b + c = 15. Total cases = (15 + 3 - 1) choose (3 - 1). 17 choose 2 = 136 total cases.

3. The twins both get 2 candies.

a + b + c = 13. Total cases = 15 choose 2. 15 choose 2 = 105 total cases. 

4. The twins both get 3 candies.

a + b + c = 11. Total cases = 13 choose 2 = 78 total cases. (You can see the pattern).

In the end, we have 19 choose 2 + 17 choose 2 + 15 choose 2... + 3 choose 2 (this is the each twin gets 8 candies case) = 615 total ways.

Well you definitely want to take out the absolute value signs. And by doing so we have 4 cases:

x^2 - 30x - 1 + x^2 - 30x + 29 = 30

-x^2 + 30x + 1 + x^2 - 30x + 29 = 30

-x^2 + 30x + 1 - x^2 + 30x - 29 = 30

x^2 - 30x - 1 - x^2 + 30x - 29 = 30

Let's solve each one. 

1. 2x^2 - 60x + 28 = 30 => 2x^2 - 60x - 2 = 0 => x^2 - 30x - 1 = 0 => \(x = {15\pm\sqrt{226}}\)

2. 30 = 30.

3. -2x^2 + 60x - 58 = 0 => x^2 - 30x + 29 = 0 => (x - 29)(x - 1) = 0 => x = 29, 1

4. - 30 = 30

we have four solutions that are bolded