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In right triangle DBC, the sides are: BC==78, BD==30,  DC==72. Angle BDC==90, Angle DBC==67.38, Angle BCD ==22.62 degrees

In right triangle ABD, the angles are: BDA==90,  DBA==22.62,  DAB==67.38

Sine(67.38) ==30 / AB

0.923076 ==30 / AB

AB ==30 / 0.923076

AB ==32.50

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.

1 - 15 x 5c==75c
2 - 13 x 5c + 10c==75c
3 - 11 x 5c + 2 x 10c==75c
4 - 9 x 5c  +  3 x 10c==75c
5 - 7 x 5c  +  4 x 10c==75c
6 - 5 x 5c  +  5 x 10c==75c
7 - 3  x 5c  +  6 x 10c==75c
8 - 1 x 5c   +  7 x 10c==75c

Number of ways ==8

There are 54 possible orders.

You can just use Binet's formula: https://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html

A computer code written in C++ gives:

10,626 ways

Using combinatorics, we have:

[20 + 5 - 1] C [5 - 1] ==24 C 4 == 10,626 ways

A ccomputer code written in C++ gives:
372 ways.

Using combinatorics, we have the following:
1 - Each of the twins receives 0 candies. That leaves: [17 - 1] C [17 - 3] ==120 ways
2 - Each of the twins receives 1 candy. That leaves: [15 - 1] C [15 - 3]==91 ways
3 - Each of the twins receives 2 candies. That leaves: [13 - 1] C [13 - 3]==66 ways
4 - Each of the twins receives 3 candies. That leaves: [11 - 1] C [11 - 3] ==45 ways
5 - Each of the twins receives 4 candies. That leaves: [9 - 1] C [9 - 3] ==28 ways
6 - Each of the twins receives 5 candies. That leaves: [7 - 1] C [7 - 3] ==15 ways
7 - Each of the twins receives 6 candies. That leaves: [5 - 1] C [5 - 3]==6 ways
8 - Each of the twins receives 7 candies. That leaves: [3 - 1] C [3 - 3]==1 way.

Grand Total: 120 + 91 + 66 + 45 + 28 + 15 + 6 + 1==372 ways.

Use inverse operations:

8x = 7x + 10

-7x   -7x

x = 10

By sine law,

AB = AD*sin(angle BAD)/sin(ABD) = BD/sin(BCD)*sin(BDC) = 30/72*24 = 10

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

Well there is. Just find any integer less than 100 that is a 1, 2, 3 or have their prime factorization be 2^a * 3^b.

You can do easy counting from there. No need to brute force.

Say the perimiter of DGV is P, then from the given, we know that FPM's Perimiter : DGV's Perimiter = 8:5

Substituing in, we have 72:P = 8:5. In the scale factor 8:5, we know that each unit is 9 for FPM, likewise, each unit for 5 would also be nine. In other words, solving it as an equation 72/p = 8/5 you obtain p = 45. 

Perimiter of DGV is 45 units.

Aww was hoping for a more clever less tedious answer... Thank you anyways!

[7^0 / 100^1]  +  [7^1 / 100^2]  +  [7^2 / 100^3]  + ..........and so on. Is it right?

(1 / 100, 7 / 10000, 49 / 1000000, 343 / 100000000, 2401 / 10000000000, 16807 / 1000000000000, 117649 / 100000000000000, 823543 / 10000000000000000, 5764801 / 1000000000000000000, 0.0000000000 0040353607, 0.0000000000 0002824752 49.......and so on

sumfor(n, 0, 10000, (7^n / (100^(n+1))))==0.01075268817 - sum of infinite G. S.

OR:

Sum ==F / [1  -  R], where F==First term,  R==Common Ratio

Common Ratio ==7 / 100

Sum ==[1 / 100] / [1  -  7/100]==1/100  x  100/93==100 / 9,300 ==0.01075268817 - sum of infinite G. S.

(a)

sumfor(k, 0, 3, (4  nCr k))==15 ways - without any restrictions

(b)

[4 + 5 - 1] C [4] ==8 C 4 ==70 ways - without any restrictions

a + b + c + d + e + f + g + h + i + j = 10

a, b, c, d, e, f, g, h, i, j = 1, 2 or 3.

Try stars and bars with removal of some cases. Brute forcing it will take too long and isn't smart

(a) TRUE The opposite of a number is basically that number's negative. For example, 2 + (-2) = 0. 

(b) FALSE Not always! It depends if the first number is greater or less than the second number. For example, 2 - 3 = -1

(c) FALSE Actually, It's the quite opposite! When a positive integer is multiplied to a negative integer, the product is ALWAYS negative. For example. (-2)*3=-6

(d) FALSE NOOOOOOOOOOOO. Just because the sum of a postive and its oppositie integer is 0, it doesn't mean the product is 0. Remember, the only way to get a product to equal 0 is by multiplying by 0!