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ABC  will form a triangle  and the circle is the  circumcumcircle of the triangle

radius =   [product of the  sides ] /  [4 area ]

semiperimeter  =  [ 8 + 15 + 12 ] / 2=   17.5

Area =  sqrt  [ 17.5 ( 9.5) (2.5) (5.5) ] 

radius =   [ 8 * 12 * 15 ] / [ 4 sqrt [ 17.5 (9.5)(2.5)(5.5) ] ]  ≈  7.53

Draw a rectangle in the middle like so:

The width is r-13, while the length is r - 10, and the diagonal is r:

Therefore, use the pythagorean theorem.

\({(r-10)}^{2}+{(r-13)}^{2}={r}^{2}\)

\(r=\frac{46\pm \sqrt{1040}}{2}=23 \pm \sqrt{260} = 23\pm 2\sqrt{65}\)

We know because the distance of 13 to the wall, r > 13, therefore the only solution is \(23+2\sqrt{65}\).

First we fix the psychology book in the middle. There are 3 choices we can make for this book.

For the remaining 6 slots (3 on either side of the psychology book), we can put any of the 7 textbooks in the first slot.

Once we put a book in the first slot, we have 6 choices left for the second, 5 choices left for the third, and so on.

So, following the fundamental counting principle, we might be tempted to say there are 7⋅6⋅5⋅4⋅3⋅2=(7−6)!7!​=5040 ways of arranging the books.

However, this overcounts all the arrangements where we just move all the math books to one side and all the psychology books to the other side.

There are 2! ways of doing this (since we can switch the two sides), so we have to divide our previous answer by 2.

Therefore, there are 7!/(7- 6)!​ * 1/2! = 2520​ ways of arranging the textbooks.

Simplify as

8a - 5b =   12

9a +(6 + m)b = k

Multiply the first  equation by 9  and the  second by 8

72a  - 45b =  108

72a + 8(6+m) =  8k

To have infinite solutions  both of these must be true

8 (6 + m)  =  -45

48 + 8m = -45

8m = - 93

m = -93 / 8  

and

8k = 108

k = 108 / 8  =   27 / 2

There are 2 ways to start with the letter A: either from the first A in the array, or from the second A.

To reach the second A, we must move down once. Once we have reached the first or second A, we can move in any direction to reach the letter R.

There are 3 ways to do this: we can move down, right, or down and then right.

Once we have reached the letter R, we can again move in any direction to reach the letter C.

There are 3 ways to do this: we can move down, left, or left and then down.

Once we have reached the letter C, the last step is to move up to reach the letter H, and there is only 1 way to do this.

Therefore there are 2×3×3×1 = 18​ ways to spell the word "ARCH" in the array provided.

A

4                  5

B           D               C

             3

Let BD = 3- x     and DC =  x

By the Angle Bisector Theorem

BD/AB = DC / AC

(3-x) / 4  = x / 5

5(3-x)  = 4x

15 - 5x  =  4x

15 = 9x

x = 15/9 =  5/3

[ADC ] =   (1/2)DC * AB =  (1/2)(5/3)(4)  = 10 / 3

The lowest prime that satisfies this is 5, because 5 is right after 4.

To find the highest, do 200/5 = 40, and round down which is 37.

There are a total of 10 primes between these 2:

5, 7, 11, 13, 17, 19, 23, 29, 31, 37

Lets count the total number (excluding those that have already been counted)

Groups with 5: 7, 11, 13, 17, 19, 23, 29, 31, 37 total : 9

Groups with 7: 11, 13, 17, 19, 23 total : 5

Groups with 11: 13, 17 total : 2

No more after this ~

There are a total of 9 + 5 + 2 = 16 total products.

Lets use complementary counting, it is much each to count all the cases where the freshman don't stand next to each other.

There are 7! ways to arrange 7 distinct people.

To have no two freshman stand next to each other, we must start with a freshman and alternate between freshman and sophomore like this:

F S F S F S F.

We can arrange the freshman in this arrangement 4! ways, and the sophomores 3! ways, so there are \(4!*3!\) ways for this arrangement.

Therefore there are 7!-3!*4! = 4896 ways to do this.

The word "ARCH" has to contain the letter A.

There is only one letter A, therefore it much start there.

We can go hoizontally, diagonally, or vertically all once.

Therefore there are 3 ways to do this.

\(a+ar+a{r}^{2}+a{r}^{3} + \dots=4\\{a}^{3}+{a}^{3}{r}^{3}+{a}^{3}{r}^{6}+{a}^{3}{r}^{9}+\dots=10\\\text{Using the sum of infinite geometric series formula}:\\\begin{cases} \frac{a}{1-r}=4 \\ \frac{{a}^{3}}{1-{r}^{3}}=10 \end{cases} \\\text{Isolate a:}\\ \begin{cases} a=4(1-r) \\ {a}^{3}=10(1-{r}^{3}) \end{cases} \\\text{Substitute:}\\ 64{(1-r)}^{3}=10(1-{r}^{3})\\ 54{r}^{3}-192{r}^{2}+192r-54=0\\ (r-1)(9{r}^{2}-23r+9)=0\\ r=1, r=\frac{23+\sqrt{205}}{18}, r=\frac{23-\sqrt{205}}{18}\\ \text{take the range} -1 <1 \;\text{therefore}\; r=\frac{23-\sqrt{205}}{18} \)

simplifies to

\(x^7+4x^6+8x^5+16x^4+22x^3+19x^2+15x+7\)

and the coefficient will be 15

This is easy

\(\frac{1}{4}+\frac{1}{4*7}+\frac{1}{7*10}\)

is

\(\frac{35}{140}+\frac{5}{140}+\frac{2}{140}\)

which becomes

\(\frac{42}{140}\)

which simplifies to:

\(\frac{3}{10}\)

In class, we derived that

\(\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}.\)

Fill in the blanks to make a true equation:

\(\frac{5x}{(x - 1)(x^2 + 2)(x + 7)^3)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 2} + \frac{D}{x + 7} + \frac{E}{(x + 7)^2} + \frac{F}{(x + 7)^3}\)

P  O   P  =   (5/8) (3/7) ( 4/6)  = 5 / 28  =  10/56

O  P   O  =  (3/8) (5/7) ( 2/6)  = 5 / 56  =    5/56

15 / 56 

Area  = (1/2) (1^2) sin 90°  =  1/2

https://web2.0calc.com/questions/trapezoid_132

Simplify as

2x^2  + sx  +  38

We will have a square of a  binomial when the discriminant =   0

s^2  -  4(2)(38)  =  0

s^2  =  304

s  =  ± sqrt [304]    =  ± 4sqrt (19)

Draw AE , BF  perpendicular to CD

Let DE =  x

Let CF =  104 - 78 - x  =  26 - x

We can solve this  to  help find  the height

AD^2  - DE^2   = BC^2  -  CF^2

10^2  - x^2  = 24^2 - ( 26-x)^2

100 - x^2  = 576  - x^2 + 52x  - 676

100 = 52x - 100

200   = 52x

x  =  200 /  52  =   50  / 13

The height is   sqrt   [ 24^2  - (50/13)^2 ]    ≈  23.7

Area =   (1/2) height (sum of bases)  =  (1/2) (23.7) (78 + 104)  =   2156.7

https://web2.0calc.com/questions/geometry_93750

Simplify as 

3x^2 - 32x  +  c

We will have a square of a binomial when the  discriminant  = 0

32^2  -  4(3)c  =  0

1024  =  12c

c = 1024/ 12 =  256/3

Law of Cosines

PR^2  = PQ^2 +QR^2   -2(PQ * QR)cos ( PQR)

23^2  =18^2 + 25^2  - 2(18 * 25) cos (PQR)

[23^2 - 18^2 - 25^2 ] / [ -2 * 18 * 25 ] =  cos (PQR)

7/15 = cos (PQR)

Law of Cosines again

PM^2  = PQ^2 + (QR/2)^2 - 2(PQ * QR/2) cos (PQR)

PM^2  = 18^2 +(12.5)^2  - 2 (18 * 12.5)(7/15)

PM^2 = 270.25

PM = sqrt (270.25)  ≈  16.44

The probability of Raina getting alternating colors depends on whether she picks an orange ball or a purple ball first. Let's analyze both scenarios:

Scenario 1: Orange - Purple - Orange (OPO)

Probability of picking an orange ball first: 3 out of 8 total balls (3 orange + 5 purple).

Probability of picking a purple ball after an orange: 5 out of 7 remaining balls (1 orange left + 5 purple).

Probability of picking another orange ball after a purple: 2 out of 6 remaining balls (2 orange left).

So, the probability of OPO is (3/8) * (5/7) * (2/6) = 5/56.

Scenario 2: Purple - Orange - Purple (PPO)

Probability of picking a purple ball first: 5 out of 8 total balls.

Probability of picking an orange ball after a purple: 3 out of 7 remaining balls (3 orange left + 4 purple).

Probability of picking another purple ball after an orange: 4 out of 6 remaining balls (4 purple left).

So, the probability of PPO is (5/8) * (3/7) * (4/6) = 5/56.

Total Probability

Since these scenarios (OPO and PPO) are mutually exclusive (they can't happen at the same time), to get the overall probability, we simply add the probabilities of each scenario:

Total Probability = Probability(OPO) + Probability(PPO) = (5/56) + (5/56) = 10/56 = 5/28

Therefore, the probability of Raina drawing three balls with alternating colors (orange-purple-orange or purple-orange-purple) is 5/28.