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Between the 5 different mathematics textbooks, there is 5! ways to order them.

Between the 4 different psychology textbooks, there is 4! ways to order them.

Treating each of these as a group, we can order the group of of mathematics textbooks, and psychology textbooks in \(\frac{2!}{1!1!}\) = 2 different ways. (this is because either the group of mathematics textbooks can be first, or the group of psychology textbooks can be first).

There are 5!*4!*2 = 5760 different ways to order the textbooks.

19x^2  - 6x + 31   =   0

x =    [ 6  +  sqrt [ 36 - 4 * 19 * 31 ] ] /  38   =

[ 6 + sqrt [ -2320] ] /  38  =

[6 + i [-16 * 145 ] / 38   = 

[ 6 + 4i sqrt [145] ] / 38  = 

[ 3 + 2sqrt [ 145 ] i ] / 19  =

3 /19   +  (2/19)sqrt (145)  i

a =  3/19       b = (2/19) sqrt (145)

ab =   (3/19) (2/19) sqrt (145)  =  

(6 / 361) sqrt (145)

(15 - 16i) /2  - 3i (2) / 2   =

[ 15 - 16i - 6i ]   / 2  =

15/2  - (16i -6i ) / 2  =

15/2  - (22i)  / 2  = 

15/2 - 11i

We simplify, into -8x²-kx+47=0.

If the equation has no real solutions then, b²-4ac < 0.

So, k²-4(-8)(47)<0. or k²+1184<0

k² must be greater than or equal to zero, so k²+1184 must be greater than zero, so there is no possibility of the discriminant being negative, so k can be infinitely big.

mn = m + 11n  - 18

mn - m  =  11n - 18

m ( n -1)  = 11 n  - 18

m =  (11n - 18) / ( n -1)

m will not be < 12  as n approaches  - inf

m  will not be > 10  as  n  approaches inf

m       n

4        2

10      8

18      0

12     -6

This is a classic case of Simon's favorite factoring trick.

Supposed the mystery number is m.

We group like terms, getting ab-18a+8b+m.

We can take the a out of the first two terms, a(b-18) + 8b + m. We know b - 18 should also factor into 8b + m.

So, 8b+m = 8(b-18), and m = -144. The mystery number is -144.

This can be factored into (a+8)(b-18).

A lot of casework, but relatively simple.

We find the digits for each number, find the number of ways to arrange them, to represent the different integers.

Here are the possible cases:

5 = 

C1) 1, 1, 1, 1, 1

C2) 1, 1, 1, 2

C3) 1, 2, 2

C4) 1, 1, 3

C5) 2, 3

For C1, there is 1 way to order this.

For C2, there is \(\frac{4!}{3!1!}\), or 4 ways to order this.

For C3, there is \(\frac{3!}{2!1!} \), or 3 ways to order this.

For C4, there is \(\frac{3!}{2!1!}\), or 3 ways to order this.

For C5, there is 2 ways to order this.

There are a total of 1+4+3+3+2 = 13 numbers.

Looking at the terms of both sides, we see that a quadratic * f(x) is a quartic, so f(x) must be a quadratic. 

We additionally know that the coefficient of the quadratic term of f(x) is 2, because the only way 2x⁴ can be formed, is the multiplying of the quadratic terms, and since on quadratic term is x², the other term must be 2x⁴/x² = 2x².

Similarly, applying this same trick, we know that the constant term of f(x) must be 1/5, because only constant terms * constant terms can create constant terms, and on constant term is -5, so the other must be 1/5, so that -5*1/5 = -1. Another way to visualize this is by plugging in 0 as x.

Now we need the linear term.

We know f(x) so far as 2x²+ax+1/5.

To find the linear term, we consider that (x²-2x-5)*f(x) is the quartic.

(x²-2x-5)*(2x²+ax+1/5)=2x⁴+19x³ ... 

From here we can directly get a. We see that the cubic term 19x³ on the right, should also be equal to the cubic term of the expansion on the left.

We notice that cubic terms in the expansion can only be formed from quadratic terms * linear terms.

The cubic terms on the expansion on the left is -2x*(2x²) = -4x^3 and x²*ax, so the cubic term in the expansion is (-4+a)x³,

We equate the two cubic terms getting, -4+a = 19. a = 23.

But we see that (x²-2x-5)(2x²+23x+1/5) is not the correct expansion, and polynomials can't have negative exponents, I believe that this question is not possible. 

Combine like terms: \(19x^2-6x+31=0\)

By vieta's formulas, the sum of the roots of a quadratic ax^2 + bx + c = -b/a, and their product = c/a.

Here, our roots, A and B, have a product of the term c/a = 31/19, thus A * B = 31/19 (capitalized such that it isn't confused for the coefficients of the quadratic.)

(a + 2) / (a + 3)  = b / 7

7 (a + 2) = b(a + 3)

7a + 14  = ab + 3b

7a - ab  =  3b  - 14

3b + ab  = 7a + 14

b ( 3 + a)  = 7a -+14

b =  (7a +14) / (3 + a)

a        b

-10    8

-4    14

-2      0

4       6

We treat the 4 freshmens as a group. 

In the group of freshman there are 4! ways to order the freshmans.

To order the remaining objects, 4 sophomores, one freshman, and one group of freshmans, there is 6! ways to order this.

So there are a total of 4! * 6! ways to order

But we must be careful, because if there is a group of 5 freshmanm, we overcount. The conbination S₂F₁F₄F₃F₅F₂S₃S₄S₁, can be formed as S₂F₁ + a group + S₃S₄S₁, or S₁ + a group + F₂S₃S₄S₁. 

So we overcount each possiblity of five people in a group twice, so we need to subtract the number of possibilities with 5 people in a row.

The possibilities for 5 people in a row, treating the 5 freshmen as a group is 5! (the arrangments within the group) * (5!) (the ways to order 1 group + 4 sophomores). 

So the answer should be 4!*6!-5!*5! = 2880 ways

Combine like terms: \(x^2+23x+k\)

Because this is a quadratic, and the problem states it has a double root, we know the expression can be factored to \((x-r)^2\).

When expanded, it is expanded to \(x^2-2xr+r^2\). Setting these expressions equal to each other, we get -2xr = 23x, r = -23/2.

We also get that r^2 = k, so k = (-23/2)^2 = 529/4.

If there are 3 skaters, including 3 Americans, in how many ways can the medals be awarded to three of the 3 skaters if exactly one of the Americans wins a medal...

As you can see, there are only three competitors of which all are American, so assuming none get disqualified and complete it, we will have 3 winners, each American, so the question conradicts itself saying that only exactly one American can win a medal...

0 ways.

A coefficient of x can only be formed, by only a constant term times, a linear term.

We see that this can occur in two ways:

1*(-23x) = -23x

x(14) = 14x

Summing these up, we get -23x+14x=-9x.

The Coefficient of x is -9.

x^2  + 3x + p

p =  [ (coefficient on x)  / 2 ] ^2 =  [ 3/2]^2 =  9/4

x^2  + 3x + 9/4   =

(x + 3/2)^2

Law of Cosines

AE^2 = AF^2  + FE^2  - 2 (AF * FE) cos (60°)

AE^2  = 2^2 + 5^2  - 2(2* 5) (1/2)

AE^2  = 19

AE = sqrt (19)

x^3 ( x^2 + tx)   =   x^5 + tx^4

This has no x^2 term anyway

18 / 16  = r^3

9 / 8  = r^3

r =  (9/8)^(1/3)

24th term  = 16 * (9/8)^(1/3) ≈   16.64

Here :   https://web2.0calc.com/questions/circles_146

Let D be the center of the  circle and let its radius = r

Let E be the center of one of the small semi-circles

We can construct right triangle BDE  with EBD  = 90°

BE = 1

BD = 2 - r

ED =  1 + r

BE^2 + BD^2  = ED^2

1^1 + (2 -r)^2  = (1 + r)^2

1 + r^2 - 4r + 4  = r^2 + 2r + 1

6r =  4

r = 4/6  = 2/3

There are two cases to consider:

Case 1: Two cards of the same color come from the same pile.

Choose the color of the two cards in 4 ways (red, green, blue, or yellow).

Choose two different numbers from the chosen color in 7C2 ways (7 ways to choose the first number, then 6 ways to choose the second number given the first one has already been chosen).

Choose the third card, which can be any of the remaining 21 cards in 21 ways.

The total number of hands in this case is 4 * 7C2 * 21 = 504.

Case 2: Two cards of the same color come from different piles.

Choose two different colors in 6C2 ways (6 ways to choose the first color, then 5 ways to choose the second color given the first one has already been chosen).

Choose one number from each of the chosen colors in 7 * 7 = 49 ways.

Choose the third card, which can be any of the remaining 19 cards in 19 ways.

The total number of hands in this case is 6C2 * 49 * 19 = 1716.

Total number of hands:

Adding the number of hands from both cases, we get 504 + 1716 = 2220.

Therefore, Yunseol can draw 2220​ three-card hands with exactly two cards of the same color.

Let $O_1$ and $O_2$ be the centers of circles $P$ and $Q$ respectively. Let $A$ and $B$ be the points where line $PQ$ intersects the circles $P$ and $Q$ respectively.

Let $C$ be the midpoint of $\overline{AB}$. Let $D$ be the foot of the perpendicular from $O_2$ to $\overline{PQ}$, and let $E$ be the foot of the perpendicular from $O_1$ to $\overline{O_2D}$.

Since $\overline{O_1E}$ and $\overline{O_2D}$ are both perpendicular to $\overline{PQ}$, $O_1EDO_2$ is a rectangle. Therefore, $O_1E=O_2D$.

Since $O_1O_2=12+8=20$, we know $O_1E=O_2D=10$. Because $\angle O_1AQ=\angle O_2BQ=90^\circ$, angles $O_1EA$ and $O_2DE$ are both right angles.

Thus, by the Pythagorean Theorem on $\triangle O_1AE$, \[AE=\sqrt{12^2-10^2}=2\sqrt{7}\]

and by the Pythagorean Theorem on $\triangle O_2DE$, \[ED=\sqrt{20^2-10^2}=10\sqrt3.\]

Finally, we find the length of $\overline{QR}$. By the Pythagorean theorem on $\triangle CDE$,

\[QR = CD = \sqrt{AE^2+DE^2}=\sqrt{(2\sqrt7)^2+(10\sqrt3)^2} = \boxed{4\sqrt{10}}.\]