Boseo

Here are the answers to your question:

e^(pi*i/2) = 1/2 + 1/2*i.

e^(-2*pi*i/3) = 1/sqrt(2) - 1/sqrt(3)*i

e^(9*pi*i/4) = 1/3 + sqrt(2)/4*i

Hope this helps!

Let (x,y) be a point on the graph of y=1/2​(x2−18). The distance between this point and the origin is given by the distance formula:

x2+y2​=x2+(1/2​(x2−18))2​.

Squaring both sides, we get:

x2+21​(x2−18)2=x4−4x2+81=(x2−2)2+77.

To minimize the distance, we want to minimize x2−2. Since the square of a real number is non-negative, its minimum value is 0. Therefore, the minimum value of x2−2 is 0, which occurs when x=±2. Substituting x=±2 back into the equation for the distance, we get:

sqrt(2^2 + (1/sqrt(2)((2)^2 - 18))^2) = sqrt(4 + 1/2) = sqrt(9/2) = 3/sqrt(2) = 3*sqrt(2)/2

Therefore, the answer is 3*sqrt(2)/2.

We know that since f(x) is divisible by x3, it can be written in the form f(x)=ax5+bx4+cx3+dx2+ex+f.

We are also given that f(x)+3 is divisible by (x+1)3, which can be expanded as (x+1)(x+1)(x+1)=x3+3x2+3x+1.

This tells us that there exist polynomials q(x) and r(x) such that:

f(x)+3=(x3+3x2+3x+1)q(x)+r(x)

Since (x3+3x2+3x+1) has a degree of 3, and we are given that f(x)+3 is divisible by it, then the remainder r(x) must be a polynomial of degree at most 2

(because any higher degree term in r(x) would be canceled out by the division). This means r(x)=mx2+nx+p for some constants m, n, and p.

Substituting the form of f(x) and expanding the equation, we get:

ax5+bx4+cx3+dx2+ex+f+3=(x3+3x2+3x+1)q(x)+mx2+nx+p

Matching coefficients of like terms on both sides, we get the following system of equations:

a=0 (since the coefficient of x5 on the right side is 0)

b=m+3q(1)

c=n+3q(0)+1 (since the constant term on the right side is 1)

d=p+3q(−1)

e=3q(−2)

f=p+3q(−3)

Since we are looking for the smallest possible value of f, we want to minimize the constant term p. From equation (6), we see that f is minimized when p is

minimized.

Setting p=0 and solving the remaining equations, we get:

a=0

b=m+3

c=n+1

d=−3

e=−6

f=−9

Therefore, the polynomial f(x) that satisfies the given conditions is:

f(x)=4x5+(m+3)x4+(n+1)x3−3x2−6x−9

The final step is to find the values of m and n that minimize f(x). Since we can choose any values for m and n as long as p=0, we can simply set them to 0.

This results in the polynomial:

f(x) = 4x^5 + 3x^4 + x^3

We can rewrite the given equation as:

ab−c2a+b−c​+bc−a2b+c−a​+ca−b2c+a−b​=1.

Notice that the left-hand side of the equation resembles the sum of the reciprocals of three numbers, but with some extra terms. Let's manipulate the equation to exploit this similarity.

First, we can factor the quadratic terms in the denominators:

(a−c)(b)a+b−c​+(b−a)(c)b+c−a​+(c−b)(a)c+a−b​=1.

Then, we can multiply both sides of the equation by the common denominator, which is the product of all the denominators:

(a−c)(b)⋅(a−c)(b)a+b−c​+(b−a)(c)⋅(b−a)(c)b+c−a​+(c−b)(a)⋅(c−b)(a)c+a−b​=(a−c)(b)⋅(b−a)(c).

Expanding both sides, we get:

a(b+c−a)+b(c+a−b)+c(a+b−c)=(a−c)(b)(c−a).

This simplifies to:

3abc=abc−a2b−ab2−a2c−ac2−b2c.

Since a+b+c=1, we can substitute:

3abc=abc−(a2+b2+c2)−(ab+ac+bc).

We know that (a+b+c)2=a2+b2+c2+2(ab+ac+bc), so:

3abc=abc−(a+b+c)2+2(ab+ac+bc).

Substituting a+b+c=1, we get:

3abc=abc−1+2(ab+ac+bc).

Finally, we can factor out abc to find:

2abc=abc−1.

Solving for abc, we get:

abc = 1/3.

Therefore, the answer is abc = 1/3.

First, we expand the product in the given equation: k = (6x+12)(x−8) = 6x2−36x+96.

We see that a=6, and b=−36, so the least possible value of k occurs when x = 2a−b ​ = 2⋅6−(−36) ​= 3​.

Setting x = 3 into the given equation, we see that this value does indeed result in the least possible value of k:

k = 6(3)+(−36)+96 = 150−108 = 42​.

Therefore, the answer is 42.

Here's how to find the solution for z:

Multiply each equation by its denominator:

xy = x + y (Equation 1)

xz = 2(x + z) (Equation 2)

yz = 4(y + z) (Equation 3)

Rearrange equations to isolate z:

From Equation 2: z = (xz - 2x) / 2 (Divide both sides by x)

Substitute this expression for z in Equation 3: yz = 4(y + (xz - 2x) / 2)

Simplify and solve for y:

yz = 2y + 2xz - 4x

Group terms with y: yz - 2y = 2xz - 4x

Factor out y: y(z - 2) = 2x(z - 2)

Divide both sides by (z - 2) and solve for y: y = 2x

Substitute y back into Equation 1:

xy = x + 2x

Combine like terms: xy = 3x

Divide both sides by x: y = 3

Combine information about y from steps 3 and 4:

We know y = 3 and y = 2x.

Therefore, 3 = 2x.

Solve for x: x = 3/2.

Substitute x and y back into the expression for z from step 2:

z = (x(3/2) - 2(3/2)) / 2

Simplify: z = (3/2 - 3) / 2

Therefore, z = -3/4.

Therefore, in the solution where the system of equations has exactly one solution, z is -3/4.

Here are the steps for problem 2:

Find the equation of the perpendicular line passing through A: The slope of the given line is 21​, and its negative reciprocal is −2.

Thus, the equation of the perpendicular line passing through A will be of the form y=−2x+b. We can plug A's coordinates into this equation to solve for b: $1=−2(7)+b⟹b=15.$

Therefore, the equation of the perpendicular line is y=−2x+15.

Find the intersection point of the two lines: The intersection point of the two lines will be the midpoint of the segment connecting A and its reflection B.

To find it, we can solve the system of equations: ${y=21​x+4y=−2x+15​$

Substituting the first equation into the second equation, we get: $21​x+4=−2x+15.$

Solving for x, we find x=35​. Substituting this value back into either equation, we find y=317​.

Therefore, the intersection point is (35​,317​).

Find the reflection point B: Since the reflection point B is the same distance from A as the intersection point is from A, we can find B by calculating the vector from A to the intersection point and multiplying it by 2.

The vector from A to the intersection point is (35​−7,317​−1)=(−316​,314​).

Multiplying this vector by 2, we get (−332​,328​). Finally, adding this vector to A's coordinates gives us the coordinates of B: (7,1) + (-32/3, 28/3) = (1/3, 31/3).

Therefore, B = (1/3, 31/3).

Factoring the equation, we get a2(b−c)+b(b2−c2)+c(c2−a2)=0. This can be rewritten as (b−c)(a2−ab+c)+b(b−c)(b+c)=0.

Therefore, (b−c)(a−b)(a−c)+b(b−c)(b+c)=(b−c)(a−b)(a−c)(b+c)=0. So b−c or a−b, a−c, or b+c must equal 0.

If b−c=0, then b=c. Since 0≤b,c≤5, it follows that b=c=0,1,2,3,4,5. In each case, we can choose a to be any integer between 0 and 5, inclusive.

This gives us 6⋅6=36​ possibilities.

Therefore, the answer is 36.

Let r be the radius of the circumcircles of triangles ABX and ACX. Since the circumcircles have the same radius, we know ∠A=∠C.

Since AX=13 and CX=4, we can use the Law of Sines on triangle ACX to find AC:

\begin{align*} \frac{\sin C}{AC} &= \frac{\sin A}{AX} \\

\sin C &= \frac{AC}{AX} \cdot \sin A \\

\sin C &= \frac{AC}{13} \cdot \sin C \\

AC &= 13 \end{align*}

Similarly, using the Law of Sines on triangle ABX to find AB:

\begin{align*} \frac{\sin B}{AB} &= \frac{\sin A}{AX} \\ \sin B &= \frac{AB}{AX} \cdot \sin A \\ \sin B &= \frac{AB}{13} \cdot \sin A \\ AB &= 13 \end{align*}

Now that we know all side lengths of triangle ABC, we can use Heron's formula to find the area:

\begin{align*} \text{Area of } ABC &= \sqrt{s(s-AB)(s-AC)(s-BC)} \\ &= \sqrt{(13+13+4)/2 \cdot (13+4-13)/2 \cdot (13-4-10)/2 \cdot (4-10-13)/2} \\ &= \sqrt{30 \cdot 0 \cdot -3 \cdot -17} \\ &= \boxed{91} \end{align*}

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.