ABJeIIy

The answer is f(17) = -21.

For this problem, you must consider magnitude and argument of complex numbers geometrically.

The answers are as follows:

r_1 = 16

theta_1 = 3*pi/2

r_2 = sqrt(3)

theta_2 = 5*pi/4

r_3 = 12

theta_3 = pi/3

For this problem, you must consider magnitude and argument of complex numbers geometrically.

The answers are as follows:

r_1 = 16

theta_1 = 3*pi/2

r_2 = sqrt(3)

theta_2 = 5*pi/4

r_3 = 12

theta_3 = pi/3

For this problem, you must consider magnitude of complex numbers geometrically.

For the first part, the answers are |v| = 2, |w| = 3, |z| = 3.

For the second part, the answers are |v + w| = sqrt(2), |w + z| = 2, |z + v| = sqrt(3).

For this problem, you must consider magnitude of complex numbers geometrically.

For the first part, the answers are |v| = 2, |w| = 3, |z| = 3.

For the second part, the answers are |v + w| = sqrt(2), |w + z| = 2, |z + v| = sqrt(3).

For this problem, you must consider addition and subtraction of complex numbers geometrically.  The answer turns out to be s, t, u.

The coordinates of the point are (2/3*(-6) + 1/3*(15), 2/3*19 + 1/3*(-5)) = (1,11).

Let's break down the expression step by step.

sqrt(49 - 20*sqrt(6)) = ?

First, we can start by evaluating the expression inside the parentheses:

49 - 20*sqrt(6) = ?

We can use the distributive property to multiply 20 by sqrt(6):

49 - 20*sqrt(6) = 49 - 20(sqrt(6))

Now, we can simplify this expression by recognizing that 49 is a perfect square:

49 = 7^2

So, we can rewrite the expression as:

49 - 20(sqrt(6)) = 7^2 - 20(sqrt(6))

Next, we can use the difference of squares formula to simplify further:

a^2 - b^2 = (a + b)(a - b)

In this case, a = 7 and b = sqrt(6), so we get:

7^2 - sqrt(6)^2 = (7 + sqrt(6))(7 - sqrt(6))

Now, we can simplify the expression by combining like terms:

(7 + sqrt(6))(7 - sqrt(6)) = ?

To combine these terms, we can use the fact that sqrt(x)^2 = x:

7 + sqrt(6) and 7 - sqrt(6) are both factors of (7)^2 - (sqrt(6))^2

So, we can factor out a common factor of (7 - sqrt(6)):

= (7 - sqrt(6))(7 + sqrt(6))

And that's the simplified form of the original expression!

sqrt(49 - 20*sqrt(6)) = 7 - sqrt(6)

Now, we want to find the ratio PQ/FH. We can find the lengths of PQ and FH by using similar triangles. Let's look at triangle HFG and triangle GNF.

Since HM is parallel to GN, we have angle NFG = angle GFM. Additionally, since FM/MG = 1/2, we have FG = 2GF. Similarly, using triangle HEN and triangle EMF, we find that EH = 3HE.

Now we see that FG:GF = EH:HE = 3:2. Therefore, triangles HFG and GNF are similar.

From here, we know that PQ/FH = QF/FH = HN/HF. By letting HF = a and HN = b, we have a+b = FH. Since HN:NE = 1:3, b/3y = 1/3, b = y. Similarly, we know that a/x = 2x/x = 2, a = 2x.

Now we have PQ/FH = HN/HF = y/(a+b) = y/(2x+y). Let Y = PQ/FH = y/(2x+y). Since y = HN = 1 and x = FM = 1, we have Y = 1/(2+1) = 1/3.

Therefore, the ratio PQ/FH is 1/3.

There are 25,480 ways of coloring the squares.

Problem 2:

We can solve this problem by analyzing the factored expressions and using casework to explore the possible values of x and y.

Factoring the Expressions:

(2x^2 + 4x + 8) can be factored as 2(x^2 + 2x + 4). We can further factor (x^2 + 2x + 4) by noticing it's a perfect square trinomial. So, 2(x^2 + 2x + 4) = 2(x + 2)^2.

(3y^2 - 6y + 7) can be factored as 3(y^2 - 2y + 7/3). This trinomial cannot be factored further with integer coefficients.

Setting Up the Cases: The equation is: 2(x + 2)^2 * 3(y^2 - 2y + 7/3) = 24

We can rewrite it as: (x + 2)^2 * (y^2 - 2y + 7/3) = 4 ---------(1)

Since 4 can be factored as 1 * 4 or 2 * 2, we will consider two cases based on these factorizations:

Case 1: (x + 2)^2 = 1 and (y^2 - 2y + 7/3) = 4

Case 2: (x + 2)^2 = 4 and (y^2 - 2y + 7/3) = 1

Solving Case 1:

(x + 2)^2 = 1 --> x + 2 = ±1 --> x = -1 or x = -3

(y^2 - 2y + 7/3) = 4 --> This quadratic factors as (y - 1)(y - 7/3) = 0. So, y = 1 or y = 7/3.

Therefore, in Case 1, we have four possible ordered pairs: (-1, 1), (-1, 7/3), (-3, 1), and (-3, 7/3).

Solving Case 2:

(x + 2)^2 = 4 --> x + 2 = ±2 --> x = 0 or x = -4

(y^2 - 2y + 7/3) = 1 --> This quadratic has no real number solutions because the discriminant (b^2 - 4ac) is negative.

Therefore, Case 2 has no real number solutions for (x, y).

Total Ordered Pairs: Combining the results from both cases, we have a total of 4 possible ordered pairs from Case 1: (-1, 1), (-1, 7/3), (-3, 1), and (-3, 7/3).

Answer: There are 4 possible ordered pairs (x, y) that satisfy the given equation.

Problem 1:

Let's analyze each statement and see if it's necessarily true given the conditions:

(A) a + c > b + d: Since a > b and c > d, adding them together would maintain this inequality. So, A is true.

(B) 2a + 3c > 2b + 3d: Multiplying both sides of a > b and c > d by positive constants (2 and 3) won't change the direction of the inequality. So, B is true.

(C) a - c > b - d: Subtracting the same number (c) from both sides of a > b doesn't necessarily guarantee a - c > b - d. It depends on the relative values of a, b, and c. So, C is not necessarily true.

(D) ac > bd: Since a > b and c > d, multiplying them together would maintain the inequality if both a and c have the same sign (both positive or both negative). However, if a and c have opposite signs, the product could be negative. So, D is not necessarily true.

(E) a^2 + c^2 > b^2 + d^2: Squaring a positive number increases its value, and squaring a negative number makes it positive. Since a > b and c > d, squaring them preserves the inequality as long as a and c have the same sign. So, E is true.

(F) a^3 + c^3 > b^3 + d^3: Cubing a positive number increases its value significantly, and cubing a negative number makes it much more negative. However, the behavior depends on the signs of a and c relative to b and d. It's not guaranteed that a^3 + c^3 will always be greater than b^3 + d^3. So, F is not necessarily true.

Therefore, the statements that must be true are (A), (B), and (E). The answer is A, B, E.

I found the answer: (a,b,c,d,e) = (4,2,-3,-1,8).