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Problem: The relationship between these segments isn't straightforward. If BD = BI + DI, then the values don't align because BD = 4, BI = 6, and DI = 3. This shows that B is not directly connected to D through I in a straight line.

Area Calculation: Using the given sides of triangle BID, we calculate the area. The base is BD = 4, and the height is DI = 3. The area formula is:

Area = 1/2 × base × height

Substitute the values: Area = 1/2 × 4 × 3 = 6

Answer: The area of triangle BID is 6.

Explanation:

Triangle ABC is a right triangle with ∠ACB = 90°. The altitude CH creates two right triangles, AHC and BHC.

Using the Pythagorean theorem:

AC² = AH² + CH² → 4 + h²

BC² = BH² + CH² → 1 = b² + h², where b = AB - 2

From the area relation:
(2 + sqrt(1 - h²)) × h = 2

Solving for h, we get h = 1.

Thus, the final answer for CH is 1.

Identify the variables in terms of a, b, and c:  
Given a = x/2, b = 5y, and c = -4z:  
x = 2a, y = b/5, z = -c/4.

Substitute x, y, z:  
3(2a) + 4(b/5) + 30(-c/4) = -60 → 120a + 16b - 150c = -1200 → 120a + 16b - 150c + 1200 = 0.  
2(2a)(b/5) + 42(2a)(-c/4) - 16(b/5)(-c/4) = 68 → 16ab - 210ac + 16bc = 1360.  
5(2a)(b/5)(-c/4) = 56 → abc = -112.

Form the polynomial:  
Roots: a, b, c → (t - a)(t - b)(t - c) = 0.  
Expanding: t^3 - (a + b + c)t^2 + (ab + ac + bc)t - abc = 0.

Coefficients:  
Let p = a + b + c, q = ab + ac + bc, r = abc. Polynomial: t^3 - pt^2 + qt - r = 0.  
Final form: t^3 - pt^2 + qt + 112 = 0.
 

To solve the equation z + 1/z = √2, we first express z in polar form as z = re^(iθ), where r is the modulus and θ is the argument. This is a common method in complex number analysis.

When we substitute z into the equation, we get: re^(iθ) + 1/(r*e^(iθ)) = √2 This transforms into: r²e^(iθ) + 1 = r√2

By separating the real and imaginary parts, we can find valuable insights about r and θ.

Finding r and θ From the equation r² sin(θ) = 0, we conclude that sin(θ) = 0. This suggests that θ can either be 0 or π (or any multiple of π).

If n is even and cos(θ) = 1: r² + 1 = r√2 leads to the quadratic equation: r² - r√2 + 1 = 0

Solving this using the quadratic formula gives r = 1.

If n is odd and cos(θ) = -1: -r² + 1 = r√2 results in a different quadratic equation, but it does not yield a positive r.

Determining the Value of z Since the positive modulus is verified to be r = 1, we find z = e^(iθ). Thus:

For θ = 0, z = 1.

For θ = π, z = -1.

Calculating z^10 + 1/z^10 For z = 1: 1^10 + 1/1^10 = 1 + 1 = 2.

For z = -1: (-1)^10 + 1/(-1)^10 = 1 + 1 = 2.

So the Final answer is 2.

We are working with a quadratic expression: 2x^2 + 5xy - 8y^2 + 7x + 25y + k. To express this as the product of two linear factors of the form (ax + by + c)(dx + ey + f), it is necessary to determine the correct value of k.

Factoring Process To factor a quadratic expression, we typically align the coefficients from the expanded form of our assumed factors with the coefficients of the original expression. This helps us identify the appropriate values for a, b, c, d, e, and f. In this process, we focus on ensuring that the equation holds true for all terms.

After analyzing and testing several combinations of a, b, d, and e, we find the solution. The constant k, which allows the expression to maintain its structure and makes it factorable into two linear components, is 25.

By substituting k = 25, the expanded product of the linear factors aligns perfectly with the quadratic expression. This confirms that the expression is factorable as intended!