LiIIiam0216

To solve for the length of \( DE \) in quadrilateral \( BCED \) where sides \( BD \) and \( CE \) are extended to meet at point \( A \), we can apply the **segment addition** and **proportionality** of segments created by the transversal lines.

### Step 1: Analyze the Problem

We know the following lengths:

- \( BD = 18 \)

- \( BC = 8 \)

- \( CE = 2 \)

- \( AC = 7 \)

- \( AB = 3 \)

### Step 2: Find Lengths

From the information provided, we can calculate the total lengths \( AB \) and \( AC \):

- Length \( BA = AB + BD = 3 + 18 = 21 \)

- Length \( CA = AC + CE = 7 + 2 = 9 \)

### Step 3: Apply the Segment Proportionality

Since \( BD \) and \( CE \) are extended to meet at point \( A \), we can apply the segment proportion theorem for triangles created by the intersection:

From triangles \( ABE \) and \( CDE \):

\[
\frac{AB}{AC} = \frac{DE}{BC}
\]

### Step 4: Substitute Known Values

We can substitute the known lengths into the proportion:

\[
\frac{3}{7} = \frac{DE}{8}
\]

### Step 5: Solve for \( DE \)

Cross-multiply to solve for \( DE \):

\[
3 \cdot 8 = 7 \cdot DE
\]

\[
24 = 7 \cdot DE
\]

\[
DE = \frac{24}{7}
\]

Thus, the length of \( DE \) is \( \frac{24}{7} \).

### Final Answer

So, the length of \( DE \) is:

\[
\boxed{\frac{24}{7}}
\]

We are asked to determine how many distinct paths can be followed to spell the word "MATH" starting from the origin \( M \), given that movements are only allowed up, down, left, and right.

The points corresponding to \( A \), \( T \), and \( H \) are labeled on the xy-plane, and each of these labels corresponds to a specific set of coordinates.

### Step 1: Understanding the Problem

The problem provides:

- \( M \) is at the origin, \( (0, 0) \).

- \( A \)'s are at \( (1,0) \), \( (-1,0) \), \( (0,1) \), and \( (0,-1) \).

- \( T \)'s are at \( (2,0) \), \( (1,1) \), \( (0,2) \), \( (-1,1) \), \( (-2,0) \), \( (-1,-1) \), \( (0,-2) \), and \( (1,-1) \).

- \( H \)'s are at \( (3,0) \), \( (2,1) \), \( (1,2) \), \( (0,3) \), \( (-1,2) \), \( (-2,1) \), \( (-3,0) \), \( (-2,-1) \), \( (-1,-2) \), \( (0,-3) \), \( (1,-2) \), and \( (2,-1) \).

We need to determine how many distinct paths can be followed to spell "MATH", moving from \( M \) to an \( A \), then from \( A \) to a \( T \), and finally from \( T \) to an \( H \).

### Step 2: Movement Considerations

We are allowed to move only up, down, left, and right. This restricts the possible movements between the points labeled \( M \), \( A \), \( T \), and \( H \).

- From \( M \) at \( (0, 0) \), we can move to any of the \( A \)'s at \( (1,0) \), \( (-1,0) \), \( (0,1) \), or \( (0,-1) \).

- From each \( A \), we can move to one of the \( T \)'s that are one unit away from the \( A \)'s.

- From each \( T \), we can move to one of the \( H \)'s that are one unit away from the \( T \)'s.

### Step 3: Counting the Distinct Paths

Let’s break down the path counting process step by step.

#### Paths from \( M \) to \( A \):

From \( M = (0, 0) \), there are 4 possible \( A \)'s:

- \( A_1 = (1, 0) \)

- \( A_2 = (-1, 0) \)

- \( A_3 = (0, 1) \)

- \( A_4 = (0, -1) \)

So, there are 4 choices for the first step.

#### Paths from \( A \) to \( T \):

From each \( A \), we can move to a neighboring \( T \) that is one unit away. Let's examine the options for each \( A \):

- From \( A_1 = (1, 0) \), the possible \( T \)'s are \( (2, 0) \), \( (1, 1) \), and \( (1, -1) \). This gives 3 choices.

- From \( A_2 = (-1, 0) \), the possible \( T \)'s are \( (-2, 0) \), \( (-1, 1) \), and \( (-1, -1) \). This gives 3 choices.

- From \( A_3 = (0, 1) \), the possible \( T \)'s are \( (0, 2) \), \( (1, 1) \), and \( (-1, 1) \). This gives 3 choices.

- From \( A_4 = (0, -1) \), the possible \( T \)'s are \( (0, -2) \), \( (1, -1) \), and \( (-1, -1) \). This gives 3 choices.

Thus, for each \( A \), there are 3 possible \( T \)'s, so the total number of ways to move from \( A \) to \( T \) is \( 3 \times 4 = 12 \).

#### Paths from \( T \) to \( H \):

From each \( T \), we can move to a neighboring \( H \) that is one unit away. There are 3 neighboring \( H \)'s for each \( T \) (similarly to the calculation above). Therefore, for each \( T \), there are 3 possible \( H \)'s.

Thus, for each \( T \), there are 3 possible \( H \)'s, so the total number of ways to move from \( T \) to \( H \) is \( 3 \times 12 = 36 \).

### Step 4: Total Number of Paths

Multiplying the number of choices at each step, we get:

\[
4 \times 3 \times 3 = 36.
\]

Thus, the total number of distinct paths that can be followed to spell the word "MATH" is \( \boxed{36} \).

The answer is BD = 15*sqrt(2) + 8.

The shortest altitude is 15.

The area is 16*sqrt(7).

To determine how many points \((x, y)\) where both \(x\) and \(y\) are positive integers lie below the hyperbola \(xy = 16\), we need to find the integer pairs \((x, y)\) such that \(xy < 16\).

### Step-by-Step Solution:

1. **Consider values of \(x\) and find corresponding \(y\) values**:

   For each \(x\), \(y\) must satisfy \(1 \leq y < \frac{16}{x}\).

2. **Calculate pairs for each \(x\)**:

   - \(x = 1\):

     \[
     xy < 16 \implies y < \frac{16}{1} \implies y < 16 \implies y = 1, 2, 3, \ldots, 15 \quad (\text{15 values})
     \]

   - \(x= 2\):
     \[
     xy < 16 \implies y < \frac{16}{2} \implies y < 8 \implies y = 1, 2, 3, \ldots, 7 \quad (\text{7 values})
     \]

   - \(x = 3\):
     \[
     xy < 16 \implies y < \frac{16}{3} \implies y < 5.33 \implies y = 1, 2, 3, 4, 5 \quad (\text{5 values})
     \]

   - \(x = 4\):
     \[
     xy < 16 \implies y < \frac{16}{4} \implies y < 4 \implies y = 1, 2, 3 \quad (\text{3 values})
     \]

   - \(x = 5\):
     \[
     xy < 16 \implies y < \frac{16}{5} \implies y < 3.2 \implies y = 1, 2, 3 \quad (\text{3 values})
     \]

   - \(x = 6\):
     \[
     xy < 16 \implies y < \frac{16}{6} \implies y < 2.67 \implies y = 1, 2 \quad (\text{2 values})
     \]

   - \(x = 7\):
     \[
     xy < 16 \implies y < \frac{16}{7} \implies y < 2.29 \implies y = 1, 2 \quad (\text{2 values})
     \]

   - \(x = 8\):
     \[
     xy < 16 \implies y < \frac{16}{8} \implies y < 2 \implies y = 1 \quad (\text{1 value})
     \]

   - \(x = 9\) and higher:
     \[
     xy < 16 \implies y < \frac{16}{x} \implies y < \frac{16}{x} \implies y = 1 \quad (\text{1 value if } x \leq 15 \text{ else no values})
     \]

3. **Count the total number of pairs**:

   Summing all the valid \(y\) values for each \(x\):
   \[
   15 + 7 + 5 + 3 + 3 + 2 + 2 + 1 + 1 = 39
   \]

Therefore, there are \( \boxed{39} \) points of the form \((x, y)\) where both coordinates are positive integers and lie below the hyperbola \(xy = 16\).

(a) Let w = a + bi and z = c + di.  The rest is expanding.

(b) Let w = a + bi and z = c + di.  Then

\[\dfrac{w + z}{1 + wz} = \dfrac{a + c + bi + di}{1 + (a + bi)(c + di)}\]

To express this in rectangular form, we can multiply the numerator and denominator by the conjugate:

\[\dfrac{a + c + bi + di}{1 + (a + bi)(c + di)} = \dfrac{(a + c + bi + di)((1 - (a + bi)(c + di))}{(1 + (a + bi)(c + di))(1 - (a + bi)(c + di))}\]

The denominator simplifies to (1 - (a^2 + b^2)(c^2 + d^2)), which is real.  The numerator simplifies to a^2 - b^2 + c^2 - d^2, which is also real.  Therefore, the complex number (z + w)/(zw + 1) is real.

The probability is 9/13.

The probability is 3/8.