AUnVerifedTaxPayer

The answer is (8/5, -1/5).

The answer is BC = 15/2.

Angle BID = 32 degrees.

CE = 5*sqrt(3).

Let's analyze the problem step by step:

Understanding the Problem:

We have a 4x4 grid.

We need to place 8 counters.

Each row and column must have exactly 2 counters.

Approach: We can solve this problem using a combinatorial approach. We'll break it down into smaller subproblems.

Choosing Rows and Columns:

We need to choose 2 rows out of 4 to place counters. This can be done in C(4,2) ways.

Similarly, we need to choose 2 columns out of 4. This can also be done in C(4,2) ways.

Placing Counters:

Once we've chosen the rows and columns, we need to place the counters within those specific rows and columns.

For each chosen row, there are C(4,2) ways to choose the two columns where the counters will go.

Similarly, for each chosen column, there are C(4,2) ways to choose the two rows.

Calculating the Total Number of Ways:

Multiplying the number of ways for each step, we get: C(4,2) * C(4,2) * C(4,2) * C(4,2)

Calculating the Combinations:

C(4,2) = 4! / (2! * 2!) = 6

Final Calculation:

Total number of ways = 6 * 6 * 6 * 6 = 1296

Therefore, there are 1296 ways to place the 8 counters in the 4x4 grid, satisfying the given conditions.

For part (a), the number of ways is 540.

For part (b), the number of ways is 838.

We can approach this problem by systematically checking prime numbers and their corresponding values of A and B.

1. Small Prime Numbers:

For P = 2:

We need to find A and B such that A^2 + 2^3 = B^4.

This simplifies to A^2 + 8 = B^4.

By trial and error, we find that A = 1 and B = 3 satisfy this equation.

2. Larger Prime Numbers:

As P increases, the value of B^4 grows much faster than A^2 + P^3.

This means that for larger primes, it becomes increasingly difficult to find integer solutions for A and B.

In fact, it can be shown that there are no other solutions for larger prime numbers.

Therefore, the only possible value of A is 1, corresponding to the prime number P = 2.