Algebra

To solve this problem, we need to use the binomial theorem to find the expansion terms and apply that information to solve for the coefficient in another expansion.

The expression given is , and we're interested in the equality of its 17th and 18th terms.

The general term in the binomial expansion of is given by:

For , the 17th term will be:

The 18th term will be:

According to the problem, these two terms are equal:

Dividing both sides by , we get:

Simplifying, we find:

Using the identity , we simplify:

Now we use this value of for the expansion of , specifically the coefficient of . The expression is .

The general term for is:

For , the general term is:

Using the formula for negative indices:

We need (since ):

Calculate:

This simplifies through recognizing the negative factorial:

This gives:
hence the coefficient is .

Thus, the coefficient of is .

To find the ratio given that the arithmetic mean of two distinct positive real numbers and is twice their geometric mean, let's first define the arithmetic and geometric means:

The arithmetic mean (AM) of and is given by:

The geometric mean (GM) of and is given by:

According to the problem, the arithmetic mean is twice the geometric mean:

Multiply both sides by 2 to eliminate the fraction:

Square both sides to eliminate the square root:

Expand the left-hand side:

Rearrange the equation:

Now, we need to find the ratio . Let , which implies . Substitute into the equation:

Simplify:

Factor out :

Divide both sides by (since ):

Rearrange the quadratic equation:

Use the quadratic formula where , , and :

Only positive ratio makes sense with , so we select .

Thus, the ratio .

However, note that this ratio must be expressed by simplifying and checking which one it matches with options. Rationalize by dividing the ratio of calculated roots, at various trials:

Given and , calculate:

Calculate , multiply numerator and denominator by :

Thus, the correct answer is .

Step 1: {Find the domain restrictions}
For the function to be defined, the expression inside the square root must be non-negative: Also, the denominator must not be zero, i.e.,
Step 2: {Find the zeros of the numerator}
Solving: Factoring:
Step 3: {Find the zeros of the denominator}
Solving: Factoring:
Step 4: {Analyze sign changes using a number line}
The critical points partition the number line into intervals: By testing values in each interval, the function is non-negative in:
Step 5: {Conclusion}
Thus, the domain is:

Step 1: {Substituting }
Let Thus, the equation becomes:
Step 2: {Squaring both sides}
Expanding:
Step 3: {Solving for }
Rejecting as it violates the equation.
Step 4: {Finding }

Step 5: {Conclusion}
Thus, there are real solutions: .

Step 1: {Recall the property of orthogonal matrices}
A matrix is orthogonal if its transpose is equal to its inverse , i.e., . This implies that , where is the identity matrix.
Step 2: {Find the transpose of matrix }

Step 3: {Multiply and }

Step 4: {Set }
For , we must have:
Step 5: {Solve the resulting equations}
From the above equation, we get the following equations:

Step 6: {Solve for and }
Adding equations (1) and (2), we get , which gives . Substituting into equation (1), we get , which gives , so .
Step 7: {Verify the solution}
Substituting and into equation (3), we get , which is true. Therefore, and .

Step 1: Calculate the percentage of candidates who failed in either English or Mathematics or both
Percentage failed in English or Mathematics or both = Percentage failed in English + Percentage failed in Mathematics - Percentage failed in both = 62% + 42% - 20% = 84%
Step 2: Calculate the percentage of candidates who passed in both subjects
Percentage passed in both subjects = 100% - Percentage failed in English or Mathematics or both = 100% - 84% = 16%
Step 3: Assume the total number of candidates
Let the total number of candidates be .
Step 4: Calculate the number of candidates who passed in both subjects
Number of candidates who passed in both subjects = 16% of =
Step 5: Relate the number to the options
Since the options are whole numbers, we can assume for simplicity. Then, the number of candidates who passed in both subjects = .

This problem involves simplifying two terms raised to the eighth power. The terms inside the parentheses are structured such that they effectively represent hyperbolic cosine functions.
Step 1: Identify the simplification strategy




This corresponds to the highest power and matches option (C).

The problem involves maximizing the objective function subject to the constraints , with and .

Step 1: Identify the constraints and the feasible region.
We rewrite the constraint as the line equation . The feasible region is bounded by this line, the x-axis, and the y-axis, forming a triangle with vertices at the origin (0,0), a point on the x-axis, and a point on the y-axis.

Step 2: Find the intersection points with the axes.
For the x-axis , solve to get . Thus, the point is (4, 0).
For the y-axis , solve . Thus, the point is (0, 8).

Step 3: Evaluate the objective function at each vertex.
1. At (0, 0): .
2. At (4, 0): .
3. At (0, 8): .

The maximum value of at these points is 20.
Thus, the maximum value that can reach in the feasible region given the constraints is 20.

To find the distance between the two points on the ground, we use trigonometry involving angles of depression. Let's denote the points on the ground as and , with being the point closer to the building and the further point. Let the distance from the building to point be , and the distance from the building to point be .

The height of the building is . For the angles of depression, we use the fact that from the top of the building, the angle of depression is equal to the angle of elevation from the ground.

Step 1: Angle of depression to point

For point , the angle of depression is . Let :

Step 2: Angle of depression to point

For point , the angle of depression is . Let :

Step 3: Find the distance between points and

The distance between two points and is :

Thus, the distance between points and on the ground is .

• is a line with intercepts at and .
• The inequalities and indicate the feasible region is in the first quadrant.

To solve the problem, we must find the distance between two points on the ground, given the angles of depression from a 50 m tall building. We will use trigonometry to determine the horizontal distances from the building to each point.

Let's assume:

• Point A is the point with a angle of depression.
• Point B is the point with a angle of depression.

1. For Point A, with an angle of depression of , the horizontal distance from the building to Point A can be calculated using the tangent function: . Thus, the equation is:
   .
2. For Point B, with an angle of depression of , the horizontal distance to Point B is determined by: , giving us the equation:
   .
3. The total distance between the two points on the ground is the difference between these distances:
   .
4. Simplifying this, we compute:
   .

The distance between the two points is therefore: .