Solution Of System Of Linear Inequalities In Two Variables

Step 1: Knowing , the points and with form a right triangle where and are perpendicular. Step 2: Use the circle theorem that states the angle in a semi-circle is a right angle. Thus, lies on the circle with diameter .
Step 3: Calculate the equation of the circle having diameter . - The midpoint of is the average of the coordinates of and : - The radius is the distance from to (or to , which will be the same): - The equation of the circle with center and radius is: Step 4: Expand and simplify the equation of the circle: Expanding: Multiply through by 4 to eliminate the fractions: Simplifying further: Finally, divide through by 4 to simplify:

We are given 8 men and 6 women, and we need to form a committee of 10 members in such a way that the men are in the majority. Since the committee must consist of 10 members and the men must be in the majority, the committee must have at least 6 men and at most 7 men.
Step 1: We will consider two cases based on the number of men in the committee:
Case 1: 6 men and 4 women.
The number of ways to select 6 men from 8 is given by: The number of ways to select 4 women from 6 is given by: Thus, the total number of committees in this case is: Case 2: 7 men and 3 women.
The number of ways to select 7 men from 8 is given by: The number of ways to select 3 women from 6 is given by: Thus, the total number of committees in this case is:
Step 2: The total number of ways to form the committee with the men in majority is the sum of the results from the two cases: Conclusion:
Thus, the number of different possible committees in which the men are in the majority is .

We are given the equation: The general solution set of this equation involves finding values of that satisfy it.
Step 1: We use the known identity for the sum of sines: Applying this identity to the terms , we get:
Step 2: Now, the equation becomes: Factor out :
Step 3: This gives us two cases to solve: 1. , 2. . For the first case, , the solutions are: For the second case, , which has no real solutions.
Step 4: Thus, the general solutions for are of the form .
Now, we calculate the possible values of for these solutions. Since the sine function takes values between and , the possible values of are:

We are given that we have 3 men and 3 women, and we need to arrange them in a row of 6 seats. The first and last seats must be filled by men.
Step 1: The first and last seats must be occupied by men. Since we have 3 men, we can choose a man for the first seat in 3 ways, and then we can choose a man for the last seat in 2 ways (since one man has already been seated). Therefore, the number of ways to arrange men in the first and last seats is:
Step 2: After placing the men in the first and last seats, we have 4 seats remaining. These 4 seats must be filled by the remaining 1 man and 3 women. The number of ways to arrange these 4 people in the remaining 4 seats is:
Step 3: Thus, the total number of ways to arrange the 3 men and 3 women, with the first and last seats occupied by men, is:

Step 1: Start by identifying the transformation that eliminates the first degree terms in the equation . - The first-degree terms are . To eliminate these terms, we need to shift the origin to a new point where the first degree terms vanish. Step 2: To remove the linear terms , we complete the square for and to find the coordinates of the new origin . The shift required for and will be: Thus, the new coordinates are and .
Step 3: Substitute the new coordinates into the second equation , where and . Expanding and simplifying this equation will give the transformed equation in the new coordinate system.
Step 4: After substituting and simplifying the terms, we obtain the transformed equation:

Step 1: Calculate the total number of different possible answers a student can give.
- Since there are 3 questions each with 4 options, the total possible combinations are .
Step 2: Adjust for the constraint that no student can answer all questions correctly.
- Subtract the one combination where all answers are correct, leaving possible ways to answer.
Step 3: Given that no two students answer identically, the maximum number of students who could have written the test without any of them having all correct answers is therefore .