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The radius of the circle is 35.

The area of triangle ABC is 50 \sqrt{3}.

There are two cases to consider:

Case 1: Two cards are from the same color and one card from a different color.

Choose one of the four colors for the two matching cards: There are 4 ways to do this.

Choose two out of the seven cards of that chosen color: There are ⁷C₂ ways to do this (combination, order doesn't matter).

Choose one card from each of the remaining three colors: There are 3 * 3 * 3 = 27 ways to do this (since there are 3 colors left and 3 cards of each color).

Case 2: Three cards are from different colors.

Choose one card from each of the four colors: There are 4 ways to do this.

Adding the number of ways for each case:

Total hands = (Ways in Case 1) + (Ways in Case 2)

Total hands = (4 * ⁷C₂ * 27) + 4

We can calculate ⁷C₂ using the combination formula: ⁷C₂ = ⁷! / (2! * (⁷ - 2)!) = 21

Total hands = (4 * 21 * 27) + 4

Total hands = 2520 + 4

Total hands = 2524​

The answer is 2524.

We can think about this problem in two steps:

Step 1: Arrange the mathematicians and scientists as groups.

Since the mathematicians must sit together and the scientists must sit together, we can treat them as single units for now. This leaves us with 3 units (mathematicians, scientists, journalists).

There are 3! ways to arrange these 3 units around the circular table.

Step 2: Arrange the people within each group.

The mathematicians can arrange themselves in 4! ways within their designated section.

The scientists can arrange themselves in 3! ways within their designated section.

However, we've overcounted the arrangements because rotating the entire table doesn't create a new seating arrangement.

Step 3: Correct for overcounting due to circular symmetry.

Since there are 9 people sitting around the table, there are 9 ways to rotate the table so it appears as a new arrangement.

Total Arrangements:

To get the final number of arrangements, we take the number of arrangements from steps 1 and 2 and divide by the number of times we've overcounted in step 3:

Arrangements = (Arrangements of groups) * (Arrangements within groups) / (Overcounting)

Arrangements = (3!) * (4! * 3!) / (9)

Arrangements = 6 * 24 * 6 / 9

Arrangements = 864 / 9

Arrangements = 96​

The answer is 96.

We can solve this problem by considering the two unwanted scenarios and subtracting them from the total number of possibilities.

Total number of ways:

There are 3 choices (flavors) for each child, so there are 36=729 total ways for all 6 children to choose their ice cream flavors with no restrictions.

Unwanted Scenario 1: No flavor chosen by exactly 3 children

Choose the flavor that isn't picked by exactly 3: There are 3 ways to choose which flavor will be picked by only 2 or none of the children.

Choose the 3 children who get the non-chosen flavor: There are (36​)=20 ways to choose 3 children out of 6.

Choose the flavors for the remaining 3 children: Each of these children has 2 remaining flavors to choose from, for a total of 23=8 possibilities.

There are 3⋅20⋅8=480​ ways for all flavors to be chosen by either 0, 1, 2, or all 6 children.

Unwanted Scenario 2 (not necessary but can be solved for completeness):

All 3 flavors are chosen by exactly 2 children each. This can be solved similarly to scenario 1, but the calculations are slightly more involved.

Final Answer:

Since we only care about scenarios where exactly one flavor is chosen by 3 children, we subtract the unwanted scenarios from the total:

Total ways - Unwanted Scenario 1 (or Scenario 2) = 729−480 = 249​ ways.

There are two ways to tackle this problem:

Method 1: Counting Valid Triangles

Total Points: We have 3 choices for x (1, 2, or 3) and 8 choices for y (3, 4, 5, ..., 10), resulting in a total of 3 * 8 = 24 points.

Choosing 3 Points: We need to choose 3 out of these 24 points to form a triangle.

The number of ways to do this can be calculated using combinations: nCr = n! / (r! * (n-r)!) where n is the total number of elements (24) and r is the number of elements to choose (3). 24C3 = 24! / (3! * (24 - 3)!) = 2280

Not All Triangles are Valid: However, not all combinations of 3 points will form valid triangles.\

The triangle inequality states that the sum of any two sides of a triangle must be greater than the third side. We need to check how many of the 2280 combinations violate this rule.

Loop through each combination of 3 points.

Calculate the distances between all three pairs of points using the distance formula (square root of the sum of squared differences in x and y coordinates).

If any of the distances violate the triangle inequality (i.e., the sum of two distances is less than the third distance), discard that combination.

Valid Triangles: After checking all combinations, count the remaining ones that satisfy the triangle inequality. This will be the number of valid triangles.

Method 2: Faster Approach (Counting Invalid Triangles)

Total Combinations: Same as method 1, there are 2280 total ways to choose 3 points from 24.

Invalid Triangles: We can count the number of triangles that violate the triangle inequality and subtract them from the total to get the valid ones. There are two main cases for invalid triangles:

All points on a straight line: If all three chosen points fall on the same line (e.g., (1, 3), (2, 3), (3, 3)), they cannot form a triangle.

Degenerate Triangle: If two points have the same x-coordinate or the same y-coordinate, and the third point does not lie above the line connecting them, they cannot form a valid triangle. (Imagine a straight line segment connecting the two points with the same coordinate. The third point must be above this line segment for a triangle to form).

Count the number of ways to choose 3 points that fall on the same horizontal or vertical line (3 choices for x-coordinate * 8 choices for y-coordinate = 24)

Count the number of ways to choose 2 points with the same x-coordinate (3 choices for x-coordinate * 7 choices for y-coordinate for the first point * 6 choices for y-coordinate for the second point, excluding the chosen y-coordinate = 126).

Do the same for points with the same y-coordinate.

Valid Triangles: The total number of valid triangles is then: Total Combinations - Invalid Triangles 2280 - (24 + 126 + 126) = 2004

Both methods will give you the answer: there are 2004​ triangles that can be formed using 3 of the points.

The key to this problem is finding the least common multiple (LCM) of 360, 450, and 540. This represents the smallest number of days that will take for all three planets to complete one full rotation around the sun relative to their starting positions.

Here's why:

If all planets complete their rotations in multiples of their individual periods (e.g., 360 days for X, 900 days for Y, etc.), they might not be lined up

again.

The LCM ensures that after that specific number of days, X will have completed exactly m rotations (for some integer m), Y will have completed exactly n rotations (for some integer n), and Z will have completed exactly o rotations (for some integer o).

To find the LCM, we can use various methods. In this case, since the numbers are relatively close, you might be able to find it by inspection.

However, a more general approach is to use the Euclidean Algorithm:

Find the greatest common divisor (GCD) of the largest two numbers (here, 450 and 540).

Divide the LCM of the initial three numbers (which doesn't exist yet) by the GCD you just found. This gives you the LCM of the first two numbers.

Now, find the GCD of the LCM you obtained in step 2 and the remaining number (here, 360).

The final LCM is the product of the GCDs you found in steps 1 and 3.

Following these steps, you'll find the GCD of 450 and 540 to be 90. The LCM of 450 and 540 is then (450 * 540) / 90 = 300.

Finally, the GCD of 300 and 360 is 60. Therefore, the LCM of 360, 450, and 540 is (300 * 60) = 18000.

However, 18000 is not the minimum positive number of days. Why? Because all three planets might be lined up again at a smaller common multiple. We need to find the least common multiple.

Looking closer, we see that 540 (Z's rotation period) is divisible by both 360 (X's period) and 450 (Y's period).

This means that every time Z completes a full rotation, X will have completed some integer number of rotations and Y will have completed some other integer number of rotations. In other words, they will all be lined up again after every 540 days.

Therefore, the minimum positive number of days before they are on the same line again is 540​ days.

Understanding the Slopes:
 

Slope of line segment OP = 1: This means for every 1 unit you move up on the y-axis, you also move 1 unit to the right on the x-axis.

Since O is the origin (0,0), point P will be at (x, x) for some positive value x.

Slope of line segment PQ = 3: This means for every 3 units you move up on the y-axis, you move 1 unit to the right on the x-axis.

Finding Coordinates of P:

As mentioned earlier, P lies at (x, x).

Connecting P and Q:

We know the slope of PQ is 3. We need to find the y-coordinate of point Q relative to P (i.e., how much we move up from P). Let the y-coordinate of Q be y.

Relating Coordinates using Slope:

The difference in y-coordinates (y - x) is 3 times the difference in x-coordinates (which is just x in this case).

Setting up the Equation:

y - x = 3x

Combining Like Terms:

y = 4x

Finding the Slope of PQ:

Slope of PQ = (Change in y) / (Change in x) = (y_Q - y_P) / (x_Q - x_P)

Since both P and Q lie on the positive x-axis (first quadrant), x_Q and x_P are both positive values. Therefore, we can divide both sides of the equation y = 4x by x (assuming x is not zero) to get:

y/x = 4

This gives us the slope of line segment PQ, which is 4.

The degree of the polynomial p + q depends on the terms of both p and q. There are two main scenarios to consider:

No common factors: If p and q have no common factors (i.e., no variable raised to the same power in both polynomials), then the degree of p + q will simply be the larger of the two individual degrees.

In this case, p has a degree of 11 and q has a degree of 7. So, the degree of p + q would be 11.

Common factors: If p and q share some common factors (i.e., a variable raised to the same power appears in both), then the degree of p + q can be lower than the highest individual degree.

However, the degree can only be lowered by at most 1. This happens when the terms with the common factors cancel out when added together.

In this scenario, the possible degree of p + q would be between 10 (1 less than the highest degree) and 11.

Therefore, depending on the specific form of p and q, the possible degrees of the polynomial p + q are:

11: If there are no common factors between p and q.

10: If there is a common factor that cancels out upon addition.