All Answers 358504 Answers

To find the number of different sequences of rolls that could have resulted in a product of 2700, we can consider the prime factorization of 2700.

The prime factorization of 2700 is:

2700 = 2^2 * 3^3 * 5^2

Since we are rolling a standard 6-sided die, the possible outcomes for each roll are numbers 1 to 6. We need to distribute the prime factors among the six rolls.

Let's consider the factor 2. Since 2^2 is one of the prime factors, we need to assign it to two of the rolls. There are 6C2 ways to select two rolls out of six, which can be calculated as:

6C2 = 6! / (2! * (6 - 2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15

Similarly, for the factor 3, we have 3^3. We need to assign three 3's to three of the rolls. There are 6C3 ways to select three rolls out of six, which can be calculated as:

6C3 = 6! / (3! * (6 - 3)!) = 6! / (3! * 3!) = (6 * 5 * 4) / (3 * 2 * 1) = 20

For the factor 5, we have 5^2. We need to assign two 5's to two of the rolls. There are 6C2 ways to select two rolls out of six, which is the same as the selection for the factor 2, so it is also 15.

To find the total number of different sequences, we multiply these counts together:

Total number of sequences = 15 * 20 * 15 = 4500

Therefore, there could have been 4500 different sequences of rolls that would result in a product of 2700.

There are no restrictions on the numbers in the first and last boxes, so they can be filled in in 5 ways each. However, the number in the middle box must be 3, and the number in the middle box must be either 2 or 4.

If the number in the middle box is 2, then the number in the third box must be 1, and the number in the fourth box must be 4. This can be done in 3 ways.

If the number in the middle box is 4, then the number in the third box must be 3, and the number in the fourth box must be 2. This can also be done in 3 ways.

Therefore, there are a total of 5⋅5+3+3=31​ ways to fill in the boxes so that all the inequalities are true.

2700 = 2^2 * 3^3 * 5^2

There are a total of 6 combinations as follows:

125569==6! / 2! ==360 permutations

134559==6! / 2! ==360 permutations

135566==6! / 2!2! ==180 permutations

223559==6! /2!2! ==180 permutations

233556==6! /2!2! ==180 permutations

333455==6! /3!2! ==60    permutations

Total ==360+360+180+180+180+60==1,320 different sequences of rolls.

The prime factorization of 2700 is 2^2⋅3^3⋅5^1. This means that Catherine rolled a 2 twice, a 3 three times, and a 5 once. The order of the rolls matters, so there are 6!/2!3!=30 possible sequences of rolls.

Here are the 30 possible sequences:

1. 2 2 3 3 3 5 2. 2 2 3 3 5 3 3. 2 2 3 5 3 3 4. 2 2 5 3 3 3 5. 2 3 2 3 3 5 6. 2 3 2 3 5 3 7. 2 3 2 5 3 3 8. 2 3 5 2 3 3 9. 2 3 5 3 2 3 10. 2 3 5 3 3 2 11. 2 5 2 3 3 3 12. 2 5 3 2 3 3 13. 2 5 3 3 2 3 14. 2 5 3 3 3 2 15. 3 2 2 3 3 5 16. 3 2 2 3 5 3 17. 3 2 2 5 3 3 18. 3 2 3 2 3 5 19. 3 2 3 2 5 3 20. 3 2 3 5 2 3 21. 3 2 3 5 3 2 22. 3 2 5 2 3 3 23. 3 2 5 3 2 3 24. 3 2 5 3 3 2 25. 3 5 2 2 3 3 26. 3 5 2 3 2 3 27. 3 5 2 3 3 2 28. 3 5 3 2 2 3 29. 3 5 3 2 3 2 30. 3 5 3 3 2 2

I hope this helps!

To find the values of a and b, we can use the relationship between the roots and coefficients of a quadratic equation.

In a quadratic equation of the form p(x) = x^2 + ax + b, the sum of the roots is equal to the negation of the coefficient of x (a), and the product of the roots is equal to the constant term (b).

Given that the distinct roots are 2a and 2b, we have the following relationships:

Sum of the roots: 2a + 2b = -a (1) Product of the roots: (2a)(2b) = b (2)

From equation (1), we can simplify it by moving all terms to one side:

2a + 2b + a = 0 3a + 2b = 0

Now, let's solve equations (1) and (2) simultaneously:

From equation (2), we have (2a)(2b) = b, which can be rewritten as 4ab = b. We can divide both sides by b (assuming b is nonzero) to obtain 4a = 1. Dividing both sides by 4, we get a = 1/4.

Thank you... but the answer is actually -1/4.. still appreciate the effort!