To find the largest possible value of CF, we need to consider the relationship between altitudes and sides of a triangle. By using the formula CF = 2 * area(ABC) / AB, the largest possible value of CF occurs when AB is minimized. Therefore, CF will be maximum when AB is minimized to its smallest possible value.
To prove that a number is irrational, you need to show that it cannot be expressed as the quotient of two integers (i.e., it cannot be written in the form a/b, where a and b are integers and b is not equal to zero). One way to do this is to assume the number is rational and derive a contradiction.
To show that none of r, r^2, or r^3 is irrational, we can use a proof by contradiction. Assume that one of them (let's say r) is irrational. Then, we can write r as a/b, where a and b are integers and b is not equal to zero. We can also assume that a and b have no common factors, since if they did, we could simplify the fraction.
Now, let's substitute r into the equation x^3 - 2x + 5 = x^3 - x^2 + 9:
(r)^3 - 2(r) + 5 = (r)^3 - (r)^2 + 9
Simplifying this equation, we get:
(r)^3 - (r)^2 - 2(r) + 4 = 0
Multiplying both sides by b^3, we get:
a^3 - a^2b - 2ab^2 + 4b^3 = 0
This is a polynomial equation with integer coefficients. Since r is a root of this equation, a/b must also be a root. But we assumed that a and b have no common factors, so a^3 and b^3 have no common factors either. Therefore, by the rational root theorem, any rational root of this equation must be of the form ±p/q, where p is a factor of 4 and q is a factor of a^3.
However, we know that r is not rational, so it cannot be of this form. Therefore, we have a contradiction. This means that our assumption that r is irrational must be false, and r must be rational.
Using the same argument, we can show that r^2 and r^3 must also be rational. Therefore, none of r, r^2, or r^3 is irrational.
Let's label the six children A, B, C, D, E, and F. We can assume that the three pairs of siblings are AB, CD, and EF (without loss of generality).
First, let's consider the ways in which the first row can be filled. We have three choices for the leftmost chair: A, C, or E. Without loss of generality, let's say we choose A. Then, there are two choices for the middle chair: either D or F. Finally, there is only one choice for the rightmost chair, which is either C or E (whichever one was not chosen for the leftmost chair).
So there are 3 × 2 × 1 = 6 ways to fill the first row.
Now, let's consider the second row. We can't put any siblings directly in front of each other, so we have to be a bit more careful. Without loss of generality, let's say we start by putting B in the leftmost chair of the second row. Then, we can't put A in the middle chair directly in front of B, so we have to choose either C or E. Let's say we choose C. Then, we can't put D in the rightmost chair directly in front of C, so we have to put F there.
So the second row must be filled with the children B, C, and F, in that order. There are 3! = 6 ways to do this.
Putting it all together, there are 6 ways to fill the first row and 6 ways to fill the second row, for a total of 6 × 6 = 36 ways to seat the six children as described.
To solve this problem, first, you need to find the area of the right triangle XYZ using the formula A = 1/2 * base * height. Then, you can set up an equation to find the range of values for the height of triangle XYD, which will allow its area to be at most 20. Once you have the range of values, you can find the probability that a point chosen randomly in triangle XYZ lies within this range. The probability will be the ratio of the area of the range of heights to the area of the entire triangle XYZ. Dunkinrunsonyou
The smallest possible average of ten distinct, positive integers with a median of 5.5 is 4.95. This can be achieved by having the first five integers be 1, 2, 3, 4, and 5, and the last five be 6, 7, 8, 9, and 10. To calculate the average, we add all ten integers together and divide by 10: (1+2+3+4+5+6+7+8+9+10)/10 = 55/10 = 5.5. Since we want the smallest possible average, we need to minimize the sum of the ten integers. By having the first five integers start at 1 and increase by 1 up to 5, and having the last five integers start at 6 and increase by 1 up to 10, we achieve the smallest possible sum of 1+2+3+4+5+6+7+8+9+10 = 55, and therefore the smallest possible average of 4.95.