bader

The scale factor of a dilation is equal to the ratio of the length of the image of a segment to the length of the pre-image of that segment.

Since D is the midpoint of BC, the scale factor of the dilation is the ratio of the length of DE to the length of BC. Since the length of DE is half the length of BC, the scale factor of the dilation is 1/2.

We simplify the expression by simplifying each radical term separately. [ \begin{aligned} \frac{1}{\sqrt{36}} - \sqrt{27} - \frac{1}{\sqrt{27}} - \sqrt{18} + \frac{1}{\sqrt{18}} - \sqrt{9} =& \frac{1}{6} - 3\sqrt{3} - \frac{1}{3\sqrt{3}} - 3\sqrt{2} + \frac{1}{3\sqrt{2}} - 3\ =& \frac{1}{6} - 3(\sqrt{3} + \sqrt{2}) + \frac{1}{3}(\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{2}}) - 3\ =& \frac{1}{6} - 3(\sqrt{3} + \sqrt{2}) + \frac{1}{3}(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2}\sqrt{3}}) - 3\ =& \frac{1}{6} - 3(\sqrt{3} + \sqrt{2}) + \frac{\sqrt{2} + \sqrt{3}}{3} - 3\ =& \frac{1}{6} + \frac{\sqrt{2} + \sqrt{3}}{3} - 3 - 3(\sqrt{3} + \sqrt{2})\ =& \boxed{\frac{1}{6} + \frac{\sqrt{2} + \sqrt{3}}{3} - 6(\sqrt{3} + \sqrt{2})} \end{aligned} ]

We can simplify this expression by multiplying the numerator and denominator by the conjugate of the denominator. The conjugate of the denominator is (2+i)(3−i)=6+5i−3i−i2=6+2i−1=5+2i​.

For the system to have infinitely many solutions, the left-hand sides must be equal to each other, regardless of the values of a and b. Thus, 3a + 2b = 6a + 2b, which implies a = 0. Substituting this value into the first equation yields 2b = 2. Hence, b = 1. Then, the second equation becomes 0 - mb = k - 3(0) - m(1), which simplifies to k = m.

Therefore, k = m = \boxed{1}.

To solve this problem, we can use the principle of inclusion and exclusion.

Let A be the set of all ways to distribute 8 indistinguishable balls among 15 distinguishable boxes.

Let B_i be the set of all ways to distribute the balls such that box i is empty.

We want to count the number of ways to distribute the balls such that at least 3 boxes are empty. This is the same as counting the number of ways to distribute the balls such that 0, 1, or 2 boxes are empty.

We can use the following formula:

|A - B_1 - B_2 - ... - B_n| = |A| - |B_1| - |B_2| - ... - |B_n| + |B_1 \cap B_2| + |B_1 \cap B_3| + ... + |B_{n-1} \cap B_n| - |B_1 \cap B_2 \cap B_3| - ... - |B_{n-2} \cap B_{n-1} \cap B_n| + ... + |B_1 \cap B_2 \cap B_3 \cap ... \cap B_n|

where |A| is the cardinality of set A, and |B_i| is the cardinality of set B_i.

In our case, we have:

|A| = \binom{15}{8} = 6435

|B_i| = \binom{14}{7} = 3432

|B_1 \cap B_2| = \binom{13}{6} = 1771

|B_1 \cap B_3| = |B_2 \cap B_3| = \binom{12}{5} = 792

|B_1 \cap B_2 \cap B_3| = \binom{11}{4} = 330

Therefore, the number of ways to distribute the balls such that at least 3 boxes are empty is:

6435 - 3 * 3432 + 3 * 1771 - 3 * 792 + 1 * 330 = \boxed{1818}

We have 8 slots for the digits of our number, and we need to place a 1 in 4 of those slots. This can be done in C(8,4) ways.

Once we have placed the 1's, we can fill in the remaining slots with 3's. This can be done in 4! ways.

Therefore, the total number of positive integers that satisfy both of the given conditions is C(8,4)*4! = 416.

Problem 1

If the 4 balls are indistinguishable and the boxes are distinguishable, then the only way to count the number of ways to place the balls in the boxes is to consider the number of balls in each box.

There are 3 ways to place 4 balls in 3 boxes such that there are 4 balls in one box and 0 balls in the other two boxes.

There are 3 ways to place 4 balls in 3 boxes such that there are 3 balls in one box and 1 ball in another box.

There are 1 way to place 4 balls in 3 boxes such that there are 2 balls in each box.

Therefore, there are a total of 7​ ways to place 4 indistinguishable balls in 3 distinguishable boxes.

Problem 2

If the balls are distinguishable, but the boxes are indistinguishable, then we can treat the balls as if they were numbered 1, 2, 3, and 4.

There are 4 ways to place the ball numbered 1 into one of the three boxes.

Once we have placed the ball numbered 1, there are 3 ways to place the ball numbered 2 into one of the two remaining boxes.

Once we have placed the balls numbered 1 and 2, there are 2 ways to place the ball numbered 3 into one of the remaining boxes.

Finally, there is only 1 way to place the ball numbered 4 into the remaining box.

Therefore, there are a total of 4⋅3⋅2⋅1=24​ ways to place 4 distinguishable balls into 3 indistinguishable boxes.

In order for the expansion of (4x2−7y3)n to contain a term of the form cx2y3, there must be a combination of 4x2 and −7y3 in the expansion that gives a product of cx2y3.

Since n is positive, the only way for this to happen is if the coefficient of x2y3 in the expansion of (4x2−7y3)n is nonzero.

The coefficient of x2y3 in the expansion of (4x2−7y3)n can be found using the binomial theorem:

\binom{n}{3}(4x^2)^3(-7y^3)^3 = \binom{n}{3}(-2192x^6y^9) = -2192\binom{n}{3}x^6y^9

For this coefficient to be nonzero, (3n​) must be nonzero. The smallest positive integer n for which (3n​) is nonzero is n=4, so the only positive integer n such that the expansion of (4x2−7y3)n contains a term of the form cx2y3 is 4​.

1

The expansion of (1+ax)n can be written using the binomial theorem as:

(1 + ax)^n = 1 + nax + \frac{n(n - 1)}{2!}(ax)^2 + \frac{n(n - 1)(n - 2)}{3!}(ax)^3 + \dotsb + (ax)^n.

The first few terms in this expansion are:

1 + nax + \frac{n(n - 1)}{2!}(ax)^2 + \frac{n(n - 1)(n - 2)}{3!}(ax)^3.

Simplifying the first three terms, we get:

1 - 6x + 12x^2 + cx^3.

We are given that the first three terms are:

1 - 6x + 12x^2

Comparing the coefficients of x3 on both sides of the equation, we get:

\frac{n(n - 1)(n - 2)}{3!} = c

To find c, we need to find (3n​) and a. We can use the first two terms of the expansion to solve for a and n.

From the given first two terms, we have:

1 - 6x = 1 + nax

This implies that nax=20x or a=n20​.

Substituting this value of a in the expression for the second term, we get:

nax = \frac{20}{n} \cdot x = 20x

Simplifying this equation gives:

n = 20

Therefore, n=20 and a=n20​=1.

Now that we know n and a, we can find c:

c = \frac{n(n - 1)(n - 2)}{3!} = \frac{20 \cdot 19 \cdot 18}{3 \cdot 2 \cdot 1} = 4840

Therefore, the value of c is 4840​.