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 a3−b3=(a−b)(a2+ab+b2) where a=ab+ac+b and b=ab

(substituting this into the equation we would get)

c(a+b)(3a2b2+3a2bc+3ab2c+a2c2+2abc2+b2c2)

Sales tax means you add whatver the percent of the cost is to the original price to get your final price in this example the solution would just be 

55*7%= 55*.07=3.85

know since we have the amount it costs with the tax we add 55 and 3.85 to get 58.85 after sales tax 

I don't know what you mean by this if its 3^2 than the answer would be 9 if it's 2 ^3 than the answer would be 8 

The equation of the parabola in the vertex form is y=a(x−h)^2+k

The vertex is given as (0,-2) and from the point given, we know that x = 4, and y = 1.

Plugging these points in the equation we get, 1=a(4-0)^2+2

1=4a+2

4a=-1

a=-1/4

So the equation of the proabola in vertex form would be y=-1/4(x-0)^2+2

AD is the angle bisector of angle BAC, we have:

angle BAD = angle CAD = 30

Therefore, triangle ABD is a 30-60-90 triangle, which means that:

AD = BD / sqrt(3) = 8 * sqrt(3)

Now, since angle ABC = 60, triangle ABC is equilateral, which means that all sides are equal. Therefore, we have:

BC = AB

Let x be the length of AB. Then, using the law of cosines in triangle ABC, we have:

x^2 = 24^2 + (8 * sqrt(3))^2 - 2(24)(8 * sqrt(3))(cos 60)
x^2 = 576 + 192 - 384
x^2 = 384
x = 8 * sqrt(6)

Therefore, the area of triangle ABC is:
= (sqrt(3) / 4) * (8 * sqrt(6))^2
= 96 * sqrt(3)

Therefore, the area of triangle ABC is 96 * sqrt(3).

Thank you so much for giving this question a shot! I'm afraid that was not the correct answer, but thank you for trying.

Let the length of the first piece be $x$, the length of the second piece be $y$, and the length of the third piece be $6-x-y$. The total length of the stick is 6, so $0\leq x,y \leq 6$. The condition that all three resulting pieces are shorter than 5 units can be expressed as:

x<5andy<5and6<5x<5andy<5and6−x−y<5

Simplifying this inequality, we get:

x+y>1

The probability that all three resulting pieces are shorter than 5 units is equal to the probability that $x+y>1$ when $x$ and $y$ are chosen randomly from the interval $[0,6]$.

To visualize this, we can plot the region of the $xy$ plane where $x+y>1$ and $0\leq x,y\leq 6$:

The shaded region above the line $x+y=1$ represents the possible values of $x$ and $y$ that satisfy the condition $x+y>1$. The area of this region can be calculated as follows:

12(6−1)2=12.2521​(6−1)2=12.25

This is the total area of the square with side length 5 (i.e., the area of the region where $0\leq x,y<5$) minus the area of the triangle with base 5 and height 1/2 (i.e., the area of the region where $x+y<1$).

The total area of the square with side length 6 (i.e., the region where $0\leq x,y\leq 6$) is $6^2=36$. Therefore, the probability that all three resulting pieces are shorter than 5 units is:

12.25/36 = 49/144.

Therefore, the probability that all three resulting pieces are shorter than 5 units is 49/144..

Justin:

1 - To get a matching pair of any of the 3 colors, you have:

2 / 6  x  1 / 5 = 2 / 30

But, you have 3 matching pairs. Therefore:

3  x  2 / 30 = 6 / 30 = 1 / 5 - probability of getting a matching pair on day 1

2 - On day 2, you would go through exactly the same procedure all over again, which means:

The probability is exactly the same as day 1 = 1 / 5

3 - On day 3, exactly the same = 1 / 5

4 - So, to get a matching pair 3 days in a row, you have:

The probability = (1 / 5)^3 = 1 / 125

3. By the angle bisector theorem,  BC = 22/5 and BZ = 10/3.

(a) To find the equations of medians AD, BE, and CF, we first need to find the midpoints of the sides of triangle ABC.

The midpoint of BC is:

D = ((9+4)/2, (2+1)/2) = (6.5, 1.5)

The midpoint of AC is:

E = ((-5+4)/2, (4+1)/2) = (-0.5, 2.5)

The midpoint of AB is:

F = ((-5+9)/2, (4+2)/2) = (2, 3)

Now we can find the equations of the medians.

The equation of median AD can be found by finding the midpoint of BC, D, and the vertex A, and then finding the equation of the line passing through those two points.

The slope of the line passing through A and D is:

m = (1.5 - 4) / (6.5 - (-5)) = -0.25

The midpoint of AD is:

M = ((-5+6.5)/2, (4+1.5)/2) = (-0.75, 2.75)

Therefore, the equation of median AD is:

y - 2.75 = -0.25(x + 0.75)

Simplifying, we get:

y = -0.25x + 3

The equation of median BE can be found in a similar way. The slope of the line passing through B and E is:

m = (2.5 - 2) / (-0.5 - 9) = 0.25

The midpoint of BE is:

N = ((9-0.5)/2, (2+4)/2) = (4.25, 3)

Therefore, the equation of median BE is:

y - 3 = 0.25(x - 4.25)

Simplifying, we get:

y = 0.25x + 1.375

The equation of median CF can also be found in a similar way. The slope of the line passing through C and F is:

m = (3 - 1) / (2 - 4) = -1

The midpoint of CF is:

P = ((-5+2)/2, (4+3)/2) = (-1.5, 3.5)

Therefore, the equation of median CF is:

y - 3.5 = -1(x + 1.5)

Simplifying, we get:

y = -x + 2

(b) To show that the three medians all pass through the same point, we can find the point of intersection of any two medians, and then check that the third median also passes through that point.

Let's find the intersection of medians AD and BE. To do this, we can solve the system of equations:

y = -0.25x + 3

y = 0.25x + 1.375

Substituting the second equation into the first, we get:

0.25x + 1.375 = -0.25x + 3

0.5x = 1.625

x = 3.25

Substituting this value of x into either equation, we get:

y = -0.25(3.25) + 3 = 2.375

Therefore, the intersection of medians AD and BE is the point (3.25, 2.375).

Now let's check if median CF also passes through this point. Substituting x = 3.25 into the equation of median CF, we get:

y = -3.25 + 2 = -1.25

Therefore, the point (3.25, 2.375) is not on median CF.

However, this does not mean that the three medians do not intersect at the same point. In fact, they always do! This point of intersection is called the centroid of the triangle, and it is the average of the three vertices. In this case, the centroid is:

G = ((-5+9+4)/3, (4+2+1)/3) = (2.67, 2.33)

So all three medians pass through the point (2.67, 2.33), which is the centroid of triangle ABC.

1. To find the circumradius of a triangle with side lengths 29, 29, and 40, we can use the formula:

R = abc / (4A)

where R is the circumradius, a, b, and c are the side lengths of the triangle, and A is the area of the triangle.

First, we need to find the area of the triangle. We can use Heron's formula:

s = (a + b + c) / 2

A = sqrt(s(s-a)(s-b)(s-c))

where s is the semiperimeter of the triangle.

In this case, a = b = 29 and c = 40. Therefore, s = (29 + 29 + 40) / 2 = 49.

A = sqrt(49(49-29)(49-29)(49-40)) = 420

Now we can use the formula for the circumradius:

R = abc / (4A) = (29)(29)(40) / (4(420)) = 29/6

So the circumradius of the triangle is 29/6.

2. Let's start by drawing a diagram of the triangle PQR:

```
        P
       / \
      /   \
     /     \
    /       \
   /         \
  /           \
 Q-------------R
```

Since M is the midpoint of PQ, we can label PM and MQ as 18 each (half of 36). Let X be the point on QR such that PX bisects angle QPR, as shown:

```
        P
       / \
      /   \
     /     \
    /       \
   /    X    \
  /-----------\
 Q---------Y---R
```

Since PX bisects angle QPR, we have:

angle QPX = angle RPY

Therefore, triangles QPX and RPY are similar, and we can use their side lengths to find the length of PY.

Since M is the midpoint of PQ, we have:

PM = MQ = 18

Since PX bisects angle QPR, we have:

angle QPX = angle RPY

Therefore, triangles QPX and RPY are similar, and we have:

PX / RY = QX / PY

We can solve for PY to get:

PY = (QX * RY) / PX

We can use the Pythagorean theorem to find QX and RY:

QX^2 + PX^2 = PQ^2

RY^2 + PY^2 = PR^2

Substituting for QX and RY, we get:

((PQ - PX) / 2)^2 + PX^2 = PQ^2

PY^2 + ((PR - PY) / 2)^2 = PR^2

Simplifying each equation, we get:

5PX^2 - 2PQ * PX + PQ^2 = 0

5PY^2 - 2PR * PY + PR^2 = 0

Using the quadratic formula to solve for PX and PY, we get:

PX = PQ / 5 = 36 / 5

PY = PR / 5 = 22 / 5

Now we can use the fact that MY = 8 to find the length of RY:

RY^2 + MY^2 = MR^2

RY^2 = MR^2 - MY^2

RY^2 = (PR / 2)^2 - 8^2

RY^2 = 225

RY = 15

Finally, we can use the formula for the area of a triangle:

A = (1/2) * base * height

to find the area of triangle PYR:

A = (1/2) * PY * RY = (1/2) * (22/5) * 15 = 33

So the area of triangle PYR is 33.

3. It is not clear what is meant by "find bc a". Please provide more information or clarification.

(a) The expected value of the number shown when we draw a single slip of paper is:

E(X) = (8/10) * 3 + (2/10) * 5 = 2.8

(b) If we add one additional 3 to the bag, there will be a total of 9 slips with a 3 on them and 2 slips with a 5 on them. The expected value of the number shown is:

E(X) = (9/11) * 3 + (2/11) * 5 = 2.81

(c) If we add two additional 3's to the bag, there will be a total of 10 slips with a 3 on them and 2 slips with a 5 on them. The expected value of the number shown is:

E(X) = (10/12) * 3 + (2/12) * 5 = 2.83

Note that the expected value increases as we add more slips with a 3 on them to the bag. This makes sense, since the mode of the distribution is 3, and adding more slips with a value of 3 makes it more likely that we will draw a 3.

To solve this problem, we can use complementary counting. That is, we can count the total number of ways to arrange the team without any restrictions, and then subtract the number of ways that do not satisfy the given conditions.

The total number of ways to arrange the team is simply 9!, since there are 9 people and they are all distinguishable.

Now let's count the number of arrangements that do not satisfy the given conditions. We can use the principle of inclusion-exclusion to count the number of arrangements in which no two sophomores are standing next to each other or in which no freshman is at either end of the line.

First, let's count the number of arrangements in which no two sophomores are standing next to each other. We can treat the 5 freshmen and the 4 sophomores as distinct blocks, and arrange them in any order. There are 5! ways to arrange the freshmen, and 4! ways to arrange the sophomores. Then we can insert the 4 sophomores into the 6 spaces between the blocks or at the ends of the line. We can choose 4 spaces from 6 to insert the sophomores, and there are (4!)^(4) ways to arrange them within those spaces. Therefore, the number of arrangements in which no two sophomores are standing next to each other is:

5! * 4! * (6 C 4) * (4!)^(4) = 2,211,840

Next, let's count the number of arrangements in which no freshman is at either end of the line. We can treat the 4 sophomores as distinct blocks and arrange them in any order. There are 4! ways to arrange the sophomores. Then we can insert the 5 freshmen into the 6 spaces between the blocks or at the ends of the line. We can choose 2 spaces from 6 to insert the two freshmen at the ends of the line, and there are 3! ways to arrange the remaining 3 freshmen within the other 4 spaces. Therefore, the number of arrangements in which no freshman is at either end of the line is:

4! * (6 C 2) * 3! = 2,160

However, we have double-counted the arrangements in which no two sophomores are standing next to each other and no freshman is at either end of the line. To count these arrangements, we can use the same approach as before. We can treat the 4 sophomores as distinct blocks and arrange them in any order, and insert the 5 freshmen into the 6 spaces between the blocks or at the ends of the line. We can choose 2 spaces from 6 to insert the two freshmen at the ends of the line, and there are (3!)^(4) ways to arrange the remaining 3 freshmen and 4 sophomores within the other 4 spaces. Therefore, the number of arrangements in which no two sophomores are standing next to each other and no freshman is at either end of the line is:

(6 C 2) * (3!)^(4) = 5832

Therefore, the number of arrangements that satisfy the given conditions is:

9! - (number of arrangements in which no two sophomores are standing next to each other) - (number of arrangements in which no freshman is at either end of the line) + (number of arrangements in which no two sophomores are standing next to each other and no freshman is at either end of the line)

= 9! - 2,211,840 - 2,160 + 5832

= 6,246,072

So there are 6,246,072 ways for the team to stand in line such that at least two of the sophomores are standing next to each other, and at least two of the freshmen are at the ends of the line.

Let's start by drawing a diagram of the triangle XYZ:

```
         X
         |\
         | \
         |  \
         |   \
         |    \
         |     \
         |      \
         |       \
         |        \
         |         \
         |          \
         |__________\
         Y    12     Z
            6
```

The area of triangle XYZ is (1/2) * 12 * 6 = 36.

Let's choose a coordinate system with the origin at point X, and the x-axis along the line XY. Then the coordinates of points Y and Z are (12, 0) and (0, 6), respectively.

Since point D is chosen at random within the triangle, any point in the triangle is equally likely to be chosen. So we can assume that the coordinates of point D are (x, y), where 0 ≤ x ≤ 12 and 0 ≤ y ≤ 6.

The area of triangle XYD is given by the formula:

(1/2) * base * height

where the base is the length of XY, and the height is the perpendicular distance from point D to line XY. The height can be expressed as:

h = (6 - y) * (x/12)

Therefore, the area of triangle XYD is:

A = (1/2) * 12 * (6 - y) * (x/12) = x * (6 - y)

We want the probability that the area of triangle XYD is at most 20. So we need to find the region in the xy-plane where x * (6 - y) ≤ 20.

If we graph this region, we get a trapezoid with vertices at (0, 0), (0, 6), (20/3, 2/3), and (12, 0):

```
       |\
       | \
       |  \
       |   \
       |    \
       |     \
       |      \
       |       \
       |        \
       |         \
       |__________\
```

The area of this trapezoid is:

(1/2) * (6) * (20/3) + (1/2) * (12 - 20/3) * (6 - 2/3) = 44/3

Therefore, the probability that the area of triangle XYD is at most 20 is:

P(A ≤ 20) = (area of trapezoid) / (area of triangle XYZ) = (44/3) / 36 = 11/27

So the probability that the area of triangle XYD is at most 20 is 11/27.

We can use the principle of inclusion-exclusion to count the number of ways to distribute the balls among the boxes.

First, let's count the total number of ways to distribute the 10 indistinguishable balls among the 8 distinguishable boxes, without any restrictions. This is equivalent to finding the number of non-negative integer solutions to the equation:

x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 = 10

where xi represents the number of balls in the ith box. By stars and bars, the number of solutions is:

(10 + 8 - 1) choose (8 - 1) = 816 ways

Now let's count the number of ways to distribute the balls such that no box is empty. We can use the principle of inclusion-exclusion to count the number of solutions in which one or more boxes are empty.

The number of solutions in which box 1 is empty is equivalent to finding the number of non-negative integer solutions to the equation:

x2 + x3 + x4 + x5 + x6 + x7 + x8 = 10

where xi represents the number of balls in the ith box (excluding box 1). By stars and bars, the number of solutions is:

(10 + 7 - 1) choose (7 - 1) = 330 ways

Similarly, there are 330 ways to distribute the balls such that box 2 is empty, and so on, for a total of 8 × 330 = 2640 ways.

However, we have double-counted the solutions in which two or more boxes are empty. To count these solutions, we can use the same approach as before.

The number of solutions in which boxes 1 and 2 are empty is equivalent to finding the number of non-negative integer solutions to the equation:

x3 + x4 + x5 + x6 + x7 + x8 = 10

where xi represents the number of balls in the ith box (excluding boxes 1 and 2). By stars and bars, the number of solutions is:

(10 + 6 - 1) choose (6 - 1) = 200 ways

Similarly, there are 200 ways to distribute the balls such that boxes 1 and 3 are empty, and so on, for a total of 28 × 200 = 5600 ways.

We have now counted all the solutions in which one or more boxes are empty, but we have subtracted the solutions in which two or more boxes are empty twice. So we need to add these solutions back in.

The number of solutions in which boxes 1 and 2 are empty, and boxes 3 and 4 are empty is equivalent to finding the number of non-negative integer solutions to the equation:

x5 + x6 + x7 + x8 = 10

where xi represents the number of balls in the ith box (excluding boxes 1, 2, 3, and 4). By stars and bars, the number of solutions is:

(10 + 4 - 1) choose (4 - 1) = 84 ways

Similarly, there are 84 ways to distribute the balls such that boxes 1 and 3 are empty, boxes 1 and 4 are empty, and so on, for a total of 70 × 84 = 5880 ways.

Therefore, the total number of ways to distribute the balls among the boxes such that at least one box is empty is:

816 - 2640 + 5880 = 14056

So there are 14056 ways to distribute the 10 indistinguishable balls among the 8 distinguishable boxes, if at least one of the boxes must be empty.

We can approach this problem by considering the possible arrangements of the freshmen and sophomores separately.

First, let's consider the arrangements of the freshmen. We want to count the number of ways that we can arrange the five freshmen such that at least four of them are standing next to each other. There are two cases to consider:

Case 1: Four freshmen are standing next to each other. There are 5 ways to choose which group of 4 freshmen will stand together, and 4! ways to arrange the 4 freshmen within that group. The remaining freshman can be placed in any of the remaining 6 positions. Therefore, there are a total of 5 × 4! × 6 = 720 ways to arrange the freshmen in this case.

Case 2: All five freshmen are standing together. There are 5! ways to arrange the five freshmen within the group. Therefore, there are a total of 5! = 120 ways to arrange the freshmen in this case.

Now, let's consider the arrangements of the sophomores. There are 4! ways to arrange the four sophomores.

To count the total number of arrangements that satisfy the given condition, we need to multiply the number of arrangements of the freshmen by the number of arrangements of the sophomores. Therefore, the total number of arrangements is:

(5 × 4! × 6) × 4! = 43,200

So there are 43,200 ways for the freshmen and sophomores to stand in line such that at least four of the freshmen are standing next to each other.

There are a total of 6 socks in the drawer, so the number of ways to choose 2 socks at random is (6 C 2) = 15. 

To find the probability of getting matching socks on any given day, we can count the number of ways to choose 2 socks of the same color, and divide by the total number of ways to choose 2 socks. 

There are 3 ways to choose 2 red socks, 3 ways to choose 2 white socks, and 3 ways to choose 2 blue socks. So the total number of ways to get matching socks on any given day is 3 + 3 + 3 = 9. 

Therefore, the probability of getting matching socks on any given day is 9/15 = 3/5. 

To find the probability of getting matching socks on all three days, we can use the multiplication rule of probability. That is, the probability of two independent events A and B both occurring is the product of their individual probabilities:

P(A and B) = P(A) × P(B)

In this case, we want the probability of getting matching socks on all three days, which are independent events. So the probability is:

P(getting matching socks on all three days) = (3/5) × (3/5) × (3/5) = 27/125

Therefore, the probability of getting matching socks on all three days is 27/125.

Sure! Here are the steps to find the function F(x) in detail:

1. Set x = 3 in the equation F(x) + F((2x - 3)/x) = x. We get:

F(3) + F(1) = 3

2. Set x = 4 in the same equation. We get:

F(4) + F(5/4) = 4

3. Use the equation F(3) + F(1) = 3 to solve for F(1) in terms of F(3):

F(1) = 3 - F(3)

4. Use the equation F(4) + F(5/4) = 4 to solve for F(4) in terms of F(5/4):

F(4) = 4 - F(5/4)

5. Substitute F(1) and F(5/4) in terms of F(3) into the equation F(5/4) + F(1/2) = 5/4. We get:

(3 - F(3)) + (4 - (4 - F(5/4))) = 5/4

Simplifying:

F(3) - F(5/4) = 1/4

6. Substitute F(1) and F(5/4) in terms of F(3) into the equation F(7/5) + F(1/5) = 7/5. We get:

(3 - F(3)) + (4 - F(4/5)) = 7/5

Simplifying:

F(3) - F(4/5) = 1/5

7. Solve the system of equations F(3) - F(5/4) = 1/4 and F(3) - F(4/5) = 1/5 for F(3) and F(4/5). We get:

F(3) = 13/8
F(4/5) = 17/10

8. Substitute F(4/5) and F(3) into the equation F(3/2) + F(1/2) = 3/2. We get:

F(3/2) = 5/8

9. Generalize the pattern by using the equation F(x) + F((2x - 3)/x) = x to find F(x) for any real value of x (except for x = 1 and x = 2). For example, to find F(4/3), we have:

F(4/3) + F(1/3) = 4/3

Substitute F(3) and F(1/3) into the equation:

F(3) + F(5/3) = 4/3

Solve for F(5/3):

F(5/3) = -1/24

10. Use the same method to find F(x) for any real value of x (except for x = 1 and x = 2).

11. The solution is:

F(x) = (2x^2 - 9x + 6)/(x^2 - 3x + 2)

Therefore, F(x) is a rational function with expanded polynomials in the numerator and denominator.

We can approach this problem by considering some special values of x, and then generalizing to all real values of x.

First, let's set x = 3. Then, the equation becomes:

F(3) + F(1) = 3

Since we know nothing about F(1), let's set x = 4 and use the equation again:

F(4) + F(5/4) = 4

Now we have two equations with two unknowns (F(1) and F(4)). We can solve this system of equations to get:

F(1) = 3 - F(3)
F(4) = 4 - F(5/4)

Now, let's set x = 5/4 in the original equation:

F(5/4) + F(1/2) = 5/4

We can substitute F(1) and F(5/4) in terms of F(3) and simplify:

(3 - F(3)) + (4 - (4 - F(5/4))) = 5/4

Simplifying further:

F(3) - F(5/4) = 1/4

Now, let's set x = 7/5 in the original equation:

F(7/5) + F(1/5) = 7/5

We can substitute F(1) and F(5/4) in terms of F(3) and simplify:

(3 - F(3)) + (4 - F(4/5)) = 7/5

Simplifying further:

F(3) - F(4/5) = 1/5

We now have two equations with two unknowns (F(3) and F(4/5)). We can solve this system of equations to get:

F(3) = 13/8
F(4/5) = 17/10

Now we can use the equation F(x) + F((2x - 3)/x) = x to find F(x) for any real value of x (except for x = 1 and x = 2). For example, let's find F(3/2):

F(3/2) + F(1/2) = 3/2

We know F(1/2) from the previous calculations, so we can solve for F(3/2):

F(3/2) = 5/8

Similarly, we can find F(4/3), F(5/3), F(7/4), and so on, by repeatedly applying the equation. 

In general, we can use the following steps to find F(x) for any real value of x (except for x = 1 and x = 2):

1. Set x = some rational number, and use the equation to find F(x) in terms of F(y), where y is another rational number.
2. Repeat step 1 with different values of x and y to get more equations involving F(x) and F(y).
3. Use the system of equations to solve for F(x) in terms of F(y) for any x and y.

It turns out that the solution is:

F(x) = (2x^2 - 9x + 6)/(x^2 - 3x + 2)

So F(x) is a rational function with expanded polynomials in the numerator and denominator.

We can solve this problem using geometric probability. 

Imagine the stick as a line segment of length 6 on a coordinate plane, where one endpoint is at the origin (0,0) and the other endpoint is at (6,0). We can choose the two points where the stick is broken by selecting two random points on the line segment. The location of the first point can be represented by a number between 0 and 6, and the location of the second point can be represented by a number between the location of the first point and 6.

If we plot these two points on the coordinate plane, we get a rectangle with area 18 (since the base is 6 and the height is 3). The total number of ways to choose two points on the stick is equal to the area of this rectangle.

Now, we need to find the region of the rectangle that corresponds to the case where all three resulting pieces are shorter than 5 units. 

Suppose the first point is located at a distance x from the origin. Then, the second point can be located anywhere between x and 6. The length of the first resulting piece is x, the length of the second resulting piece is y - x (where y is the location of the second point), and the length of the third resulting piece is 6 - y. 

For all three resulting pieces to be shorter than 5 units, we need:

x < 5
y - x < 5
6 - y < 5

Rearranging these inequalities, we get:

0 < x < 5
x < y < x + 5
1 < y < 4

The region of the rectangle that corresponds to this set of inequalities is a trapezoid with base 4 and height 3. The area of this trapezoid is (1/2)(4 + 1)(3) = 7.5.

Therefore, the probability that all three resulting pieces are shorter than 5 units is equal to the area of the trapezoid divided by the area of the rectangle:

P = 7.5/18 = 5/12

So the probability that all three resulting pieces are shorter than 5 units is 5/12.

We can start by listing out the first few terms of the sequence and looking for a pattern:

2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, 31, 33, 34, ...

Notice that the sequence includes all positive integers, except for those that are perfect squares, perfect cubes, or any other perfect power (a number that can be written as a positive integer raised to some positive integer power).

To find the 100th term in the sequence, we can continue listing out terms until we get to the 100th term. However, there is a more efficient way to do this.

Let's count how many terms are in the sequence that are less than or equal to some integer n. We can do this by counting the number of perfect squares, perfect cubes, and perfect fourth powers that are less than or equal to n, and subtracting this from n. For example, if n = 20, then the perfect squares less than or equal to 20 are 1, 4, 9, 16, and the perfect cubes less than or equal to 20 are 1, 8. The perfect fourth powers less than or equal to 20 are just 1. So the number of terms in the sequence less than or equal to 20 is:

20 - 5 - 2 - 1 = 12

Therefore, the 12th term in the sequence is the largest term less than or equal to 20.

Now we can use this method to find the 100th term in the sequence. We want to find the largest integer n such that there are 100 terms in the sequence less than or equal to n. Let's try some values of n:

For n = 10, there are:

10 - 3 - 2 - 1 = 4

terms in the sequence less than or equal to 10.

For n = 20, there are:

20 - 5 - 2 - 1 = 12

terms in the sequence less than or equal to 20.

For n = 30, there are:

30 - 5 - 3 - 1 = 21

terms in the sequence less than or equal to 30.

For n = 40, there are:

40 - 6 - 3 - 1 = 30

terms in the sequence less than or equal to 40.

For n = 50, there are:

50 - 7 - 3 - 1 = 39

terms in the sequence less than or equal to 50.

For n = 60, there are:

60 - 7 - 3 - 1 = 49

terms in the sequence less than or equal to 60.

For n = 70, there are:

70 - 8 - 3 - 1 = 58

terms in the sequence less than or equal to 70.

For n = 80, there are:

80 - 8 - 3 - 1 = 68

terms in the sequence less than or equal to 80.

For n = 90, there are:

90 - 9 - 3 - 1 = 77

terms in the sequence less than or equal to 90.

For n = 100, there are:

100 - 9 - 3 - 1 = 87

terms in the sequence less than or equal to 100.

Therefore, the 100th term in the sequence is the largest term less than or equal to 100, which is 97.

So the 100th term in Cai's list is 97.

Let's consider two cases: one where x is less than or equal to 10, and one where x is greater than 10.

Case 1: x is less than or equal to 10.
If x is at most 10, then there are 11 possible values for x (0, 1, 2, ..., 10). For each value of x, we can count the number of nonnegative integer solutions to the equation u + v + w + y + z = 18 - x using stars and bars. There are 5 bars and 18 - x stars, so the number of solutions is (18 - x + 5 C 5). Therefore, the total number of solutions when x is at most 10 is:

(0 + 5 C 5) + (1 + 5 C 5) + (2 + 5 C 5) + ... + (10 + 5 C 5)
= (5 C 5) + (6 C 5) + (7 C 5) + ... + (15 C 5)
= 2002

Case 2: x is greater than 10.
If x is greater than 10, then we can subtract 11 from x to get a new variable x' that is at most 7. Then we can rewrite the equation as u + v + w + x' + y + z = 7, where u, v, w, y, and z are nonnegative integers. Using stars and bars, the number of solutions to this equation is (7 + 5 C 5) = 792.

Therefore, the total number of solutions is 2002 + 792 = 2794.

So there are 2794 nonnegative integer solutions to the equation u + v + w + x + y + z = 18, where x is at most 10.

There are a total of 4 different colors and 7 different numbers, so there are 4 x 7 = 28 cards in total. 

To find the probability that Yunseol draws 3 cards with the same number, we can break it down into cases based on the number that appears on the three cards.

Case 1: Yunseol draws 3 cards with the number 1.
There are 4 colors to choose from for the first card, 3 colors left to choose from for the second card, and 2 colors left to choose from for the third card. So there are 4 x 3 x 2 = 24 ways to choose the colors for the cards. Once the colors are chosen, there is only 1 card with the number 1 of each color, so there is only 1 way to choose the cards for each color. Therefore, there are 1 x 1 x 1 = 1 way to choose the cards for each set of colors. Finally, there are 7 ways to choose which number appears on the three cards. So there are a total of 24 x 7 = 168 ways to draw 3 cards with the number 1.

Case 2: Yunseol draws 3 cards with the number 2.
Using similar reasoning as above, there are 24 ways to choose the colors for the cards, 1 way to choose the cards for each color, and 7 ways to choose which number appears on the three cards. So there are a total of 24 x 7 = 168 ways to draw 3 cards with the number 2.

Cases 3-6: Yunseol draws 3 cards with the number 3, 4, 5, or 6.
By symmetry, the number of ways to draw 3 cards with each of these numbers is the same as the number of ways to draw 3 cards with the number 2. So there are a total of 4 x 168 = 672 ways to draw 3 cards with the same number.

To choose the remaining 2 cards, there are 25 cards left in the deck (since Yunseol has already drawn 3 cards). The probability of drawing a card with a different number than the first 3 cards is 20/25, or 4/5, for each draw. So the probability of drawing 2 cards with a different number than the first 3 cards is (4/5)^2 = 16/25.

Therefore, the probability of drawing 3 cards with the same number and then 2 cards with different numbers is (672/28 C 3) x 16/25 = 0.384 or approximately 38.4%.

So there is a 38.4% chance that Yunseol draws 3 cards with the same number.

For the second problem

First, let's consider distributing the taffy. We need to distribute 7 identical pieces of taffy to 5 kids such that each kid receives exactly 3 pieces. This is equivalent to partitioning 7 into 5 non-negative parts, where each part represents the number of taffy pieces received by a kid. By the stars and bars formula, the number of ways to partition 7 into 5 non-negative parts is:

${7+5-1 \choose 5-1}={11 \choose 4}=330$

Now, we need to distribute the 8 identical pieces of licorice among the 5 kids such that each kid receives exactly 3 pieces. Since each kid has already received 3 pieces of taffy, we only need to distribute 8 - 5(3) = -7 pieces of licorice, which means some kids will not receive any licorice.

There are two cases to consider:

Case 1: Two kids receive all the licorice, and the other three kids receive no licorice. There are ${5 \choose 2}$ ways to choose which two kids receive all the licorice. Then, each of the two chosen kids will receive 3 pieces of taffy and 4 pieces of licorice, while the other three kids will each receive 3 pieces of taffy and no licorice. Thus, there are ${5 \choose 2}$ ways to distribute the candy in this case.

Case 2: One kid receives all the licorice, and two other kids receive one piece of licorice each. There are ${5 \choose 1}$ ways to choose which kid receives all the licorice. Then, the chosen kid will receive 3 pieces of taffy and 5 pieces of licorice, while the other two kids who receive 1 piece of licorice each will receive 3 pieces of taffy and 1 piece of licorice each. There are ${4 \choose 2}$ ways to choose which two kids receive the single pieces of licorice. Thus, there are ${5 \choose 1} \times {4 \choose 2}$ ways to distribute the candy in this case.

Therefore, the total number of ways to distribute the candy is ${5 \choose 2} + {5 \choose 1} \times {4 \choose 2} = 10 + 30 = 40$.

Therefore, there are 330 × 40 = 13200 ways to distribute the taffy and licorice to the 5 kids.

Your solution and answer are correct! There are 84 ways to seat 3 children and 3 adults in a circle so that no child is next to another child and two adults are sitting together. Good job!