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To find the number of ways to choose 3 circles from 12 arranged in a circle such that no two circles are adjacent, we can use a combinatorial approach.

First, since the circles are arranged in a circle, we'll convert the problem into a linear arrangement. We can fix one circle as a reference point, which effectively reduces our problem to a line of 11 circles (with the circle we fixed being considered as already chosen).

Once we fix a circle, we need to select 2 more circles from the remaining 11 while ensuring that no two chosen circles are adjacent. To achieve this, we can represent the chosen circles and the unchosen circles as follows:

If we choose two additional circles, there will be at least one unchosen circle between any two chosen circles. Therefore, we can denote chosen circles as 'C' and unchosen circles as 'U'. So, if we denote our arrangement with chosen circles and a required space, it will look something like this:

- C - U - C - U - C

This configuration illustrates that for each chosen circle, 1 unchosen circle must be placed between them, leading us to count the total number of positions left for unchosen circles after accounting for the chosen ones.

### Create the remaining sequence for unchosen circles:

Given:

- We have chosen 3 'C's, thus requiring 2 'U's to separate them.

- This means we use up 3 + 2 = 5 spots out of the 11 available, leaving us with 6 unchosen circles to place.

### Transform into a stars and bars problem:

Now we need to distribute the remaining unchosen circles among available spots. We can think of placing the 6 remaining unchosen circles in gaps created by the two boundaries (left and right of 'C's):

If we denote the gaps as follows (where '|' represents gaps):

- | U | U | C | U | C | U | C | U | |

We have potential gaps before the first 'C', between the 'C's, and after the last 'C'. In our case, we have fixed the first circle, which means we currently have two gaps to the left and one gap to the right (which connects back to the fixed circle in circular arrangement).

- Let G - denotes the gap available, thus a sequence could be modeled as:

- G (pre gap), C, U, C, U, C (for the already occupied spaces) plus additional G's around.

### Fill spaces with the left-over circles:

Thus, after fixing the first chosen circle, thus we have 8 available spots (in gaps) to place our remaining circles, as constructed above, calculating the arrangement with stars and bars theorem.

Using the formula for distributing \( n \) indistinguishable objects (stars) into \( k \) distinguishable boxes (spaces):

\[ \text{total ways} = \binom{n+k-1}{k-1} \]

where \( n = 6 \) (remaining unchosen circles) and \( k = 3 \) (gaps). Substitute this into the equation:

\[
\text{total ways} = \binom{6 + 3 - 1}{3 - 1} = \binom{8}{2} = 28
\]

### Final Answer:

Thus, the total number of ways to choose the circles such that no two are adjacent is \( \boxed{220} \) ways.

\( \dfrac{1}{\sqrt2+\sqrt 3}+\dfrac{1}{\sqrt 2-\sqrt 3}\\ =\dfrac{\sqrt 2-\sqrt 3+\sqrt 2+\sqrt 3}{2-3}\\ =\dfrac{2\sqrt{2}}{-1}\\ \color{blue}=-2\sqrt{2}\ or =-\sqrt{8}\)

  !

We should consider the information from both flights, put that information together, and then solve for the distance of the trip. We will take advantage of the relationship of distance, rate, and time with the formula d = rt.

Let d be the distance of the one-way trip from Penthaven to Jackson, in miles

Let r be the rate of speed of the airplane during no wind, in miles per hour

let t be the time of the one-way trip from Penthaven to Jackson, in hours

Case 1) No wind

Here, we know the total time of the round trip without wind is 3 hours and 20 minutes. Converting into hours, \(t_1 = 3 \text{ hr} + 20 \text{ min} * \frac{1 \text{ hr}}{60 \text{min}} = 3\frac{1}{3} \text{ hr} = \frac{10}{3} \text{ hr}\). Both the distance d and the rate of speed r of the trip is unknown. However, the given information concerns a round trip, which means \(2d = \frac{10}{3}r \text{ so } d = \frac{5}{3}r\).

Case 2) 70 miles per hour of wind

Here, we know the total time of the round trip with wind is 3 hours and 50 minutes. Converting into hours, \(t_2 = \text{ hr} + 50 \text{ min} * \frac{1 \text{ hr}}{60 \text{ min}} = 3\frac{5}{6} \text{ hr} = \frac{23}{6} \text{ hr}\). Both the distance d and the rate of speed r of the trip is unknown again. However, there is some relationship between rates of speeds. For example, while the plan is flying with the aide of the wind, then plane is flying at r + 70 and at r - 70 on the return flight. If d = rt, then t = d/r, so we can sum the individual times.

\(\frac{d}{r + 70} + \frac{d}{r - 70} = \frac{23}{6}\).

Now, solve for d.

\(\begin{cases} d = \frac{5}{3}r \\ \frac{d}{r + 70} + \frac{d}{r - 70} = \frac{23}{6} \end{cases} \\ r = \frac{3}{5}d \text{ so } r^2 = \frac{9}{25}d^2\\ d(r - 70)+d(r+70) = \frac{23}{6}\left(r^2 - 4900\right) \\ 6dr - 420d + 6dr + 420d = 23\left(r^2 - 4900\right) \\ 12dr = 23r^2 - 112700 \\ 12d * \frac{3}{5}d = 23*\frac{9}{25}d^2 - 112700 \\ 180d^2 = 207d^2 - 2817500\\ 27d^2 = 2817500 \\ d = \sqrt{\frac{2817500}{27}} \approx 323.035 \text{ mi}\)

The problem fails to state the number of available seats at this table, but I will assume for problem-solving purposes that there are 7 seats at this circular table. 

There are 7 guests in total, but both mathematicians must sit beside each other. Because of this, we should consider the group of 2 mathematicians as 1 entity that occupies 2 seats. The remaining 5 non-mathematician entities can freely occupy the remaining 5 seats. This effectively reduces the problem to arranging 6 entities around a circular table.

6 entities have 6! seating arrangements. However, note that table is circular. If all the entities rotated one seat to the left, then the relative seating positions would remain the same. The entities could rotate a total of 6 times before returning to their original seat, so we have effectively overcounted by a factor of 6. Account for this by dividing by 6.

Also, the mathematicians are in a group of 2, so there are 2 ways to arrange the mathematicians within its group.

Therefore, the number of arrangements is \(\frac{6!}{6} * 2 = \frac{720}{6} * 2 = 120 * 2 = 240\).

Whenever there are multiple radical terms in the denominator, the process of simplification becomes quite a challenge, but I realized after some time that the denominator \(1 - \sqrt[3]{3} + \sqrt[3]{9} = \left(\sqrt[3]{3}\right)^2 - \sqrt[3]{3} + 1\) looks to take the form of the beginnings of the factorization of the sum of cubes.

In the general factorization, \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). If we consider the case where \(a = \sqrt[3]{3} \text{ and } b = 1\), then we can multiply the denominator by \(\sqrt[3]{3} + 1\) to transform the denominator into a sum of cubes and then simplify. 

\(\frac{4}{1 - \sqrt[3]3 + \sqrt[3]9} * \frac{\sqrt[3]{3} + 1}{\sqrt[3]{3} + 1} = \frac{4\left(\sqrt[3]{3} + 1\right)}{\left(\sqrt[3]{3}\right)^3 + 1^3} = \frac{4\left(\sqrt[3]{3} + 1\right)}{3 + 1} = \sqrt[3]{3} + 1\)

          Fill in the blanks with three distinct positive integers.   
          1/___ + 1/___ + 1/___ = \frac{13}{12}   

           1 / 1  +  1 / 18  +  1 / 36  =  39 / 36  =  13 / 12     

Note that the above is the fractions 36/36 + 2/36 + 1/36 = 39/36   

_.   

In the problem, it wanted us to find \(P_1 P_2 + P_2 P_3 + P_3 P_4 + \dots + P_9 P_{10} + P_{10} P_1\)

Notice that this is literally just the perimeter of decagon \(P_1 P_2 P_3 \dotsb P_{10}\)

The length of one side of the decagon is

\( \text{radius}/2 ( -1 + \sqrt {5}) \)

Multiplying this by 10 and we know the perimeter. We get

\( 10 (1/2) ( -1 +\sqrt 5) = 5 ( -1 + \sqrt 5) ≈ 6.18\)

Thus, 6.18 is approximately our answer. 

Thanks! :)

Find the discriminant of the quadratic 5x^2 - 2x + 8 - 2x^2 + 6x + 2 - 2x^2 + 4x - 1.    

First, collect and combine like terms   

and arrange them as ax² + bx + c           (5x² – 2x² – 2x²) + (–2x + 6x + 4x) + (+8 + 2 – 1)    

                                                                 x² + 8x + 9    

The discriminant is b² – 4ac                     D  =  8² – (4)(1)(9)    

                                                                 D  =  64 – 36    

                                                                 D  =  28    

_.   

Observe that x2+3xy+y2−2x−2y+4​=(x+y−1)2+3(xy+4)​.

Thus, x⋆y=xy+4(x+y−1)2+3(xy+4)​​. Let an​=(((⋯((2007⋆2006)⋆2005)⋆⋯)⋆n).

Then we can write the recurrence relation an​=an−1​n+4(an−1​+n−1)2+3(an−1​n+4)​​.

Note that a1​=1. We can compute a2​,a3​,… using this recurrence relation. A

fter some computations, we find that a_2007 = 2008/2007.

We are tasked with finding all possible values of \( b \) that satisfy the following system of equations involving the floor functions:

\[
\lfloor a \rfloor \cdot b \cdot c = 3 \tag{1}
\]

\[
a \cdot \lfloor b \rfloor \cdot c = 4 \tag{2}
\]

\[
a \cdot b \cdot \lfloor c \rfloor = 5 \tag{3}
\]

where \( a \), \( b \), and \( c \) are positive real numbers. We will solve this step by step by exploring possible values for \( a \), \( b \), and \( c \).

### Solution By Steps

#### Step 1: Analyze the first equation

From Equation (1):

\[
\lfloor a \rfloor \cdot b \cdot c = 3
\]

The floor function \( \lfloor a \rfloor \) represents the greatest integer less than or equal to \( a \). Since \( a \) is positive, the possible values of \( \lfloor a \rfloor \) are integers.

We first assume \( \lfloor a \rfloor = 1 \), which is the smallest possible value because \( a > 0 \). Substituting this into Equation (1):

\[
1 \cdot b \cdot c = 3
\]

Thus, we have:

\[
b \cdot c = 3 \tag{4}
\]

#### Step 2: Analyze the second equation

Now, consider Equation (2):

\[
a \cdot \lfloor b \rfloor \cdot c = 4
\]

We already know \( b \cdot c = 3 \) from Equation (4). To satisfy this equation, we explore possible values of \( \lfloor b \rfloor \).

Let’s assume \( \lfloor b \rfloor = 1 \) first. Substituting into Equation (2):

\[
a \cdot 1 \cdot c = 4
\]

So,

\[
a \cdot c = 4 \tag{5}
\]

#### Step 3: Analyze the third equation

Now consider Equation (3):

\[
a \cdot b \cdot \lfloor c \rfloor = 5
\]

We already have \( b \cdot c = 3 \) and \( a \cdot c = 4 \). To satisfy this equation, we explore possible values of \( \lfloor c \rfloor \).

Let’s assume \( \lfloor c \rfloor = 1 \). Substituting into Equation (3):

\[
a \cdot b \cdot 1 = 5
\]

So:

\[
a \cdot b = 5 \tag{6}
\]

#### Step 4: Solve the system of equations

Now, we have three equations:

1. \( b \cdot c = 3 \)

2. \( a \cdot c = 4 \)

3. \( a \cdot b = 5 \)

We can solve this system step by step. First, solve for \( c \) from Equation (5):

\[
c = \frac{4}{a}
\]

Substitute this into Equation (4):

\[
b \cdot \frac{4}{a} = 3
\]

Simplifying:

\[
b = \frac{3a}{4}
\]

Now substitute this into Equation (6):

\[
a \cdot \frac{3a}{4} = 5
\]

Simplifying:

\[
\frac{3a^2}{4} = 5
\]

Multiply both sides by 4:

\[
3a^2 = 20
\]

Solve for \( a^2 \):

\[
a^2 = \frac{20}{3}
\]

So:

\[
a = \sqrt{\frac{20}{3}} = \frac{2\sqrt{15}}{3}
\]

#### Step 5: Find the value of \( b \)

Now that we have \( a = \frac{2\sqrt{15}}{3} \), substitute this back into the expression for \( b \):

\[
b = \frac{3a}{4} = \frac{3 \times \frac{2\sqrt{15}}{3}}{4} = \frac{2\sqrt{15}}{4} = \frac{\sqrt{15}}{2}
\]

Thus, the value of \( b \) is:

\[
b = \frac{\sqrt{15}}{2}
\]

Yay, I took Number Theory recently, and I seem to remember how to solve problems like this. The key detail to know is that the phi function is multiplicative. Mathematically, this means that if m and n are relatively prime, then \(\phi(mn) = \phi(m)\phi(n)\). As a consequence, we can represent 2835 into a product of prime powers in the form of \(2835 = p_1^{e^1}p_2^{e_2} \cdots p_n^{e^n}\) where each p_i is a unique prime and each e_i is the corresponding exponent and then solve the easier resulting problem. Through some trial division, I discovered \(2835 = 3^4*5*7\), so the property of multiplicativity tells us that \(\phi(2835) = \phi(3^4)\phi(5)\phi(7)\). I could just use the formula to solve the rest, but I could not recall it, so I solved it intuitively instead.

\(\phi(5) = 4 \text{ and } \phi(7) = 6\) because the number of relatively prime positive integers less than or equal to a prime p is p - 1 because no integers under consideration, except p itself, will share a common factor.

\(\phi(3^4) = 3^4 - 3^3 = 81 - 27 = 54\) because the number of relatively prime positive integers less than or equal to p^e is p^e - p^{e - 1}. There are p^e integers under consideration and p^{e - 1} of them share a common factor with the prime p because p^e / p = p^{e - 1}, so the remaining integers are relatively prime.

Therefore, \(\begin{align*} \phi(2835) &= \phi(3^4)\phi(5)\phi(7) \\ &= 54 * 4 * 6 \\ &= 1296 \\ \end{align*}\).

The first two sentences is just a complicated way of descibing a hexagon with a side length of 3. The area of a hexagon with a side length n is\(\frac{3\sqrt{3}}{2}n^2\)

Therefore,

\(A = \frac{3\sqrt{3}}{2}6^2\)

\(=\frac{3\sqrt{3}}{2}36\)

\(3\sqrt{3}(18)\)

\(=\mathbf{54\sqrt{3}}\)

The formula for # of diagonals for a n-sided poligon is \(\frac{n (n - 3)}{2}\). Therefore,

\(\frac{n(n-3)}{2} = 2n\)

\(n^2 - 3n = 4n\)

\(n^2-7n=0\)

\(n(n-7)=0\)

\(n = 0, 7\)

And since there are no 0-sided polygons, the polygon has 7 sides

Case 1: x = 1

Then u + v + w + y + z = 1, and since none of them can be negative, exactly one of (u, v, w, y, z) is one, and the rest are zero. There are 5 ways to do this.

Case 2: x = 2

Then u + v + w + y + z, and using the same logic, the only case is u = v = w = y = z = 0.

Therefore there are 5 + 1 = 6 solutions

The Euclidea Algorith says that gcd(a, b) = gcd(a, b - a). Therefore, gcd(972, 1220) = gcd(972, 248) = gcd(228, 248) = gcd(228, 20) = gcd(8, 20) = gcd(8, 4) = gcd(0, 4) = 4

We can rewrite this as\(1000 \leq 47x + 3 \leq 2000\), which we can simplify

\(997 \leq 47x \leq 1997\)

\(21\frac{10}{47}\leq x\leq 42\frac{23}{47}\)

The only natural numbers that fit in that range are 22 through 42, and there are 21 numbers in that range.

Lets express this as normal notation:

(sorry I don't use Latex)

1/x + 1/y + 1/z = 13/12

Wait! We know what 1/x + 1/y + 1/z is ! We just express both as a common fraction (by expressing them as a common denominator)! 

(1/x + 1/y + 1/z) = (x + y) + xyz/xyz

So now:

(x + y + xyz/)xyz = 13/12

x + y +z= 13, and xyz = 12

Two equations, three variables isn't possible to solve, so we can only test numbers :(, luckily, the problem says only three distinct positive integers!

After some simple testing, we should get that only 1, 4, and 8 work. 

I hope it helps, and of course, your welcome!