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We want to find the exponential form of the complex number:

\[
e^{\frac{17\pi i}{60}} + e^{\frac{27\pi i}{60}} + e^{\frac{37\pi i}{60}} + e^{\frac{47\pi i}{60}} + e^{\frac{57\pi i}{60}}
\]

### Step 1: Recognize the terms as roots of unity

The terms \( e^{\frac{17\pi i}{60}}, e^{\frac{27\pi i}{60}}, e^{\frac{37\pi i}{60}}, e^{\frac{47\pi i}{60}}, e^{\frac{57\pi i}{60}} \) are all complex numbers in exponential form. Notice that these exponents can be expressed as:

\[
\frac{17\pi i}{60}, \frac{27\pi i}{60}, \frac{37\pi i}{60}, \frac{47\pi i}{60}, \frac{57\pi i}{60}
\]

These correspond to angles \( \theta = \frac{17\pi}{60}, \frac{27\pi}{60}, \frac{37\pi}{60}, \frac{47\pi}{60}, \frac{57\pi}{60} \).

Since \( e^{i\theta} \) represents a point on the unit circle in the complex plane, each of these terms can be considered as specific roots of unity, although not all are primitive roots.

### Step 2: Consider the sum of the angles

The sum can be expressed as:

\[
S = e^{\frac{17\pi i}{60}} + e^{\frac{27\pi i}{60}} + e^{\frac{37\pi i}{60}} + e^{\frac{47\pi i}{60}} + e^{\frac{57\pi i}{60}}
\]

These angles are evenly spaced on the unit circle by an angle increment of \( \frac{10\pi}{60} = \frac{\pi}{6} \).

### Step 3: Utilize symmetry

These angles correspond to the roots of the equation \( x^5 - 1 = 0 \) rotated by a small angle \( \frac{17\pi}{60} \). The angles \( \frac{17\pi}{60}, \frac{27\pi}{60}, \frac{37\pi}{60}, \frac{47\pi}{60}, \frac{57\pi}{60} \) map to the five vertices of a regular pentagon inscribed in the unit circle but rotated slightly.

For a regular \( n \)-gon (in this case, \( n = 5 \)), the sum of vectors corresponding to the vertices is zero if the vectors are evenly distributed around the circle (because they symmetrically cancel each other out). However, here, the vertices are slightly rotated, but they are still symmetrically spaced around the circle.

### Step 4: Apply properties of roots of unity

Given the symmetry and spacing:

\[
S = e^{i\frac{\theta_0}{60}} \cdot \left(1 + e^{i\frac{2\pi}{5}} + e^{i\frac{4\pi}{5}} + e^{i\frac{6\pi}{5}} + e^{i\frac{8\pi}{5}}\right)
\]

Where \( \theta_0 \) is \( \frac{17\pi}{60} \), and the roots sum to zero because they represent the vertices of a pentagon. Thus, the sum

:

\[
S = 0
\]

### Conclusion:

The exponential form of the given complex number is:

\[
\boxed{0}
\]

Given that triangle \(ABC\) has altitudes \(AD = 14\), \(BE = 16\), and \(CF = h\), where \(h\) is a positive integer, we need to find the largest possible value of \(h\).

### Step 1: Use the area formula with altitudes

The area of triangle \(ABC\) can be expressed in terms of the altitudes and the corresponding sides:

\[
\text{Area} = \frac{1}{2} \times a \times h_a = \frac{1}{2} \times b \times h_b = \frac{1}{2} \times c \times h_c
\]

where \(h_a = AD\), \(h_b = BE\), and \(h_c = CF = h\), and \(a\), \(b\), and \(c\) are the lengths of the sides opposite the vertices \(A\), \(B\), and \(C\), respectively.

### Step 2: Express the sides in terms of the area and altitudes

Let the area of triangle \(ABC\) be denoted as \(K\). Then:

\[
K = \frac{1}{2} \times a \times 14 = 7a
\]

\[
K = \frac{1}{2} \times b \times 16 = 8b
\]

\[
K = \frac{1}{2} \times c \times h = \frac{1}{2}ch
\]

### Step 3: Equate the expressions for the area \(K\)

From the first two equations:

\[
7a = 8b \quad \Rightarrow \quad a = \frac{8b}{7}
\]

Using the equation for \(K\) involving \(h\):

\[
7a = \frac{1}{2} ch
\]

Substitute \(a = \frac{8b}{7}\) into the equation:

\[
7\left(\frac{8b}{7}\right) = \frac{1}{2}ch
\]

Simplify:

\[
8b = \frac{1}{2}ch
\]

Multiply both sides by 2:

\[
16b = ch
\]

Thus:

\[
h = \frac{16b}{c}
\]

### Step 4: Maximize \(h\)

We aim to maximize \(h = \frac{16b}{c}\), where \(b\) and \(c\) are positive integers.

Recall that \(K = 7a = 8b = \frac{1}{2}ch\). We want to maximize \(h\), so:

\[
c = \frac{16b}{h}
\]

To maximize \(h\), minimize \(c\). Since \(b = 7\), \(a = 8\), choose \(b = 7\) and \(c = 1\) which gives:

\[
h = \frac{16 \times 7}{1} = 112
\]

However, since \(b = \gcd(7,8)\), choose \(b = 7\) for simplicity. So:

\[
K = 112, h = 1
]

Thus:

\[
h \leq 112
\]

### Final Answer:

Since \(h = 112\), the maximum value for the given maximum number is \( \boxed{112} \).

We can complete this problem in two different ways. 

The first tactic is to essentially compare this root to the quadratic equation of \(5x^2+21x+v = 0\)

From this quadratic, we can identify that a = 5, b = 21, c = v. 

We have the equation

\(\frac{-21-\sqrt{201}}{10} = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ \frac{-21-\sqrt{201}}{10} = {-21 \pm \sqrt{21^2-4(5)(v)} \over 2(5)}\\ \frac{-21-\sqrt{201}}{10} = {-21 - \sqrt{441-20v} \over 10}\\ \sqrt{201} = \sqrt{441-20v}\\ 201 = 441-20v\\ v=12\)

This is a bit complicated and takes a lot of computations, but it does give us the correct answer. 

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The second tactic is to use conjugations of square roots. 

This is because the conjugate root theorem states that if a root of a polynomial is a square root \(a+\sqrt b\), then its conjugate, \(a-\sqrt b\) is also a root

We can apply that to this problem. If  \(\frac{-21-\sqrt{201}}{10}\) is a root, then \(\frac{-21 + \sqrt {201}}{ 10 }\) is also a root. 

The product of the roots is \([ (-21)^2 - 201 ] / 100 = 240/100 = 2.4\)

However, in the quadratic, we also have that \(v / 5\)

Thus, we have 

\(v/5 = 2.4\\ v = 12\)

SO 12 is the final answer. 

Thanks! :)

You are missing some values in your question, but I'm gonna assume the following was asked. 

A machine randomly generates one of the nine numbers   1,2,3...8,9   with equal likelihood. What is the probability that when Tsuni uses this machine to generate four numbers their product is divisible by 4. Express your answer as a common fraction.

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Now, since we can draw two of the same numbers and order doesn't matter, we don't do \(9\choose 4\) to find the total amount of possibilites, but instead just do

\(9 \cdot 9\cdot 9 \cdot 9 = 9^4 = 6561\)

Now, we can use complementary counting to find the number of cases that DO NOT work and then subtract it from 1. 

This is because it's MUCH easier to calculate odd numbers rather than even numbers. 

Our first case is that all 4 numbers are odd. If all 4 numbers are odd, the product is odd, therefore meaning it cannot possibly be divisble by 4. 

Since out of the 9 numbers, 5 numbers are odd, we have a \(5/9\) chance each generation that the number is odd. 

Since we need 4 odd numbers in a row, we simply have \(({5 \over 9})^4 = {625 \over 6561}\)

Our next case is where we get 3 odd numbers in a row and then a 2 or a 6. If we roll a 4 or 8, the product will automatically be divisble by 4. 

Thus, we have \({2 \times 5^3 \times {4 \choose 3} \over 6561} = {1000 \over 6561}\)

Subtracting the sum from 1, we get the final answer of \(1 - ({625 \over 6561} + {1000 \over 6561} ) = \color{brown}\boxed{4936 \over 6561}\)

I would like verification of the final answer, but I believe this is the correct answer. 

Thanks! :)

First, we have to simplify the left side of the equation. 

Expanding it out, we get

\(2x^2-x-28=-31+jx\)

Moving all terms to one side and combining all like terms, we achieve

\(2x^2-x-jx-28+31=0\\ 2x^2-\left(j+1\right)x+3=0\)

If there's only one solution, then that must mean that the descriminant is 0. 

The descriminant of this quadratic is 

\((j+1)^2 - 4(2)(3)\\ j^2+2j-23\)

Setting this to 0, we can now solve for j. We get

\(j^2+2j-23=0\)

Unfortunately, j cannot be factored conveniently, so we must use the quadratic equation. (I'm going to leave that for you). 

From that, we find that

\(j=-1+2\sqrt{6},\:j=-1-2\sqrt{6}\)

I'm pretty sure this is the correct answer, but you can check and verify. 

Thanks! :)

mmm...take a look at the points given. 

\(P_1 P_2 + P_2 P_3 + P_3 P_4 + \dots + P_9 P_{10} + P_{10} P_1\)

This is essentially just the perimeter of the decagon \(P_1P_2P_3P_4P_5...P_{10}\)

The lenngth of one side is essentially

\(radius/2 ( -1 +\sqrt 5)\)

Now that we know one side, we can find the perimeter through the equation

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

Thus, the answer is 6.18, 

Thanks! :)

Vertex (0,0) form of this parabola 

y = d ( x-0)^2  + 0 

y = dx^2          When 2   Y = 8     So 'd' = 2 

then the equation is    Y = 2x^2        Then   A + b + c = 2 

Simplify L side 

x^2 +18x +3  = ...............      complete the square 

(x +18x + 81)  +3   - 81 

(x +9)^2 - 78           Done      

We have a triangle JKL with midpoints M, N, and O of its sides. Then, we have another triangle PQR formed by the midpoints of the sides of triangle MNO.

The area of triangle PQR is given as 12. We need to find the area of triangle JQR.

Key Concept

Midpoint Theorem: The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and half its length.  

Area Scaling: If the sides of a triangle are scaled by a factor of 'k', then its area is scaled by a factor of 'k^2'.

Solution

Triangle MNO and Triangle JKL:

Since M, N, and O are midpoints of JK, KL, and LJ respectively, triangle MNO is similar to triangle JKL with a scale factor of 1/2 (due to the midpoint theorem).

Therefore, the area of triangle MNO is (1/2)^2 = 1/4 times the area of triangle JKL.

Triangle PQR and Triangle MNO:

Similarly, triangle PQR is similar to triangle MNO with a scale factor of 1/2.

So, the area of triangle MNO is (1/2)^2 = 1/4 times the area of triangle PQR.

Finding the area of triangle JKL:

We know the area of triangle PQR is 12.

So, the area of triangle MNO is 12 * 4 = 48.

Therefore, the area of triangle JKL is 48 * 4 = 192.

Finding the area of triangle JQR:

Triangle JQR is half of triangle JKL (since QR is a median).

So, the area of triangle JQR is 192 / 2 = 96.

Therefore, the area of triangle JQR is 96.

Let's set up a system of equations to solve this problem. 

Let's let b be bagels

Let's let t be toast

And let's let m be muffins. 

From the problem, we have the system

\(2T + 1B = 3.15 \\ 1M + 1B = 3.50 \\ 1T + 2B + 3M = 8.15\)

Now, we want everything to be in terms of bagels. 

From the first two equations, we get the equations

\([3.15 - B ] / 2 = T \\ M = 3.50 - B\)

Now, plugging these values into the third equation, we get

\([ 3.15 - B ] / 2 + 2B + 3 [ 3.50 - B ] = 8.15 \\ [3.15 - B ] + 4B + 6 [ 3.50 -B ] = 16.30 \\ -3B + 3.15 + 21 = 16.30 \\ -3B = 16.30 - 3.15 - 21 \\ -3B = -7.85 \\ B = 7.85 / 3 ≈ $2.62\)

This is 262 cents or about 261 and 2/3 cents. 

Thanks! :)

Let's give this problem a shot. 

Now, we know that \((a+b)^2 = a^2+2ab+b^2\)

From the problem, we ALSO know that \((a+b)^2 = 1^2=1\)

Thus, we have the equation \(a^2+2ab+b^2=1\)

Since the problem gives us the value of \(a^2+b^2=2\), we can easily figure out what ab is. 

We have

\(2ab+2=1\\ 2ab=-1\\ ab=-1/2\)

The value of ab will come in handy later. 

Now, let's focus on what we must find. a^3+b^3. From a handy equation, we know that

\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

Wait! We already know all the values needed to solve the problem. Plugging in 1, 2, and -1/2, we get

\(a^3+b^2=(1)(2-(-1/2)) = 2+1/2 = 5/2\)

Thus, our final answer is 5/2. 

Thanks! :)

Let's make some observations of the problem. 

First, note that triangle ABC is a 45-45-90 right triangle. This means that \(AB = BC = 12/\sqrt 2 = 6\sqrt 2\)

Also, we have that

\(AD = AD\\ BD = BD \\ AB = BC\)

This means that triangles  ABD  and CBD  are congruent

Thus, we have \([ ABD ] = (1/2) [ABC] = (1/2) ( 1/2) ( 6\sqrt 2)^2 = (1/4) (72) = 18\)

So our aswer is 18. 

The answer is \(5.82\)