The3Mathketeers

$5x^2 - kx + 8 - 2x^2 + 25 = 0 \\ 3x^2 - kx + 33 = 0$

After some simplification, I arrive at the following quadratic. In order to analyze when this quadratic has no real solutions, evaluate the discriminant, specifically when the discriminant is less than zero.

$\Delta = b^2 - 4ac \\ \Delta = (-k)^2 - 4 * 3 * 33 \\ k^2 - 396 < 0 \\ k^2 < 396 \\ k < \sqrt{396} \approx 19.900$

When the discriminant is less than zero, there are no real solutions, so k = 19 is the largest integer that makes the resulting quadratic have no real solutions.

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. 

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

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

Since a, b, c, d, and e are all variations of a similar question, I think it is fair just to do problems a and d and leave the rest.

Expected value is the average outcome of running a long-term experiment, which is calculated with the formula $E(X) = \sum_{i = 1}^n x_iP(x_i)$ where E(X) is the expected value, x_i is a specific outcome, and P(x_i) is the probability of x_i occurring. Let's apply this to problem a.

Upon drawing a slip of paper from the bag, 2 possible outcomes can occur: 3 or 8 with their respective probabilities. With the unmodified bag, 3 occurs with a probability of 8/10 because there are eight 3's and ten slips of paper. Likewise, 8 occurs with a probability of 2/10 because there are two 8's and ten slips of paper. Substitute this information into the formula and the answer follows naturally.

$\begin{align} E(X) &= 3 * \frac{8}{10} + 8 * \frac{2}{10} \\ &= \frac{24}{10} + \frac{16}{10} \\ &= 4 \end{align}$

Therefore, E(X) = 4 for problem a.

Next, let's consider problem d. Here, the expected value is given, so E(X) = 6. Adding 8's to the original contents of the bag increases both the number of 8's and the total number of 8's in the bag. Let a represent the number of 8's to add to the bag such that E(X) = 6. This would result in the following equation.

$\begin{align*} E(X) &= \sum_{i = 1}^n x_iP(x_i) \\ 6 &= 3 * \frac{8}{10 + a} + 8 * \frac{2 + a}{10 + a} \\ 60 + 6a &= 24 + 16 + 8a \\ 20 &= 2a \\ a &= 10 \end{align*}$

Therefore, add ten 8's to the bag. To be a proper mathematics steward, note that 10 + a appears in the denominator. This means a ≠ -10 under any circumstance. However, this is a non-issue regardless because the variable a represents the number of 8's added, so a negative value for a is nonsensical anyway.

The given quadratic equation is $ax^2 + 8x + 4 = 6x - 40$. First, simplify it to standard form.

$ax^2 + 8x + 4 = 6x - 40 \\ ax^2 + 2x + 44 = 0$

Now, we can use the discriminant of a quadratic, which gives information about the number of solutions. In order for the quadratic to have only one solution, the discriminant should be equal to 0.

$D = 0 \\ b^2 - 4ac = 0 \\ 2^2 - 4 * a * 44 \\ 4 - 176a = 0 \\ 176a = 4 \\ a = \frac{4}{176} \\ a = \frac{1}{44}$

According to the given statement, $\triangle {\text{PQR}} \cong \triangle {\text{STU}}$. Corresponding angles of congruent triangles have equal measure. Therefore, since $m \angle \text{S} = 78^{\circ}$, this means that $m \angle \text{P} = 78^{\circ}$. By the widely known Triangle Sum Angle Theorem, the sum of the interior angles of the triangle equals 180 degrees.

$m \angle {\text{P}} + m \angle {\text{Q}} + m \angle {\text{R}} = 180^{\circ} \\ 78^{\circ} + m \angle {\text{Q}} + 80^{\circ} = 180^{\circ} \\ 158^{\circ} + m \angle {\text{Q}} = 180^{\circ} \\ m \angle {\text{Q}} = 22^{\circ}$

Strictly speaking, an exponent tower is one of the few operations in mathematics that is evaluated from the "top" of the tower to the "bottom" of the tower. Unfortunately, this number is too large to write down. Although, we can attempt to write an approximation.

${{3^3}^3}^3 = {3^3}^{27} = 3^{7\;625\;597\;484\;987} \approx {10^{10}}^{12.5609}$. The exponent is too large, so this number cannot be written down realistically. This number has about one trillion digits, which is far too many.

This question is a bit too broad, but I assume you want to simplify the expression $5(3x^2 + 9x) - 14$ as much as possible? If so, there is not much to do to simplify this particular expression other than do the expansion within the parentheses. 

$5(3x^2 + 9x) - 14 = 15x^2 + 45x - 14$

Saying that the arithmetic mean of x and y is 18 mathematically means that $\frac{x + y}{2} = 18$. Also, saying that x and y have a geometric mean of $\sqrt{47}$ means that $\sqrt{xy} = \sqrt{47}$. With this information, we want to find the value of $x^2 + y^2$. We can use clever algebraic manipulation to find this value without too much trouble.

$\frac{x + y}{2} = 18; \sqrt{xy} = \sqrt{47} \\ x + y = 36; xy = 47 \\ (x + y)^2 = 36^2 \\ x^2 + 2xy + y^2 = 1296 \\ x^2 + y^2 = 1296 - 2xy \\ x^2 + y^2 = 1296 - 2 * 47 \\ x^2 + y^2 = 1202$

Sometimes, it is hard to wrap one's head around how to determine how one quantity is related to another without a visual picture. As a result, I decided to create a Venn diagram that should make the situation clearer. I have included all the given information in the Venn diagram.

The given information states that there are twice as many students in the cooking club as in the drama club. Since we are using the a-variable to represent the number of students in the drama club, this means that 2a students are in the cooking club. We are letting the b-variable be in the middle; this represents the students who are members of both clubs. Our objective is to find an expression that represents the students in the one of the two clubs but not both.

I will start with the cooking club. The cooking club has 2a members. With the help of the diagram, it is clear there are b students who are members of both. Therefore, 2a - b represents the number of students who are only apart of the cooking club.

Considering the drama club, there are a total members. There are b students who are members of both. Therefore, a - b represents the number of students only participating in the drama club.

Now, we just find the sum of these parts. $2a - b + a - b = 3a - 2b$ students who are members of one of the clubs but not both.