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I sincerely apologize for not being able to solve this and making your homework late.

You take the four Aces, four 2's, and four 3's from a standard deck of 52 cards, forming a set of 12 cards.

You then deal all 12 cards at random to four players, so that each player gets three cards.

What is the probability that each player gets an Ace, a 2, and a 3?

Speed is the rate of change of distance with respect to time, so the greatest speed will be at the point where the graph of distance against time is steepest.

It's the standard formula for solving quadratic equations (though I should have written "t = ..." not "x = ..."), which is often necessary when the quadratic equation doesn't factor nicely (as yours doesn't).

I'm not quite sure we've had this ? This looks overly complicated. 

Use the quadratic equation to get the two values of t:

$t = {-(-25) \pm \sqrt{(-25)^2-4\times 75} \over 2}$

I'll leave you to crunch the numbers.

Edited to replace x by t.

"How many distinct, non-equilateral triangles with a perimeter of 60 units have integer side lengths a, b and c such that a, b, c is an arithmetic sequence?"

Let k be a positive real number. The square with the vertices (k,0), (0,k), (-k,0), and (0,-k) are plotted on the coordinate plane.

Find conditions on a>0 and b>0 such that the ellipse 

is contained inside the square (and tangent to all of its sides)

HINTS: suppose that the line x+y=k is tangent to the ellipse 

Algebraically, what can we say about the solutions? In particular, the number of solutions?

Each week, between 30 and 50 students show up for an archery class run by Betty and Wilma.

Usually the students break up into groups of equal size for target practice.

However, this week, Betty noticed that she could not break the students up into multiple groups of equal size.

Wilma noticed that if she and Betty both joined the students in practicing,

they still could not break the archers up into groups of equal size.

How many students showed up to the archery class this week?

$\text{Let $x$ be the number of students} \\ \text{$x$ is a prime number and $x+2$ is a prime number } \\ \text{We are looking for a $\mathbf{\text{twin prime}}$ number between $30$ and $50$}$

$x=41$ and $x+2= 43$

41 students showed up to the archery class this week

The arithmetic mean of three numbers x, y and z is 24.
The arithmetic mean of x, 2y and z – 7 is 34.
What is the arithmetic mean of x and z ?
Express your answer as a decimal to the nearest tenth.

$\begin{array}{|rcll|} \hline \mathbf{ \dfrac{x+y+z}{3} } &=& \mathbf{24} \\\\ x+y+z &=& 3*24 \quad | \quad : 2 \\\\ \dfrac{x+y+z}{2} &=& 3*12 \\\\ \dfrac{x+z}{2} + \dfrac{y}{2} &=& 3*12 \\\\ \dfrac{x+z}{2} + \dfrac{y}{2} &=& 36 \\\\ \mathbf{ \dfrac{x+z}{2} } &=& \mathbf{ 36 - \dfrac{y}{2} } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{ \dfrac{x+2y+z-7}{3} } &=& \mathbf{34} \\\\ x+2y+z-7 &=& 3*24 \quad | \quad : 2 \\\\ \dfrac{x+z}{2}+ \dfrac{2y-7}{2} &=& 3*17 \\\\ \dfrac{x+z}{2}+ \dfrac{2y-7}{2} &=& 51 \quad | \quad \mathbf{ \dfrac{x+z}{2} =36 - \dfrac{y}{2} } \\\\ 36 - \dfrac{y}{2}+ \dfrac{2y-7}{2} &=& 51 \\\\ 36 + \dfrac{2y-7-y}{2} &=& 51 \\\\ 36 + \dfrac{y-7}{2} &=& 51 \\\\ \dfrac{y-7}{2} &=& 51-36 \\\\ \dfrac{y-7}{2} &=& 15 \\\\ y-7 &=& 2*15 \\\\ y-7 &=& 30 \\ \mathbf{y} &=& \mathbf{37} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{ \dfrac{x+z}{2} } &=& \mathbf{ 36 - \dfrac{y}{2} } \quad | \quad \mathbf{ y=37} \\\\ \dfrac{x+z}{2} &=& 36 - \dfrac{37}{2} \\\\ \dfrac{x+z}{2} &=& \dfrac{35}{2} \\\\ \mathbf{ \dfrac{x+z}{2} } &=& \mathbf{17.5} \\ \hline \end{array}$

The arithmetic mean of x and z is $\mathbf{17.5}$

Regions A, B, C, J and K represent ponds.

Logs leave pond A and float down flumes (represented by arrows) to eventually end up in pond B or pond C.

On leaving a pond, the logs are equally likely to use any available exit flume.

Logs can only float in the direction the arrow is pointing.

What is the probability that a log in pond A will end up in pond B?

Express your answer as a common fraction.

$\begin{array}{|rcll|} \hline \mathbf{C} &=& \mathbf{(AJ-JC) + (AJ-CJ) + (AK-KC) + AC} \\\\ C &=& \left(\dfrac{1}{3}*\dfrac{1}{3}\right) +\left(\dfrac{1}{3}*\dfrac{1}{3}\right) + \left(\dfrac{1}{3}*\dfrac{1}{2}\right) + \dfrac{1}{3} \\\\ C &=& \dfrac{1}{9} + \dfrac{1}{9} + \dfrac{1}{6} + \dfrac{1}{3} \\\\ C &=& \dfrac{2}{9} + \dfrac{1}{6} + \dfrac{1}{3} \\\\ C &=& \dfrac{2}{9} + \dfrac{1}{6} + \dfrac{3}{9} \\\\ C &=& \dfrac{5}{9} + \dfrac{1}{6} \\\\ C &=& \dfrac{5*6+1*9}{9*6} \\\\ C &=& \dfrac{39}{54} \\\\ \mathbf{C} &=& \mathbf{\dfrac{13}{18}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{B} &=& \mathbf{(AJ-JB) + (AK-KB)} \\\\ B &=& \left(\dfrac{1}{3}*\dfrac{1}{3}\right) + \left(\dfrac{1}{3}*\dfrac{1}{2}\right) \\\\ B &=& \dfrac{1}{9} + \dfrac{1}{6} \\\\ B &=& \dfrac{6+9}{9*6} \\\\ B &=& \dfrac{15}{54} \\\\ \mathbf{B} &=& \mathbf{\dfrac{5}{18}} \\ \hline \end{array}$

Let
$z = \frac{\sqrt{2 + \sqrt{3}}}{2} + i \cdot \frac{\sqrt{2 - \sqrt{3}}}{2}$
Find

$1 + z + z^2 + z^3 + \dots + z^{2016}$

$\begin{array}{|rcll|} \hline \mathbf{z} &=& \mathbf{\frac{\sqrt{2 + \sqrt{3}}}{2} + i \cdot \frac{\sqrt{2 - \sqrt{3}}}{2}} \\\\ z &=& \sqrt{ \left( \frac{\sqrt{2 + \sqrt{3}}}{2} \right)^2 + \left( \frac{\sqrt{2 - \sqrt{3}}}{2} \right)^2 } e^{ i* \arctan\left( \frac{\sqrt{2 - \sqrt{3}}} {\sqrt{2 + \sqrt{3}}} \right) } \\\\ z &=& 1* e^{ i*\frac{\pi}{12} } \\\\ \mathbf{z} &=& \mathbf{e^{ i*\frac{\pi}{12} }} \\ \hline \end{array}$

Sum of a geometric sequence

$\begin{array}{|rcll|} \hline \mathbf{1 + z + z^2 + z^3 + \dots + z^{2016}} &=& \mathbf{\dfrac{z^{2017}-1}{z-1}} \\\\ &=& \dfrac{\left( e^{ i*\frac{\pi}{12} } \right)^{2017}-1}{e^{ i*\frac{\pi}{12} }-1} \\\\ &=& \dfrac{ e^{ i*\frac{2017}{12}\pi } -1}{e^{ i*\frac{\pi}{12} }-1} \\\\ &=& \dfrac{ e^{ i*30255^\circ } -1}{e^{ i*15^\circ }-1} \\\\ &=& \dfrac{ \cos(30255^\circ) +i\sin(30255^\circ) -1}{\cos(15^\circ) +i\sin(15^\circ)-1} \\\\ &=& \dfrac{ \cos(15^\circ) +i\sin(15^\circ) -1}{\cos(15^\circ) +i\sin(15^\circ)-1} \\\\ \mathbf{1 + z + z^2 + z^3 + \dots + z^{2016}} &=& \mathbf{1} \\ \hline \end{array}$

13 ,  31 , 33 , 35 , 53 >>Total = 5

Probability =5 / 6^2 =5 / 36

"A steep mountain is inclined 70 degree to the horizontal and rises 3500 ft above the surrounding plain. A cable car is to be installed by connecting a cable from the top of the mountain to a spot on the plain that is 1000 ft from the base of the mountain. Find the shortest length of cable needed. "

Beween 1 and 488 inclusive, the number 4 appears =188 times

Between 1 and 488 inclusive, the number 8 appears =98 times

188  -  98 =90 more "4's" appear than "8's"

Convert 

$2*e^{i \frac{\pi}{6} }$

to rectangular form.

$\begin{array}{|rcll|} \hline 2*e^{i \frac{\pi}{6} } &=& 2 \Bigg(\cos\left( \frac{\pi}{6}\right) +i\cos\left(\frac{\pi}{6} \right) \Bigg) \\\\ 2*e^{i \frac{\pi}{6} } &=& 2 \Bigg(\dfrac{\sqrt{3}}{2} +i*\dfrac{1}{2} \Bigg) \\\\ \mathbf{2*e^{i \frac{\pi}{6} }} &=& \mathbf{\sqrt{3}+1} \\ \hline \end{array}$

"What is the rectangular equivalence to the parametric equations?

x(θ)=4cosθ+2,y(θ)=2sinθ−5 , where 0≤θ<2π .

Drag a term into each box to correctly complete the rectangular equation."

@heureka, what do $x_t$ and $y_t$ mean in your solution? Are they bases from logs?

no

$(x_t,y_t)$ are the coordinates of the tanget point P at the ellipse on my line $y = x+k$. See my link and the graph.

The tanget point of the circle is $(-10,10)$, so $x_t=-10$ and $y_t =10$.

The sum is 1.  See the following:

$Mass = volume\ times\ density.$

$The\ density\ of\ water\ is\ \frac{1kg}{dm^3}.$

  !

When

$x = \dfrac{3 + 5i}{2}$,

find the value of  $2x^3 + 2x^2 - 7x + 72$.

$\begin{array}{|rcll|} \hline x^2 &=& \left(\dfrac{3 + 5i}{2}\right)^2 \\\\ x^2 &=& \dfrac{(3 + 5i)^2}{4} \\\\ x^2 &=& \dfrac{9+30i-25}{4} \\\\ x^2 &=& \dfrac{-16+30i}{4} \\\\ \mathbf{x^2} &=& \mathbf{\dfrac{-8+15i}{2}} \\ \hline \end{array} \begin{array}{|rcll|} \hline x^3 &=& \dfrac{(-8+15i)}{2}*\dfrac{(3 + 5i)}{2} \\\\ x^3 &=& \dfrac{(-8+15i)(3 + 5i)}{4} \\\\ x^3 &=& \dfrac{-24-40i+45i-75}{4} \\\\ x^3 &=& \dfrac{-99+5i}{4} \\\\ \mathbf{x^3} &=& \mathbf{\dfrac{-99+5i}{4}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{2x^3 + 2x^2 - 7x + 72} &=& 2*\dfrac{(-99+5i)}{4} + 2*\dfrac{(-8+15i)}{2} - 7*\dfrac{(3 + 5i)}{2} + 72 \\\\ &=& \dfrac{-99+5i}{2} + (-8+15i)- \dfrac{7(3 + 5i)}{2} + 72 \\\\ &=& \dfrac{-99+5i}{2} -8+15i- \dfrac{21 + 35i}{2} + 72 \\\\ &=& \dfrac{-99+5i}{2} - \dfrac{21 + 35i}{2} + 64 +15i \\\\ &=& \dfrac{-99+5i-(21 + 35i)}{2} + 64 +15i \\\\ &=& \dfrac{-99+5i-21 - 35i}{2} + 64 +15i \\\\ &=& \dfrac{-120-30i}{2} + 64 +15i \\\\ &=& -60-15i + 64 +15i \\\\ &=& \mathbf{4} \\ \hline \end{array}$

@heureka, what do $x_t$ and $y_t$ mean in your solution? Are they bases from logs?

Let $k$ be a positive real number. The square with vertices  $(k, 0)$, $(0, k)$, $(-k, 0)$, and $(0, -k)$ is plotted in the coordinate plane.
Find conditions on $a > 0$ and $b > 0$ such that the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ is contained inside the square (and tangent to all of its sides).

I Suppose that the line $\mathbf{y=x+ k}$ is tangent to the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$.

Thank you, Guest !