TheXSquaredFactor

Assuming we are dealing with standard playing cards and we are ignoring jokers, then 26 out of the 52 cards are red. 12 cards (Jacks,  Queens, and Kings)  are face cards. 6 cards are both red and a face card, so we must subtract these to avoid double-counting these.

$P(\text{red} \cup \text{face})=P(\text{red})+P(\text{face})-P(\text{red}\cap\text{face})\\ P(\text{red} \cup \text{face})=\frac{26}{52}+\frac{12}{52}-\frac{6}{52}\\ P(\text{red} \cup \text{face})=\frac{32}{52}=\frac{8}{13}$

This question took me over an hour to solve, but I can now share my findings with you! I am making the assumption that you meant triangles AXB and AXC in place of APB and APC. It is quite algebra-intensive. If this is not what you meant, then I wasted a long time of my day today :(

I created a diagram, and I added line segments and circles where I found necessary. I realize that the diagram is quite large and cluttered. Here it is below:

I will be utilizing coordinate geometry for this problem. With the diagram in its current orientation, I know that B=(-14,0) and X=(0,0) and C=(4,0). This way, I know the length of the base of $\triangle ABC$ is 18. The only coordinate I need now is the y-coordinate of point A because that is the height of $\triangle ABC$. Finding the area of the triangle is trivial at that point. Unfortunately, finding this coordinate is not so easy. 

Since I don't know what the y-coordinate of point A, and I do not see an easy way to find the y-coordinate (or even the x-coordinate) of point A, I will denote them with variable letters. Let a = the x-coordinate of point A, and let b = the y-coordinate of point A. It would be nice if there was a way to relate a and b together, and I see one way of doing that. Since $XA=13$, I can use the distance formula to relate a and b together like so:

$d^2_{XA}=(a-0)^2+(b-0)^2\\ 13^2=a^2+b^2\\ b^2=169-a^2\\ b=\sqrt{169-a^2}$

b is now written in terms of a, but now I am going to introduce a theorem that seems unrelated. That theorem states the following: The perpendicular bisector of a chord of a circle passes through the center of that circle. I have slyly used that theorem to my advantage here. $\overline{AX}$ is a chord of both circumcircles of $\triangle ABX$ and $\triangle ACX$. $\overleftrightarrow{ED}$ is the perpendicular bisector of $\overline{AX}$, which, by the theorem I just stated, also pass through the circumcenters of both triangles. 

We can find the slope of $\overline{AX}$ and then use the properties of perpendicular line to find the slope of $\overleftrightarrow{ED}$.

$m​​_{\overline{AX}}=\frac{b}{a}\\ \therefore m_{\overleftrightarrow{ED}}=\frac{-a}{b}$

But we can do more than that! Point M on the diagram represents the midpoint of $\overline{AX}$, and we can find those coordinates in terms of a and b, too.

$M=(\frac{a}{2},\frac{b}{2})$.

But we can do more that that, too. We can now derive an equation of $\overleftrightarrow{ED}$ because we know the slope of the line and one point of the line. I will use point-slope form to start and simplify as much as possible from there.

Remember that theorem I mentioned earlier? Yeah, we are using that one again. $\overline{BX}$ is a chord of circle E, so E's center lies on its perpendicular bisector, which also lies on the line $x=-7$. By parallel reasoning, circle D's center lies on $x=2$. Let's use this newly discovered information to our advantage again!

We now know the x- and y-coordinates of the both circumcircles, so we can use the distance formula to write an expression for both radii.

Conveniently, both radii are of the same length, so we can set the distances we just found equal to each other. We will finally be able to isolate a variable. How exciting! At this point, the rest is simply a test of your algebraic finesse. For an added challenge, I decided to avoid the use of a calculator.

Finally! Remember that b was the y-coordinate of point A, which is also the height of $\triangle ABC$. Now, we can finally find the area of the triangle.

$A_{\triangle ABC}=\frac{1}{2}*18*12\\ A_{\triangle ABC}=108$

That's it! We are done.
 

Thanks for the recognition :)

An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant. 

The sum of n terms of an arithmetic sequence can be found with the formula $S_n=n*\frac{a_1+a_n}{2}$.

The term $a_n$ can be found with the following formula $a_n=a_1+d(n-1)$. When we put these formulas together, we can solve for d.

$S_{7}=7*\frac{a_1+a_7}{2}\\ a_7=a_1+6d\\ 63=\frac{7(2a_1+6d)}{2}\\ 9=a_1+3d\\ 9=3+3d\\ 3=1+d\\ d=2$

An arithmetic sequence is a sequence of numbers such that the difference of consecutive terms is constant. One can find the nth term of an arithmetic sequence with $a_n=a_1+d(n-1)$.

$a_n$ represents the value of the nth term of the sequence.

$a_1$ represents the value of the first term of the sequence.

$d$ is the common difference, or the difference of consecutive terms

$n$ represents the desired term, 10 in this case.

The common difference is 7, and the first term is 8, and we are looking for the 10th term of this sequence.

$a_{10}=8+7(10-1)\\ a_{10}=8+70-7\\ a_{10}=71$

This is a duplicate question, so I will link you to the URL where CPhill answered this question:

You have asked a lot about this topic lately, so I encourage you to view the other answers we all have provided as examples for your sketch for this one. I have done two already. I will provide the hyperlink to both below.

You have asked a lot about this topic lately, so I encourage you to view the other answers we all have provided as examples for your sketch for this one. I have done two already. I will provide the hyperlink to both below.

Here is a sketch:

$\frac{4\pi}{3}$ lies between $\pi$ and $\frac{3\pi}{2}$, so the terminal side lands in the third quadrant.

This is a duplicate post, so the hyperlink below leads you to the URL where I answered this already.