TheXSquaredFactor

f,g, and h are all polynomial functions, so they all adhere to the following general form shown below.

\(f(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_2x^2+a_1x^1+a_0\\ g(x)=b_nx^n+b_{n-1}x^{n-1}+b_{n-2}x^{n-2}+...+b_2x^2+b_1x^1+b_0\\ h(x)=c_nx^n+c_{n-1}x^{n-1}+c_{n-2}x^{n-2}+...+c_2x^2+c_1x^1+c_0 \)

The constant term of \(f(x)\), represented as \(a_0\) in the general form, is \(-4\). In the function \(h(x)\), the constant term is 3. One way to guarantee that a polynomial function will output its constant term is to substitute in \(x=0\) for that polynomial. 

\(h(0)=f(0)*g(0)\\ c_0=a_0*b_0\\ 3=-4b_0\\ b_0=\frac{-3}{4}\\ \therefore g(0)=b_0=\frac{-3}{4}\)

A diagram would be of great assistance for this problem, so I created one for you. I added lines and points where I saw fit to add clarity to the diagram:

In order to find the area of any triangle, we need the length of the base and the length of the perpendicular height (also known as the altitude) of the triangle. Then, we can apply the formula \(A_{\triangle ABC}=\frac{1}{2}bh\) to find the area. In the diagram above, \(\overline{AB}\) represents the base, and \(\overline{EC}\) represents the perpendicular height of the triangle. 

\(m\angle ADB=120^{\circ}\)as each central angle \(\triangle ABC\) forms divides equally in thirds. By inserting an angle bisector of \(\angle ADB\), this means that \(m\angle ADE=60^{\circ}\). This angle bisector also happens to be a perpendicular of chord \(\overline{AB}\). Combining all of this information together means that \(\triangle ADE\) is a 30-60-90 triangle.

A 30-60-90 triangle has a constant ratio of its sides of \(1:\sqrt{3}:2\). In this particular case, \(r\) is the length of the hypotenuse of \(\triangle ADE\). Let's use these ratios to find the length of the base and the height in terms of \(r\).

By parallel reasoning, \(BE=\frac{\sqrt{3}}{2}r\).

We can find \(AB\), the length of the base of the triangle in terms of r now.

\(AB=AE+BE\\ AB=\frac{\sqrt{3}}{2}r+\frac{\sqrt{3}}{2}r\\ AB=\sqrt{3}r\)

Similarly, we can do the same for \(CE\), the length of the perpendicular height of the triangle.

\(CE=CD+DE\\ CE=r+\frac{1}{2}r\\ CE=\frac{3}{2}r\)

Finally, let's solve for the area of the inscribed triangle in terms of r.

\(A_{\triangle ABC}=\frac{1}{2}*AB*CE\\ A_{\triangle ABC}=\frac{1}{2}*\sqrt{3}r*\frac{3}{2}r\\ A_{\triangle ABC}=\frac{3\sqrt{3}}{4}r^2\)

I am not sure how Guest arrived at a=5, but that answer is incorrect.

In order for a function to be continuous at x=c, the function must fulfill three criteria. They are the following:

     1) \(f(c)\) is defined.

     2) \(\lim_{x\rightarrow c}f(x)\) is defined.

     3) \(\lim_{x\rightarrow c}f(x)=f(c)\)

Right now, only one criterion is met currently. \(f(3)\) is defined in this piecewise function because \(x=3\) is included in the domain. Let's move on to the next condition.

In order for \(\lim_{x\rightarrow c}f(x)\) to be defined, we have to ensure that left-side and right-side limit approach the same value. We set both limits equal to each other and find possible values for a, if one exists.

\(\lim_{x\rightarrow 3^-}f(x)=\lim_{x\rightarrow 3^+}f(x)\\ 3*3^2+2=3a-1\\ 29=3a-1\\ 3a=30\\ a=10 \)

At \(a=10\), both limits approach the same value. Therefore, we have met the second condition.

Finally, we have to make sure that the limit and the true value are not incongruous.

\(\lim_{x\rightarrow 3}f(x)\stackrel{?}{=}f(3)\\ 29=29\)

Yes, that means that a=10 makes the function continuous.

It's quite a relief that I did not waste our time.

ElectricPavlov beat me to it, but I guess I will present this since it took me half an hour to write.

By definition, the sine function is the ratio of the opposite side of the reference angle to the hypotenuse of a triangle. 

By definition, the secant function is the ratio of the hypotenuse of a triangle to its adjacent side of the reference angle.

First, I would pay attention to the sign of the answer to \(\sin(x)\). As it is right now, it is negative, and there are two quadrants where the sine of an angle is negative: quadrant III and quadrant IV. 

Since the sine function is negative in these two quadrants, I would draw a right triangle with the given ratio of the sides and solve for the third side so that I could generate the ratio of the hypotenuse to the adjacent side of the triangle. I made a diagram so that you can follow along:

Just by inspection, the ratio of this triangle's side lengths is exactly that of a 30-60-90 triangle. The length of the third side, therefore, is \(\sqrt{3}\).

Therefore, \(\sec x=\frac{2}{-\sqrt{3}}=\frac{-2\sqrt{3}}{3}\\ \sec x=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\).

I will do the first two problems and then I will let you apply the techniques I demonstrated for the rest of them.

a) \(\frac{(n-2)!}{(n-3)!}=5\)

The key to this question is to manipulate the left-hand side of the equation by "expanding" one of the factorials until a major cancellation occurs. Here is the way I would approach it.

\(\frac{(n-2)(n-3)!}{(n-3)!}=5\)

This manipulation is very helpful as it is now clear that \((n-3)!\) is a factor of both the numerator and the denominator. We can cancel this common factor to simplify the equation significantly.

\(n-2=5\\ n=7\)

Okay, let's apply this approach to the next one, too.

b) \(\frac{(n+2)!}{n!}=42\)

The procedure is the same: Expand one of the factorial terms until a major cancellation occurs.

\(\frac{(n+2)(n+1)n!}{n!}=42\)

This time, \(n!\) was the common factor!

\((n+2)(n+1)=42\)

This is a fairly standard quadratic. Let's solve it now.

\(n^2+n+2n+2=42\\ n^2+3n-40=0\\\)

This quadratic happens to be factorable. 

\((n-5)(n+8)=0\\ n=5\text{ or }n=-8\)

Always quickly check to make sure that your solutions are valid. Here, \(n=-8\) would be considered an extraneous solution because the input of a factorial function should always be nonnegative.

Based on the appearance of the "link" you sent, this is probably a file buried deep inside of your computer. Of course, I do not have access to that, so I (nor can anyone else) can help you. Maybe try uploading the image here.

The midpoint of (7,9) and (10,2) is located at the average of x- and y-coordinates of both vertices.

\(M=(x,\frac{9+2}{2})\\ M=(x,\frac{11}{2})\)

This midpoint is the same for the other two vertices of the square. Let's let the y-coordinates of the first unknown vertex be \(y_1\) and the second unknown vertex one be \(y_2\). Then, we would calculate the average of those vertices as follows:

\(\frac{y_1+y_2}{2}=\frac{11}{2}\\ y_1+y_2=11\)

Therefore, the sum of the y-coordinates of the other two vertices is 11.

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.