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The expression an^2 - a(n - 1) a(n + 1) = (-1)^n can be rearranged to give an=(-1)^n + a(n - 1) a(n + 1). When n = 3, this gives us a3=(-1)^3 + a2 a4. Since a2 = 5 and a4 is unknown, we can substitute a2 = 5 and solve for a4. This gives us a4=5/a3. Substituting this expression into the equation for a3 gives us a3=(-1)^3 + 5/a3. Solving for a3 gives us a3=2.

I think that the probability \(WX > YZ\) equals the probability of \(WX < YZ\).

This means that each has a probability of \(\color{brown}\boxed{1 \over 2}\)

Hard problem!

We can use the given relation to find a_3, as follows:

a_3^2 - a_2 a_4 = (-1)^3 (using n = 3 in the given relation)

We know a_0 and a_1, and we can find a_2 using the relation with n = 2:

a_2^2 - a_1 a_3 = (-1)^2

Substituting the given values a_0 = 1 and a_1 = 5, we get:

a_2^2 - 5a_3 = 1 (equation 1)

We need to find a_3, so we also need to find a_4. We can use the given relation again with n = 4:

a_4^2 - a_3 a_5 = (-1)^4

Substituting a_0 = 1 and a_1 = 5, and using equation 1 to eliminate a_5 in favor of a_3, we get:

a_4^2 - a_3 (a_2 + 1/a_3) = 1

Simplifying, we get:

a_4^2 - a_2 a_3 - 1 = 0 (equation 2)

Now we have two equations with two unknowns, a_2 and a_3. We can solve for a_2 in equation 1 and substitute in equation 2, to get a quadratic equation in a_3:

(a_3^2 - 1/5)^2 - 5a_3^2 - 1 = 0

Expanding and simplifying, we get:

a_3^4 - (5/2)a_3^2 + (24/25) = 0

This is a quadratic equation in a_3^2, which we can solve using the quadratic formula:

a_3^2 = [(5/2) +/- sqrt((5/2)^2 - 4*(24/25))]/2

a_3^2 = [5 +/- sqrt(5/4)]/2

Taking the positive root, we get:

a_3^2 = (5 + sqrt(5))/4

Taking the square root, we finally get:

a_3 = sqrt((5 + sqrt(5))/4)

Therefore, a_3 is approximately 1.618, which is the golden ratio.

13.125

The equation 3x-5y=12 can be rearranged to give y=3x/5. Substituting this expression into the given equation x+y=X+Y gives us X+Y=x+3x/5. Solving for x gives us x=(5/2)(X+Y). Since the x-coordinate is three times the y-coordinate, we have 3(X+Y)/5=X+Y, which can be rearranged to give X+Y=15. Therefore, the value of X+Y at the point (x,y) is 15.

We are given that the point (x,y) lies on the line 3x - 5y = 12 and the x-coordinate is three times the y-coordinate. So we can write:

x = 3y

Substituting this into the equation of the line, we get:

3(3y) - 5y = 12

Simplifying this, we get:

4y = 4

So, y = 1

Substituting y = 1 into x = 3y, we get:

x = 3(1) = 3

Therefore, the point (x,y) is (3,1) and the sum of the coordinates is:

X + Y = 3 + 1 = 4

So the value of X + Y at the given point is 4.

Let's start by drawing a diagram:

We see that $\triangle PQR$ is a right triangle with $\angle PQR = 90^\circ$ and $PR$ as the hypotenuse. Therefore, $PR^2 = PQ^2 + QR^2$, or $22^2 = 36^2 + 26^2$, so $PR = \sqrt{36^2 + 26^2} = 10\sqrt{17}$.

Since $M$ is the midpoint of $PQ$, $PM = QM = 18$, so $PX = 18 - MY = 10$. Therefore, $\triangle PQX$ is a $9$-$10$-$17$ right triangle, so $\angle QPX = \arctan \frac{9}{10}$ and $\angle RPX = \arctan \frac{17}{10}$.

Since $PX$ bisects $\angle QPR$, we have $\angle QPA = \angle APR$, so $\triangle PQA \cong \triangle APR$ by ASA. Therefore, $PA = PR = 10\sqrt{17}$.

Since $Y$ lies on the perpendicular bisector of $PQ$, we have $PY = QY = 18$, so $AY = AP - PY = 10\sqrt{17} - 18$.

Since $\triangle APX$ and $\triangle AQY$ are similar, we have $\frac{AY}{AX} = \frac{QY}{PX}$, or $\frac{10\sqrt{17} - 18}{AX} = \frac{18}{10}$. Solving for $AX$, we get $AX = \frac{9(10\sqrt{17} - 18)}{5} = 18\sqrt{17} - 36$.

Therefore, the altitude from $P$ to $QR$ has length $MY + AY = 8 + (10\sqrt{17} - 18) = 10\sqrt{17} - 10$. Thus, the area of $\triangle PQR$ is $\frac{1}{2} \cdot PQ \cdot PR = \frac{1}{2} \cdot 36 \cdot 10\sqrt{17} = 180 \sqrt{17}$.

Now I understand how to do it, thanks for the help Melody

I think AX is meant to be PX ? 

Please proofread your question.

ok now I understand what you have done.

In this case order does not count.  I am glad you worked that out for yourself :)

The expression 2a + 3c represents the cost in dollars for a family to go to the game. When an adult ticket is 17 dollars and a child ticket is 12 dollars, the cost would be (2 x 17) + (3 x 12) = 68 dollars.

do not encourage cheaters

5 choose 2 is the number of ways to pick the order of the ways you pick a dog and cat, but now I realize that's probably wrong.

The value of cos(2a + b) can be found by using the double angle formula, which states that cos(2a + b) = 2cos^2(a) - 1 + 2cos(a)sin(b). Substituting a = sin^(-1)(4/5) and b = tan^(-1)(12/5) gives us cos(2a + b) = 2(4/5)^2 - 1 + 2(4/5)sin(tan^(-1)(12/5)) = 14/25.

Let the square be ABCD and the other vertices of the equilateral triangles be E, F, G, and H. (So that the four equilateral triangles are ABE, BCF, CDG, and ADH.) Now consider the angles HAE, EBF, FCG, and GDH. (It would definitely help to draw the diagram yourself.) We know that these angles are 360 - 90 - 60 - 60 = 150 degrees. Now, let the side length of the outer square be x. By Law of Cosines, we have

\(x^2 = 2^2 + 2^2 - 2 \cdot 2 \cdot 2 \cdot \cos150^\circ = 8-8 \cdot (-\frac{\sqrt{3}}{2}) = 8+4\sqrt{3}\)

so we see that the side length of the outer square is \(\sqrt{8+4\sqrt{3}}\)

GUys, this is a russian hacker.