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The answer is 19.5 degrees. Hope this helps!

Here is a link to someone who had the same question as you. Hope it helps!

https://web2.0calc.com/questions/help_87049

Thank you so much for the help! The diagram really helped me understand the answer more throughly! 

The angle just below 'x' is 75^o

'x' and 75⁰ form a straight line = 180^o

As follows:

Nevermind, I've figured it out. :)

Solve

\(\dfrac{x^2 + 2x}{x^2 - 4x} < 0\)

zero in numerator:

\(\begin{array}{|rcll|} \hline x^2 + 2x &=& 0 \\ x(x + 2) &=& 0 \\ \hline x= 0 &\text{or}& x=-2 \\ \hline \end{array} \)

zero in denominator:

\(\begin{array}{|rcll|} \hline x^2 - 4x &=& 0 \\ x(x -4) &=& 0 \\ \hline x= 0 &\text{or}& x=4 \\ \hline \end{array}\)

\(\begin{array}{|r|c|c|c|c|c|c|c|} \hline & (-\infty,-2) & -2 & (-2,0) & 0 & (0,4) & 4 & (4,+\infty) \\ \hline x^2 + 2x & +&0&-&0&+&+&+\\ x^2 - 4x & +&+&+&0&-&0&+\\ \hline \text{fraction} & + & 0 & - & / & - & / & + \\ \hline \end{array}\)

\(-2 < x < 0\) and \(0 < x < 4\)

How many integers \(n\) are there such that \(\dfrac{15 - n}{3 - n}\) is also an integer?

\(\begin{array}{|rcll|} \hline \dfrac{15 - n}{3 - n} &=& \dfrac{12+3-n}{3 - n} \\\\ &=& \dfrac{12}{3 - n}+\dfrac{3-n}{3 - n} \\\\ \mathbf{\dfrac{15 - n}{3 - n}} &=& \mathbf{\dfrac{12}{3 - n}+1} \\ \hline \end{array} \)

Divisors of 12:
1 | 2 | 3 | 4 | 6 | 12   (6 divisors)

\(\begin{array}{|rcll|} \hline 3-n &=& 1 \qquad \text{or} \qquad \mathbf{n=2} \\ 3-n &=& 2 \qquad \text{or} \qquad \mathbf{n=1} \\ 3-n &=& 3 \qquad \text{or} \qquad \mathbf{n=0} \\ 3-n &=& 4 \qquad \text{or} \qquad \mathbf{n=-1} \\ 3-n &=& 6 \qquad \text{or} \qquad \mathbf{n=-3} \\ 3-n &=& 12 \qquad \text{or} \qquad \mathbf{n=-9} \\ \hline \end{array}\)

\(n=\{2,1,0,-1,-3,-9\}\)

The sides of triangle \(XYZ\) are \(XY = XZ = 25\) and \(YZ = 40\).
A semicircle is inscribed in triangle \(XYZ\) so that its diameter lies on line \(\overline{YZ}\), and is tangent to the other two sides.
Find the area of the semicircle.

In triangle ABC, angle C = 90 degrees, angle A = 30 degrees, and D is the foot of the altitude from C to hypotenuse AB.  

What is the ratio of the area of triangle BDC to the area of triangle ADC?

\(\text{Let $\angle B = 90^\circ-30^\circ = 60^\circ$ }\)

\(\begin{array}{|rclcrcl|} \hline \tan{30^\circ} &=& \dfrac{CD}{AD} &~\qquad & \tan{60^\circ} &=& \dfrac{CD}{BD} \\ \hline \dfrac{\tan{30^\circ}}{\tan{60^\circ}} &=& \dfrac{CD}{AD}\above 1pt \dfrac{CD}{BD} \\\\ \dfrac{\tan{30^\circ}}{\tan{60^\circ}} &=& \dfrac{CD}{AD} \times \dfrac{BD}{CD} \\\\ \mathbf{\dfrac{\tan{30^\circ}}{\tan{60^\circ}}} &=& \mathbf{\dfrac{BD}{AD}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{[BDC]}{[ADC]} &=& \dfrac{BD*CD}{2}\above 1pt \dfrac{AD*CD}{2} \\\\ \dfrac{[BDC]}{[ADC]} &=& \dfrac{BD*CD}{2} \times \dfrac{2}{AD*CD} \\\\ \dfrac{[BDC]}{[ADC]} &=& \dfrac{BD}{AD} \quad | \quad \mathbf{\dfrac{\tan{30^\circ}}{\tan{60^\circ}}=\dfrac{BD}{AD}} \\\\ \dfrac{[BDC]}{[ADC]} &=& \dfrac{\tan{30^\circ}}{\tan{60^\circ}} \quad | \quad \tan{30^\circ}=\dfrac{\sqrt{3}}{3},\ \tan{60^\circ}=\sqrt{3} \\\\ \dfrac{[BDC]}{[ADC]} &=& \dfrac{\sqrt{3}}{3}\above 1pt \sqrt{3} \\\\ \dfrac{[BDC]}{[ADC]} &=& \dfrac{\sqrt{3}}{3\sqrt{3}} \\\\ \mathbf{\dfrac{[BDC]}{[ADC]}} &=& \mathbf{\dfrac{1}{3}} \\ \hline \end{array}\)

The area of triangle BDC to the area of triangle ADC is \(\mathbf{\dfrac{1}{3}}\)

The ratio of wine to water in bottle A is 5: 3, and the ratio of wine to water in bottle B is 5: 4. Now take the two bottles of solution at 5: 6 and mix them.  What is the ratio of wine to water in the new mixture? 

Area of circle       A₁ = pi            area of  A₄ = 10 pi

                             r_{1 =}  1                           r₄ = sqrt( 10pi / pi )  

A hexagon is circumscribed about a circle with circumference 8*pi.  What is the perimeter of the hexagon?

Four circles are drawn.
Let \(A_1, A_2, A_3, A_4\) be the areas of the regions, so
\(A_1 \)is the area inside the smallest circle,
\(A_2\) is the area outside the smallest circle and inside the second-smallest circle, and so on.
The areas satisfy
\(A_1 = \frac{A_2}{2} = \frac{A_3}{3} = \frac{A_4}{4}\)
Let \(r_1\) denote the radius of the smallest circle, and let \(r_4\) denote the radius of the largest circle. Find \(\frac{r_4}{r_1}\).

My attempt:

\(\text{Let $A_{\circ_{1} }= \pi r_1^2$ } \\ \text{Let $A_{\circ_{4} } = \pi r_4^2$ } \)

\(\begin{array}{|llcll|} \hline & 1*A_1 &=& A_{\circ_{1}} \\ & 2*A_1 = A_2 &=& A_{\circ_{2}}- A_{\circ_{1}} \\ & 3*A_1 = A_3 &=& A_{\circ_{3}}- A_{\circ_{2}} \\ & 4*A_1 = A_4 &=& A_{\circ_{4}}- A_{\circ_{3}} \\ \hline \text{sum}: & 10A_1&=& A_{\circ_{1}}+A_{\circ_{2}} - A_{\circ_{1}}+A_{\circ_{3}} - A_{\circ_{2}}+A_{\circ_{4}} - A_{\circ_{3}} \\ & 10A_1&=& A_{\circ_{4}} \quad | \quad A_1 = A_{\circ_{1}} \\ & 10A_{\circ_{1}}&=& A_{\circ_{4}} \\ & 10\pi r_1^2&=& \pi r_4^2 \\ & 10r_1^2 &=& r_4^2 \\\\ & \dfrac{r_4^2}{r_1^2} &=& 10 \\\\ & \mathbf{\dfrac{r_4}{r_1}} &=& \mathbf{\sqrt{10}} \\ \hline \end{array} \)

Hint:  1st      find the altitude from C to side AB 

          2nd     find angles CBD and CDB   

          3rd      find  CD    

17. What is the value of two numbers if their sum is 120 and their difference is 140?

\(\begin{array}{|rcll|} \hline \mathbf{a+b} &=& \mathbf{120} \\ \mathbf{b} &=& \mathbf{120 -a} \qquad (1) \\\\ \mathbf{a-b} &=& \mathbf{140} \\ a-(120 -a) &=& 140 \\ 2a-120 &=& 140 \\ 2a &=& 260 \\ \mathbf{a} &=& \mathbf{130} \\\\ \mathbf{b} &=& \mathbf{120 -a} \\ b &=& \mathbf{120 -130} \\ \mathbf{b} &=& \mathbf{-10} \\ \hline \end{array}\)

In triangle PQR, PQ=13, QR=14, and PR=15. Let M be the midpoint of QR. Find PM.

Let angle PRQ be  >β<

1)   Let's find β  first:             cos(β) = (15² + 14² - 13²) / 2*15*14       β = 53.13°

2)   Now we can find  PM      (PM)² = 7² + 15² - 2*7*15 * cos(β)    

                                                 PM = 12.166     

Find the sum of the x-coordinates of the solutions to the system of equations  y = |x^2-6x+5| and y = (29/4)-x .

Be careful - there are three solutions.  This graph should help you find them.  Simply equating the two functions, but ignoring the absolute value part of the first function will give you two of the solutios.  You need to be more careful to get the solution between x = 1 and x = 5 though.

How many minutes is it before their next meeting at point P?

Answered here: https://web2.0calc.com/questions/please-help_98787#r2 

Find the sum of the divisors of 6561.

sum: \(1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 + 6561 = \mathbf{9841}\)

Here's a hint.

You have 3 white balls chosen, at least 2 right. So keep in mind, OR superball

You miss all 3 white balls if your ticket contains 3 of the 7 white numbers that were not drawn, so there are \(\binom{7}{3}\) possibilities.

You hit 1 white ball and miss the others if your ticket contains 1 of the 3 white numbers that were drawn and 2 of the 7 white numbers that were not drawn, so there are \(3 \binom{7}{2}\)

You miss the SuperBall if you have one of the 9 blue numbers that were not drawn.

What can you do with this information?

:)

Remember what these numbers mean.

(possibilities of NOT losing)

Good luck, hope you can figure this one out. 

AoPS questions are indeed hard.