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:)

Good job!

Thanks a lot

Thank you so much!!!!!!!!! I got it. :)

So AIB is 109 degrees. IAB + IBA + 109 = 180, so IAB + IBA = 180 -109 = 71.

AX and BY are angle bisectors so IAB = CAB/2, and IBA = CBA/2. (CAB + CBA)/2 = 71. 

Rearrange the last equation and you should be very close to your answer.

If you don't understand anything, feel free to ask!

I agree the diagram is great, but the lighting on the problem is bad :). Sure, Week 7 Problem 7

It's not a bad diagram. AoPS folks do a good job writing Asymptope code.

If you'll be so kind as to provide the week and problem number?

Thanks!

Ok guest.

I apologize for complaining about AoPS in the past.

Let's just get over with it. If you want to please message me privately so we can resolve this privately.

Plot. 1,-3.    Draw line from origin to the point.  Positive x axis and this line form your angle.   CSC= 1/sin 

can you calculate sin?

https://web2.0calc.com/questions/help-me-asap-please_12#r1

thank you!

amazing solution asinus!!

100,000 =50,000,000 * 0.93^n
Divide both sides by 50,000,000
0.002 =0.93^n

Take the log of both sides
n =log(0.002) / log(0.93)
n=85.6 hours = ~86 hours - when the baby is no longer contagious, or just over 3 ½ days.

A small regional carrier accepted 21 reservations for a particular flight with 18 seats. 11 reservations went to regular customers who will arrive for the flight. Each of the raining passengers will arrive for the flight with a 45% chance, independently of each other.

1.  Find the probability that overbooking occurs

2.  Find the probability that the flight has empty seats

My attempt:

1. Find the probability that overbooking occurs

overbooking mean that more than 7 of the other 10 will arrive for the flight:

probability of 8 arriving is \(10C_8*0.45^8*0.55^2 = 45*0.45^8*0.55^2= 0.02288958944\)
probability of 9 arriving is \(10C_9*0.45^9*0.55^1 = 10*0.45^9*0.55^1= 0.00416174353\)
probability of 10 arriving is \(10C_{10}*0.45^{10}*0.55^0 = 1*0.45^{10}*1= 0.00034050629\)

That sum is 0.02739183926.

The probability that overbooking occurs is \(\mathbf{0.0274}\)

2.  Find the probability that the flight has empty seats

Just 7, all seats are full. The probability is \(10C_7*0.45^7*0.55^3 = 120*0.45^7*0.55^3= 0.07460310632\)
this plus the 0.02739183926 is 0.10199494558.
That is the probability no empty seats will occur.

The probalbility hat empty seats will occur is the compliment \((1-0.10199494558) = 0.89800505442\)

The probability that the flight has empty seats is \(\mathbf{0.8980}\)

Four positive integers \(p,q,r,s\) satisfy the following equations:
\(\begin{align*} pq+2p+q&=348 \\ qr+4q+3r&=373 \\ rs+8r+6s&=544 \end{align*}\)

The system of equations

\(\dfrac{xy}{x+y}=1,\ \dfrac{xz}{x+z}=2,\ \dfrac{yz}{y+z}=3 \)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{xy}{x+y}} &=& \mathbf{1} \\\\ \dfrac{x+y}{xy} &=& 1 \\\\ \dfrac{x}{xy}+\dfrac{y}{xy} &=& 1 \\\\ \mathbf{\dfrac{1}{y}+\dfrac{1}{x}} &=& \mathbf{1} \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\dfrac{xz}{x+z}} &=& \mathbf{2} \\\\ \dfrac{x+z} {xz}&=& \dfrac{1}{2} \\\\ \dfrac{x} {xz}+\dfrac{z} {xz}&=& \dfrac{1}{2} \\\\ \mathbf{\dfrac{1} {z}+\dfrac{1}{x}} &=& \mathbf{\dfrac{1}{2}} \qquad (2) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\dfrac{yz}{y+z}} &=& \mathbf{3} \\\\ \dfrac{y+z}{yz} &=& \dfrac{1}{3} \\\\ \dfrac{y}{yz}+\dfrac{z}{yz} &=& \dfrac{1}{3} \\\\ \mathbf{\dfrac{1}{z}+\dfrac{1}{y}} &=& \mathbf{\dfrac{1}{3}} \qquad (3) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \mathbf{x=\ ?} \\ \hline (1)+(2)-(3): & \dfrac{1}{y}+\dfrac{1}{x} + \dfrac{1} {z}+\dfrac{1}{x}- \left( \dfrac{1}{z}+\dfrac{1}{y} \right) &=& 1 +\dfrac{1}{2}- \dfrac{1}{3} \\\\ & \dfrac{2}{x} &=& 1 +\dfrac{1}{2}- \dfrac{1}{3} \quad | \quad * 6 \\\\ & \dfrac{12}{x} &=& 6 + 3- 2 \\\\ & \dfrac{12}{x} &=& 7 \\\\ & \mathbf{x} &=& \mathbf{\dfrac{12}{7}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \mathbf{y=\ ?} \\ \hline (1)-(2)+(3): & \dfrac{1}{y}+\dfrac{1}{x} - \left(\dfrac{1} {z}+\dfrac{1}{x}\right)+ \dfrac{1}{z}+\dfrac{1}{y} &=& 1 -\dfrac{1}{2}+ \dfrac{1}{3} \\\\ & \dfrac{2}{y} &=& 1 -\dfrac{1}{2}+ \dfrac{1}{3} \quad | \quad * 6 \\\\ & \dfrac{12}{y} &=& 6 - 3+ 2 \\\\ & \dfrac{12}{y} &=& 5 \\\\ & \mathbf{y} &=& \mathbf{\dfrac{12}{5}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \mathbf{z=\ ?} \\ \hline -(1)+(2)+(3): & - \left( \dfrac{1}{y}+\dfrac{1}{x} \right) + \dfrac{1} {z}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y} &=& -1 +\dfrac{1}{2}+ \dfrac{1}{3} \\\\ & \dfrac{2}{z} &=& -1 +\dfrac{1}{2}+ \dfrac{1}{3} \quad | \quad * 6 \\\\ & \dfrac{12}{z} &=& -6 + 3+ 2 \\\\ & \dfrac{12}{z} &=& -1 \\\\ & \mathbf{z} &=& \mathbf{-12} \\ \hline \end{array}\)

Write it as :  \(x = \sqrt{2+x}\)  

Square both sides and solve for x.

See https://web2.0calc.com/questions/slope-intercept-please-help#r1 

1.  Rewrite the equation in the form:  y = x - 4  This has slope m = 1, so any line parallel to it will have the same slope, so it will look like y = x + k where k is a constant.  To find k, use the fact that it goes through point (-6, -1). i.e.  you get k from:  -1 = 1*(-6) + k.

2.   Rewrite the equation in the form:  y = -x/3 - 2.  This has slope m = -1/3.  Any line perpendicular to this has slope equal to -1/m.  Again get the constant by substituting the specified point value to the perpendicular equation  y = (-1/m)*x + k.

Find the coefficient of \(x^3*y*z^2\) in the expandsion of \((3x - 5y + z)^6\).

Multinomial theorem:

For any positive integer m and any nonnegative integer n, the multinomial formula tells us how a sum with m terms expands when raised to an arbitrary power n:
\(\left( x_1+x_2+\cdots +x_m \right)^n = \sum \limits_{k_1+k_2+\dots+k_m=n} \dbinom{n}{k_1,k_2,\dots,k_m}\cdot x_1^{k_1}\cdot x_2^{k_2}\dots x_m^{k_m}\)
where  
\(\dbinom{n}{k_1,k_2,\dots,k_m} = \dfrac{n!}{k_1!k_2!\dots k_m!}\)

Given that point O is the center of the circle and OA=5, OC=8, and \(\angle ACO=30^\circ\), find \(\angle BOA\) in degrees.

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{\sin\left(90^\circ+\dfrac{x}{2}\right) }{8}} &=& \mathbf{\dfrac{\sin(30^\circ)}{5}} \quad | \quad \sin\left(90^\circ+\dfrac{x}{2}\right) = \cos\left( \dfrac{x}{2} \right),\ \sin(30^\circ) = \dfrac{1}{2} \\\\ \dfrac{\cos\left( \dfrac{x}{2} \right) }{8} &=& \dfrac{\dfrac{1}{2}}{5} \\\\ \dfrac{\cos\left( \dfrac{x}{2} \right) }{8} &=& \dfrac{1}{10} \\\\ \cos\left( \dfrac{x}{2} \right) &=& \dfrac{8}{10} \\\\ \cos\left( \dfrac{x}{2} \right) &=& \dfrac{4}{5} \\\\ \dfrac{x}{2} &=& \arccos\left( \dfrac{4}{5} \right) \\\\ \dfrac{x}{2} &=& 36.8698976458^\circ \\ x &=& 2* 36.8698976458^\circ \\ \mathbf{x} &=& \mathbf{73.7397952917^\circ} \\ \hline \end{array}\)

\(\mathbf{\angle BOA \approx 73.74^\circ}\)

x= - y/3

y= -3x

Since you don't define what x and y are, we don't know what the actual number is.