hairyberry

4)

Because PT and PU are both tangents, PT = PU (this is easy to prove using HL congruency), and PTO is congruent to PUO.

Because PTO = 90, then POT = 180-90-33=57. These two triangles are congruent. 

So arc TU has a length 57*2=114.

3)

Draw a tangent to the chord of P, forming a rectangle, labeling that point D.

QD = the length of the external tangent.

PD = 4

PQ = 20

\(QD^2=20^2-4^2\)

\(QD=8\sqrt{6}\).

1)

It's all ratios, so you can set BC = 3.

Then the radius of the semicircle is 1.

Label the point where the Semicircle is tangent to AB as D.

Label the center of the semicircle as O

Triangle ADO and triangle ACB are similar, and sset DA as x.

Using the properties of similar triangles:

\(\frac{x}{1}=\frac{\sqrt{x^2+1}+1}{3}\).

\(3x=\sqrt{x^2+1}+1 \)

\(x=\frac{3}{4}\).

\(AC = \frac{5}{4}+1=\frac{9}{4}\).

\(\frac{AC}{BC}=\frac{\frac{9}{4}}{3}=\frac{3}{4}\).

\(\log_2(\log_3x)=\log_3(\log_2x) \rightarrow \text{ apply change of base to } \log_3 \text{ on both sides and change to } \log_2 \text{.}\)

\(\log_2(\frac{\log_2x}{\log_23})=\frac{\log_2(\log_2x)}{\log_23}\)

\(\log_2(\log_2x) - \log_2({\log_23})=\frac{\log_2(\log_2x)}{\log_23}\)

\(\log_2(\log_2x) \rightarrow z \; \log_23 \rightarrow y\)

\(z-\log_2y=\frac{z}{y}\)

\(z=\frac{y\log_2y}{y-1}\)

\(x=2^{2^{\frac{(\log_23)(\log_2(\log_23))}{\log_2{3}-1}}}\)

for \(\frac{a}{|a|}\) we have 1 if a>0, -1, if a <0. This is easy to see because | | either return the positive version of the number or the negative version of the number.

If all three of x, y, z are positive, then xyz>0, then all four terms are one, producing 4.

If 2 of x, y, z are positive, then xyz<0 then there are two ones and two negative ones resulting in 0.

If one of x, y, z is positive, then xyz>0, and there are two ones and two negative ones, resulting in 0.

If x, y, z are all negative, then xyz<0, and we get all four terms as negative one resulting in -4.

So, all possible values are -4, 0, 4.

Set the center of the small circle to be O₁, and the center of the larger circle to be O₂.

Draw a perpendicular line passing through O₁, perpendicular to BO₂, label the foot of the perpendicular point C.

ABCO₁ is a rectangle, and so AB = CO₁.

Also, CO₂ = 4 - 1 = 3

Apply the pythagorean theorem, so (CO₁)²=(O₁O₂)²-(CO₂)²=25 - 9=16.

Therefore, CO₁=AB=4.

\(83^{\circ} \: \text{is } \frac{83}{360} \text{of the circle} \\ {r}^{2}\pi*\frac{83}{360}=60\pi\\ \underline{\text{Solving, we get}\: \bf{\frac{60\sqrt{498}}{83}}} \text{.}\)

Because BAC = 60, ABC = 60, ACB should equal 60.

Therefore, ABC is equilateral.

Because AD is an angle bisector, then ABD, is 30, and so ADB = ADC = 90.

Therefore, set BD = x.

\({x}^{2}+{24}^{2}={(2x)}^{2}\).

We get \(x = 8\sqrt{3}\). (These triangle ratios are very special, see if you can find the ratios for a 30-60-90 triangle for yourself)

Therefore the area is \(\frac{8\sqrt{3}*24*2}{2}=192\sqrt{3}\).

Set the distance to the beach as x.

\(\text{time} = \frac{\text{distance}}{\text{rate}}\).

Driving to the beach at 40mph is regularly is \(\frac{x\; \text{mile}}{40 \: \frac{\text{mile}}{\text{hour}}}= \frac{x}{40}\text{hour}\)

Driving to the beach at 15mph is \(\frac{x \; \text{mile}}{15 \: \frac{\text{mile}}{\text{hour}}} = \frac{x}{15}\text{hour}\)

Because she arrived 45 minutes or 3/4 hours later, in the second case:

\(\frac{x}{40}=\frac{x}{15}-\frac{3}{4}\).

Solving, the solution is x = 18, or 18 miles.

The prime factorization of 2000 is 2⁴*5³.

Catherine must have rolled 3 5s, to form the 5³.

The 2⁴ is more tricky. Catherine could've rolled 4's or 2's a certain number of times to form the 2⁴, but the number of rolls to acheive this must be 3, because she rolls a total of 6 rolls.

Consider all the cases:

#1: The rolls 2, 2 and 2² in some order.

#2: The rolls 1 2² and 2² in some order.

These are the only ways she coud've rolled a 4.

The number of ways to arrange 2, 2, 2², 3, 3, 3 is 60

The number of ways to arrange 1, 2², 2², 3, 3, 3 is also 60.

Therefore there could be 60+60 = 120 sequences.