tinfoilhat

Here we go: 

Because they have the same associated arc,

Let the intersection of AB and CM be P.

Since

That means AP/CP = PM/PB = AM/CB. Since we know CB, we have AP/CP = PM/PB = AM/4.

Note that CM = CP+PM.

By Pythagorean theorem, AB = sqrt(80).

By the angle bisector theorem, AP/8 = BP/4 and AP+BP = sqrt(80). Solving we get AP = 8 sqrt(5)/3 and BP = 4 sqrt(5)/3. If only SSA similarity was a thing...

Now we input these new values to get: 8 sqrt(5)/(3CP) = PM/(4 sqrt(5)/3) = AM/4. However, we can get CP as well with a little more calculation.

Let's drop a perpendicular from C to PB. Call the intersection point K. We have PK + BK = 4 sqrt(5)/3, PK^2 + CK^2 = CP^2, BK^2+CK^2=16, and finally (8sqrt(5)/3+PK)^2+CK^2=64. 

Solving, we get CP = 8sqrt(2)/3. Thus we input and get PM = 10sqrt(2)/3. Adding, we get 6sqrt(2)=CM.

As you can see, my solution involved lots of formulas (inscribed angle theorem, similar triangles, Pythagorean theorem, angle bisector theorem, etc) and tedious bashing... there's probably a better way. Follow along at your own risk.

If I'm understanding correctly, the question seems to have no answer.

The incenter is always inside the triangle; DIB must be a triangle. 3+4<8... it doesn't add up.

pq-2p-2q-2=0

p(q-2)-2q-2=0

p(q-2)-2q+4 = 6

p(q-2)-2(q-2)=6

(p-2)(q-2)=6

Factors of 6: ±1,2,3,6

According to these factors, (p,q) = (1, -4), (3, 8), (4, 5), (5, 4), (8, 3)

5 favorable / 10 possible

1/2

The right angle sign denotes a 90˚ angle. This means that 6x+4 + 4x+6 = 90, so 10x+10 = 90 and x = ...

p and q are two straight lines. The vertically positioned angles they make are called vertical angles, and so are the horizontally opposite ones. What do you notice about these two vertical angles, and about the sum of 30 and x?

What do you notice about angle DCA and CAB? What are the sum of the angles in a triangle?

Have you heard of alternate interior/exterior angles?

Based on the information provided: P(x) = x^4 + ax^3 + bx^2 + cx + d

P(1) = 1 + a + b + c + d

After a bit of educated guessing, it seems this polynomial would be \(\left(x + \sqrt{1 + \sqrt6}\right) \left(x + \sqrt{1 - \sqrt6} \right) \left(x - \sqrt{1 + \sqrt{6}}\right) \left(x - \sqrt{1 - \sqrt6}\right)\)

Which simplifies to \(x^4-2x^2-5\)

I'm not entirely sure why this works, but I think it's because it can be turned into a quadratic by setting y=x^2 (and the square roots must cancel out). Anyone?

Aside from that, 1-2-5 = -6.

I recommend drawing a diagram.

I'm not completely confident about my answer, but here's what I got.

We know that DF+FC+CE+EB = 124/2 = 62, and that 20(DF + FC) = 7(CE + EB), which are just 2 different ways to calculate the area of the parallelogram.

We need to calculate either 20(DF + FC) or 7(CE + EB). 

From our first equation, we get DF + FC = 62 - (CE + EB), so 20(62-(CE+EB)) = 7(CE+EB). Solving, CE+EB = 1240/27 and so the area is 7 * 1240/27 = 8680/27.

See https://web2.0calc.com/questions/help-geometry_74

It seems that the same question has been asked on brainly.com (https://brainly.com/question/12971062), and the solution matched your answer of $3.49. 

Expand everything in base 10 form. 1000A + 100B + 10C + D + 100B + 10C + D = 2012.

1000A + 200B + 20C + 2D = 2012

Divide by 2.

500A + 100B + 10C + D = 1006.

D has to be 6.

500A + 100B + 10C = 1000

Divide by 10.

50A + 10B + C = 100

C has to be 0.

50A + 10B = 100

5A + B = 10

(A,B) = (1,5) or (2,0)

Solutions: ABCD = 1506 or 2006.

We cannot have a solution without a 0, so there is no solution. 

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In case you mean A*B*C*D + B*C*D = 2012:

B*C*D(A+1) = 2012

One of these must include 503 (prime), which is not a digit. So this is not it either.

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I found another question that has a realistic solution at https://brainly.com/question/17335668, is this perhaps what you meant?

Without loss of generality, say Benny has $120.

He just spent 24 dollars on 4 pineapples and 8 mangoes.

Since a pineapple is 4 times a mango, Benny spent 1 dollar per mango and 4 dollars per pineapple. 

He now has 96 dollars, and he spends $48 on pineapples, accounting for 12 pineapples. 

He has bought 12+4 = 16 pineapples.

" such that AK in altitude to ABC."

I'm assuming this means that AK is an altitude to BC. I suggest drawing a diagram.

If CK = 4 and AC = 9, then by the Pythagorean Theorem, AK^2 = 9^2-4^2 = 65, so AK = √65. Since BK is √5, AB^2 = 65 + 5 = 70, so AB = √70.

The probability of selecting each club is 1/3. Let M stand for normal member and let C stand for co-president. (these are distinguishable)

Case 1: first club of 5 people is selected (MMMCC)

There are 5 choose 4 = 5 ways to select the four members, and we must get a group of MMCC in our choice. There are thus 3 choose 2 = 3 ways to choose MMCC out of MMMCC. The probability here is 3/5, and the total probability of this case happening and the co-president restriction is 1/3 * 3/5 = 1/5.

Case 2: second club of 7 people is selected (MMMMMCC)

There are 7 choose 4 = 35 ways to select the four members, and we must get a group of MMCC in our choice. There are thus 5 choose 2 = 10 ways to choose MMCC out of MMMMMCC. The probability here is 10/35 = 2/7, and the total probability of this case happening and the co-president restriction is 1/3 * 2/7 = 2/21.

Case 3: third club of 8 people is selected (MMMMMMCC)

There are 8 choose 4 = 70 ways to select the four members, and we must get a group of MMCC in our choice. There are thus 6 choose 2 = 15 ways to choose MMCC out of MMMMMMCC. The probability here is 15/70 = 3/14, and the total probability of this case happening and the co-president restriction is 1/3 * 3/14 = 1/14. 

1/5 + 2/21 + 1/14 = 11/30

See https://web2.0calc.com/questions/combination-probability

Joe, Alex and Sean shared some stamps in the ratio 3:8:5. Sean kept 

1/5 of his stamps for himself and gave the rest of his stamps to Joe and

Alex in the ratio 3:5. As a result, Alex had 96 stamps more than Joe.

How many stamps did Joe have in the end?

Let's say they all have x stamps.

Joe has x * 3/16 stamps, Alex with x * 1/2, and Sean with x * 5/16.

After giving away his stamps, Sean has 1/16 * x stamps. 

Joe got (1/4 * x) * 3/8 = 3/32 * x stamps. Joe now has (3/32 + 3/16)x = 9/32 x stamps.

Alex got (1/4 * x) * 5/8 = 5/32 * x stamps. Alex now has (5/32 + 1/2)x = 21/32 x stamps

21/32 x = 9/32 x + 96

12/32 x = 3/8 x = 96

x = 256.

256 * 9/32 = 72.

The guests apparently gave you correct answers. Is there a mistake/misinterpretation anywhere?

(in case of a mistake, let's check)

256 stamps means Joe has 48 stamps, Alex has 128 stamps, and Sean has 80 stamps. 

Sean now has 16 stamps, and 64 stamps were given away in the ratio 3/5.

24 stamps given to Joe, 40 stamps given to Alex.

Joe has 72 stamps, Alex has 168.

72+168+16 = 256

168-72 = 96

I don't see a problem...? 

Let B be the fully charged battery capacity.

Talking continuously for 1 hour drains B/3 capacity.

Standby for 1 hour drains B/72 capacity.

The standby has drained 27 * B/72 = 3B/8 capacity. 

The conversation drained B/3 * 45/60 = B/4 capacity.

In total, we've drained (3/8 + 1/4) = 5/8 B capacity.

We can talk for 3/8 B more capacity.

How many hours worth is 3/8 B on conversation mode?

B/hr = 3

3B/8hr = 3*3/8 = 9/8 hr.

9/8 hr = 67.5 min

Without loss of generality, assume they played 250 games.

Then, they've already played 200 games, and won 48/100 * 200 = 96 games.

They need to win 125 games to have 50% wins, which means they need to win 125 - 96 = 29 more games.

They have 50 more games to play, so they must win 29/50 of their games, or 58%. 

To check, take the weighted average of the two values. 48 * 80% + 58 * 20% = 50. 

This works, as we needed 50% wins/losses.

\(\frac{r-3}{\sqrt{r}} = -2 \sqrt{r}\)

Since the denominator of the left side cannot be 0, √r ≠ 0, and r≠0. 

Multiplying both sides by √r, we have r-3 = -2r.

Thus, 3r = 3 and r=1.