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There are 2*6*7 = 84 ways of coloring the diagonals.

By similar triangles, UV = 15.

y is equa to 1 - x, so the answer is b.

She has a total of 100 songs in her collection.

3 hours x 60 minutes =180 minutes

50  3-minute long songs can be played in:

50 x 3 = 150 minutes

180 minutes - 150 minutes = 30 minutes left.

30 / 5-minute songs = 6 songs can be played.

Therefore, the maximum number of songs she can play in 3 hours is:

50 + 6 = 56 songs.

g(f(-3)) = 2/17.

b + c = -17.

Let's first combine the top two fractions in the numerator.

\(\frac{x(x+1)-x(x-1)}{x^2-1}=\frac{x^2+x-x^2+x}{x^2-1}=\frac{2x}{x^2-1}\)

This is what our fraction looks like now:
\(\frac{\frac{2x}{x^2-1}}{\frac{x}{x^2-1}}\)

We can multiply top and bottom by \(x^2-1\), which cancels to \(\frac{2x}{x}=2\)

The determinant is (-3)(2)(5)(4) = -120.

det([[F_{1000}, F_{1001}, F_{1002}],[F_{1001},F_{1002},F_{1003}],[F_{1002},F_{1003},F_{1004}]) = -4.

Any three consecutive numbers you choose, one of them is ALWAYS a multiple of 2  and  one is ALWAYS a multiple of 3. Therefore: 2 x 3 = 6 will ALWAYS divide the product of 3 consecutive numbers.

So let the width, height, and length be x, y, and z. We are trying to find the volume, which is xyz. 

xy = 6

xz = 8

yz = 12

Multiply all these equations together. 

\((xyz)^2 =12*8*6\)

Take the square root of both sides, which is our final answer.

\(xyz=24\)

Here's a hint: let r be an integer root of p(x).

Let his average speed BEFORE motor problem =S 
The distance between A and B =D 
The distance between A and where he had motor problem =3/4D
The remaining distance to B =1/4D
D/S =5,      
3.75(hours) + 1.5(hours) + [1/4D/(S+10)] =6.25, solve for D, S
D = 200 km - distance between A and B
S =40 kp/h before motor problem.
40+10 = 50 kp/h - his speed the last 1/4D

The graph has an asymptote, so the range is all real numbers: (-inf,inf).

The possible degrees of p(x) + q(x) are 0, 1, 2, 3, 4.

if x=1 then it equals 1.0173001 if x=2 then it equals 1.03429032 an dif x=1000 then it equals −0.81115957 and if x=pi/4 then it equals sqrt2 which is the largest

I believe the answer is 6. After testing out a few cases, it becomes obvious that there is always a multiple of 2 and a multiple of 3 in three consecutive integers. 

I'm don't know a rigorous way to prove it though.

1. Arc EPF works out to 210 degrees.

2. There are 12 digaonals that have a length of P_1 P_2, which from the Sine Law, is sin (15 degrees).  There are 12 diagonals that have a legnth of P_1 P_3, which from the Sine Law, is sin (30 degrees).  We can appy the same reasoning to the other diagonals, which gives us a total sum of

(12 sin 15)^2 + (12 sin 30)^2 + (12 sin 45)^2 + (12 sin 60)^2 + (12 sin 75)^2 + (12 sin 90)^2 + (12 sin 105)^2 + (12 sin 120)^2 + (12 sin 135)^2 + (12 sin 150)^2 + (12 sin 175)^2 = 864.

Since we have double-counted, the answer is 864/2 = 432.

Solve this as a polynomial, complete the square, and factorize. 

First write it as \(z^4+4=0\)

Complete the square. \(z^4+4z^2+4-4z^2=(z^2+2)^2-(2z)^2=(z^2+2z+2)(z^2-2z+2)\)

We can factorize these two quadratics using \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) to find our answer (four complex numbers):
\(-1+i, -1-i, 1+i, 1-i\) or \(-1\pm i, 1\pm i\)

not correct\

I personally don't think that this question makes much sense. If you think about it - there are infinite consecutive integers, therefore, there are infinite integers that divide by P. Are there any numbers in this question that I am not aware of? If so, please add them (it'll help a lot). Thanks!