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We have this proportion....!!!!

1 minute / 3 bolcks    =   N minutes  / 50 blocks

1/3  = N/50      multiply both sides by  50

(50) (1/3) = N  =  16  2/3 min     =  16 min 40 sec

Very impressive, tertre......A+++++   !!!!!!

The vertices sum to 84 for every possible ordering of the face numbers.

$\text{there are only so many pairs of integers that satisfy the second equation}\\ \text{They are }(\pm 1, \pm 7),(\pm 7,\pm 1),(\pm 5,\pm 5)\\ \text{each of these produces 4 ordered pairs }(a,b) \text{ thus there are 12 total}$

I think this is correct....

Based on what we have the area=84 and the semiperimeter=21. Therefore, the incircle(inradius)=84/21=4.

By tangent formulas, we see that the two sides of the quadrilateral are x. So, we have following from the top of the triangle, starting with the side length of 15: x, 15-x, Side length of 14: x-1 and 15-x, Side Length 13: x-1 and x. So, x-1+x=13, 2x-1=13, 2x=14, and x=7. Now, quadrilateral AEIF is made up of two triangles, both with areas of 14, since $\frac{1}{2}*7*4=14$ . Thus, the answer is $14+14=\boxed{28}.$  

Pretty tough one... I would wait for an answer.

I think we should use Heron's here for the area of the triangle, and we have sides 13, 14, and 15.

To use Heron's, we also need to find the semiperimeter, which is $\frac{13+14+15}{2}=\frac{42}{2}=21.$

Now, we have $\sqrt{21(21-13)(21-14)(21-15)}=\sqrt{21(8)(7)(6)}=\sqrt{7056}=84.$

Hmm, I don't know how to continue from here, though...this was my best shot.

Uh, what should we solve here?

cost of an adult ticket: $7.00, let adults be (a)

cost of a student ticket: $5.00. let students be (s)

We have a+s=45 and 7a+5s=265, based on systems of equations.

Therefore, by multiplying by 5 for the first equation, we have 5a+5s=225 and 7a+5s=265.

So, 2a=40, a=20. Thus, the number of students tickets sold were 45-20=25 tickets.

Why does it mean that? It could mean the preimeter is 42.  

I expect you are right but I do not think that is any recognised notation.

If the units as in units squared were given it would be more obvious.

$\text{consider the polynomial } \\ q(x) = p(x) - (d-1-x)\\ q(x) = 0,~\forall x = k=0,1,\dots ,(d-1)\\ q(x) = \prod \limits_{k=0}^{d-1}~(x-(d-1-k))\\ \text{now let }d=2008 \\ q(x) = \prod \limits_{k=0}^{2007}~(x-2007+k)\\ p(2008) = q(2008) +(2007-2008) = q(2008)-1\\ q(2008) = \prod \limits_{k=0}^{2007}~k+1 = 2008!\\ p(2008)=2008!-1$

Many thanks to Walagaster over at MHF for the idea of q(x)

Thanks Chris,

I probably should have explained that

8P6 is the number of ways 6 things can be chose from 8 when order matters.

I 'tied' the 2 friends together.  Which one was on the right and which on the left had to be accounted for, that is why i multiplied by 2.

Question 1

.4 (.91)^2

The .4 means that .4 (100) = 40 pounds were originally in stock

The    .91 means that  the stock decreased by 9%  every hour

Question 2

This is an increasing exponential function

% change = [240 - 100]  = 140%

Question 3

Theta is in the third quadrant.....the sine and cosine are negative....so...tan > 0

cos (- theta)  =   -cos(theta)

And cos theta   =  [- sqrt [ 5^2 - 3^2] / 5   =  - sqrt (16)/ 5  =  -4/5

So

cos (-theta) = -cos(theta )   =  - (-4/5)   = 4/5

Question 4

tan  ( pi +  x)   =

sin (pi + x)             sin pi cos x  + cos pi sin x                        -sinx

_________   =       _________________________  =         ______ =    tan x   =   2/5

cos ( pi + x)            cos pi cos x  -  sin pi  sin x                     -cos x

The answer is 2008!-1

I don't know why yet.

An interior angle BAC of the the heptagon will measure  [(5/7)180] °

Draw  BC

And because BA = AC.....then in triangle ABC,  angle ABC = angle  ACB

And angle ACB  =  (180 - mBAC) / 2 =     ( 180 - (5/7)180 ) / 2   =  [ ( 2/7) 180 ] / 2  =  [180/7 ]°

Draw  BD......and the interior angle BAD of the 15-gon  =  [ (13/15)180 ]°

So angle DAC  =

360 - m BAD - m BAC   =   360 - [ 180 (5/7 + 13/15) ] °=  ( 360 - [ 180 (166/ 105)] )° =

[ 360 -   1992 / 7 ]°    =   [ 528 / 7 ]°

Draw DC

And in triangle  DAC....since AC = AD.....then angle ADC  = angle ACD

And angle ACD =    [ 180 - m DAC] / 2  =   [ 180 - 528/ 7 ] / 2   =

[732/ 14] °   =  [ 366/ 7]°

So......angle BCD  =  angle ACB + angle ACD = [ 180/ 7  + 366/ 7]°  =

[ 546/ 7]°   =  78°

"why do we reflect across the x axis????"

Let $z=p+iq$ , then $\overline{z}=p-iq$

So, if z is the resultant obtained by adding two complex numbers, the real part of the complex conjugate of z has the same magnitude and the same sign, but the imaginary part of its complex conjugate has the same magnitude but opposite sign i.e. its complex conjugate is just a reflection in the real (or x) axis.

Of course the problem could be tackled by finding both com plex conjugates of a pair and then simply adding them, but the result would be the same.

Let's say the the balls are

B₁  B₂  R  G

The number of permuted sets created by choosing any 2 of the 4 balls is  12

Only  1  of these sets contain 2 blue balls....and they can be permuted in two ways

{ B₁, B₂ } or {B₂, B₁}

So

2 / 12   =

1/ 6

One question, alan....why do we reflect across the x axis????

The total amount is 8 choose 3 or 56 triangles.

"Here's a list of pairwise sums of the conjugates of these complex numbers:

$\overline{z}_1+\overline{z}_2, \overline{z}_1 + \overline{z}_3, \overline{z}_1 + \overline{z}_4, \overline{z}_2 + \overline{z}_3, \overline{z}_2 + \overline{z}_4, \overline{z}_3 + \overline{z}_4$

Find the number of the quadrant each of these pairwise sums is in, and answer with the ordered list, such that your first number corresponds to the quadrant that $\overline{z}_1+\overline{z}_2$  is in, your second number corresponds to the quadrant that $\overline{z}_1+\overline{z}_3$

is in, etc."

Add two of the given complex numbers, $z_1+z_2$ say, and reflect the result in the x-axis (i.e. the Real axis).

So, for example, it looks like  $z_1+z_2$ when added will result in a complex number in the 2nd quadrant.  Reflect this in the x-axis and the result will be in quadrant 3. i.e. $\overline{z}_1+\overline{z}_2$will be in quadrant 3.  

Repeat this process for all the specified pairs.

Note that Jane and Joe can occupy any of 8 positions

12

23

34

45

56

67

78

89

And they can be seated in 2 ways  for each of these

And note that, for each of these positions, we will have 7  other  seats  available for the other 5 people to sit  in

Call these other  empty seats A B C D E F G

So....we can assign any 5 of these for the  other people to  sit  in 

And we can permute these in  P(7, 5) ways   = 2520 ways

So....the total ways =

8 * 2 * 2520   =  40,320  ways

1. We subtract one from each power: 9, 25, 36, 64. Now, taking the prime factorizations, we have $3^2, 5^2, 2^2*3^2, 2^6$ . By scanning, we can see that 64 is a cube number, therefore 65 is our answer. 

We know that 21 has factors of 1, 3, 7, 21. Remember, when you want to find the number of factors a number has, you just add one to each power, then multiply. Therefore, by picking 7 and 3, we should subtract one from each power, yielding us with 6 and 2. For it to be the smallest integer, we just plug in 2 with the 6 and 3 with the 2. Thus, the answer is $2^6*3^2=64*9=\boxed{576}.$

For the second one, answer is 18, because 3^2=9, 3^3=27, and 3^4=81.

For the first one, answer is 65, because 64=4^3.

In order for a number to have an ODD number of divisors, it must be a "Perfect Square", and the product of its exponents must be 21. So, the smallest such number must be: 2^6 x 3^2 =576. The product of its exponents is:

[6 + 1] x [2 + 1] =7 x 3 =21. So, the divisors of 576, which is a Perfect square are:

576 = 1 | 2 | 3 | 4 | 6 | 8 | 9 | 12 | 16 | 18 | 24 | 32 | 36 | 48 | 64 | 72 | 96 | 144 | 192 | 288 | 576 (21 divisors)