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For this to be true   the discriminat   b^2 - 4ac  > 0

m^2 -4(1)(4) > 0

m^2 > 16

m <-4       m >4

(-inf,-4) U (4, + inf)

3. Let's first start off with the internal angle measurement of a regular hexagon. Each angle is equal to:

\(180-\frac{360}6 = 120\) degrees. If we drop a perpendicular to the smaller hexagon from a point like A, then we get two 30-60-90 triangles(since the smaller hexagon has interior angle measurement of 120, and the angles must add up to 180, so 120 + 30 + 30 = 180).

Let's then define the length of a side of the larger hexagon as x, meaning that the midpoint segments have length \(\frac{x}2\). 

First, let's find the area of the larger hexagon. Because a regular hexagon is comprised of 6 equilateral triangles(they all have the same side length, with 60 degree interior angle measurements), the area is equal to the area of one of the equilateral triangles multiplied by 6. Given a side length x of an equilateral triangle, the area is equal to:

\((x^2*\sqrt{3})/4\), which can be derived from heron's formula or pythagoras' theorem( just name an arbitrary side of the equilateral triangle as x). With this in mind, the area of a regular hexagon with side length x would then be:

\((6*x^2*\sqrt{3})/4 = (3*x^2*\sqrt3)/2\)

Now that we have the area of our larger hexagon in terms of x, we need to find the area of the smaller hexagon in terms of x. We can accomplish this by dropping an altitude from point A to the smaller regular hexagon, forming a perpendicular bisector with that side.

We then have two 30 - 60 -90 right triangles, with the hypotenuse having length \(\frac{x}2\), and the 60 degree angle being at the angle bisector of \(\angle FAB\). The 60 degree leg(the side adjacent to the smaller hexagon) is then \(x/2 * \sqrt3/3\) (remember the ratios of a 30-60-90 triangle!).

We then have that the entire side of the smaller hexagon is \(2*x/2 * \sqrt3/3\)(because two 30-60-90 triangles make up the side of the smaller hexagon). The smaller side is then \(x* \sqrt3/3\). Using our area formula from earlier, we have that the area is equal to:

\(3*(x*\sqrt3/3)^2*\sqrt3/4 = x^2*\sqrt3/4\)

Our fraction is then equal to:

\((x^2*\sqrt3/4) / ((3*x^2*\sqrt3)/2) = (x^2*\sqrt3)/4 * 2/(3x^2*\sqrt3)\)

Cross multiplying and canceling out terms, we get:

\(2/12 = 1/6\) As our answer.

This method of finding the fraction is certainly more extreme in my opinion than other much faster solutions. One such way is just assuming the side length of the regular hexagon because the fraction will stay constant no matter the side length.

2. Please reupload your question, the side lengths of the rhombus appear to be missing from the problem.

1. First, we notice that the cosine of \(\angle ABC = 4/5\)(adjacent / hypotenuse).

With this, we can use the law of cosines to find the length of the median \(\overline {AM}\).

We get the equation:

\(\overline {AM}^2 = 4^2 + (2.5)^2 - 2*4*2.5* \cos(\angle ABC)\)

We combine like terms and get:

\(\overline {AM}^2 = 22.25 - 20* \cos(\angle ABC)\)

We realize we already have \(\cos( \angle ABC)\)= \(\frac45\)

We then substitute, and get:

\(\overline{AM}^2 = 22.25-20*\frac45 = 22.25 - 16 = 6.25\)

To get the length of the median \(\overline {AM}\), we just take the square root of this result, to get:

\(\overline{AM} = \sqrt{6.25} = 2.5\)

A different perspective...just an FYI

7⁵  possible numbers      4 out of 7 are odd     7⁵  x  4/7 = 9604  (as Chris calculated)

2⁶⁴  = 1.844674 x 10¹⁹  different  64 bit words are possible

64 is  2⁶

2 terabytes is 2 000 000 000 000 bytes

     each byte is 8 bits    = 16 000 000 000 000 bits         ^{log(16000000000000,2) = 43.86313713864835}

2^43.863137 / 2⁶       =  2^37.863137      64 bit words

That is much better now :)

Oh I see where! Thanks for reminding me melody! 

3(3)^2 + 3(3) + 12 means 3*3² + 3*3 + 12 or 3*9 + 9 + 12 or  27 + 9 + 12  or  48.

Nice work Jfan17 .... except you forgot some brackets.  Can you work out where and fix it?

Brackets are very important.  

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You can use the identity 

\(\boxed{cos(a)+cos(b)=2\;cos\frac{a+b}{2}\;cos\frac{a-b}{2}}\\~\\\)

\(y = \cos(12x) + \cos(14x)\\ \text{roots are given by}\\ \cos(12x) + \cos(14x)=0\\ 2\cos\frac{14x+12x}{2}\;\cos \frac{14x-12x}{2}=0\\ 2\cos(13x)\;\cos (x)=0\\ \cos(13x)=0\qquad \qquad  or \qquad cos(x)=0\\ 13x=\pm \frac{\pi}{2}+2\pi n \qquad or \qquad x=\pm\frac{\pi}{2}+2\pi n\\ x=\pm \frac{\pi}{26}+\frac{2\pi n}{13} \qquad or \qquad x=\pm\frac{\pi}{2}+2\pi n\\ x=\pm \frac{\pi}{26}+\frac{4\pi n}{26} \qquad or \qquad x=\pm\frac{13\pi}{26}+2\pi n\\ x=\frac{\pi}{26}(4n\pm1) \qquad or \qquad x=\pm\frac{13\pi}{26}+2\pi n\\ \)

Coding:

y = \cos(12x) + \cos(14x)\\
\text{roots are given by}\\
\cos(12x) + \cos(14x)=0\\
2\cos\frac{14x+12x}{2}\;\cos \frac{14x-12x}{2}=0\\
2\cos(13x)\;\cos (x)=0\\
\cos(13x)=0\qquad \qquad  or \qquad cos(x)=0\\
13x=\pm \frac{\pi}{2}+2\pi n \qquad or \qquad x=\pm\frac{\pi}{2}+2\pi n\\
x=\pm \frac{\pi}{26}+\frac{2\pi n}{13} \qquad or \qquad x=\pm\frac{\pi}{2}+2\pi n\\
x=\pm \frac{\pi}{26}+\frac{4\pi n}{26} \qquad or \qquad x=\pm\frac{13\pi}{26}+2\pi n\\
x=\frac{\pi}{26}(4n\pm1) \qquad or \qquad x=\pm\frac{13\pi}{26}+2\pi n\\

Hi can someone help me with this problenm?

In triangle PQR, P= Q+R. Of all three exterior angles of triangle  what is the least exterior angle, in degrees?

Exterior angle theorem

The exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.

Enterior angles:       ∠P = ∠Q + ∠R     ( ∠P  MUST be 90° )

∠P = 90°     ∠Q + ∠R = 90°

The sum of angles  Q and R is equal to the exterior angle at  P vertex.

The least exterior angle is at P vertex, and it's 90°.  

The 5 digit number will  end  in either  1, 3, 5 or 7

Since  digits  can be repeated we have 7 choices for each of  the 4 leading positions  and 4  choices for the  ending digit

So....the total number of odd five-digit countig numbers  = 

7^4  * 4  =  

9604

See the folllowing  [ not to scale] image :

GV = 5

So VH  = 21 - 5 =  16

And 

EU = 8

So UF = 15 - 8  = 7

Call  segment UX  = m  and  segment  VY = n

VX = 6 + m

UY = 6 + n

Using the intersecting  chord theorem.....we have that

GV * VH  = VX * VY

5 * 16 =  (6 + m) n      (1)

80  = 6n + mn

80 - 6n = mn

And 

EU * UF =  UY * UX

8 * 7  =  (6 + n) m

56 = 6m + mn

56 - 6m = mn

This implies  that

80 - 6n = 56 - 6m

80 - 56 = 6n - 6m

24  = 6(n - m)

4 = n - m

m + 4  = n   (2)

So using (1) and (2) we have  that

80  = (6 + m) (m + 4)

80 = m^2 + 10m + 24

m^2 + 10m - 56  =  0

(m + 14) ( m - 4) =  0

Setting both factors to 0 and solving for m  produces  m =  -14   (reject)  or  m  =  4

So

m + 4 =  n

4 + 4 =  n

8  = n

So.....the length of XY  is

m + UV +  n  =

4  +  6  +  8   =

18

∠ABC = 120°

∠YBC = 90°

∠BYC = ∠BCY = 45°

∠XBY = ∠YBX = 30°

∠AXY = 150°

∠XAY = ∠BAC = 15°

∠BAC = 15°      

Animal Farm is an allegorical novella by George Orwell, first published in England on 17 August 1945. The book tells the story of a group of farm animals who rebel against their human farmer, hoping to create a society where the animals can be equal, free, and happy.

Thanks!!!

Nice, EP!

Nice, CTG  !!!!

Agree with , CU......nice discussion....

The  directional  vector through AB  is  given by

(2 -1, 3-1, 4-2)   = ( 1, 2, 2)  = s

And

QA  =  (  1-3, 1- -7, 2- -1)  = (-2, 8, 3)

Take the cross-product of QA x s

i       j     k     i        j

-2    8    3     -2     8  =    [ 16i  + 3j  - 4k ] -  [ 8k + 6i  -4j ] =   10i  + 7j -12k  = (10, 7, -12)

1     2     2     1     2

Distance =   length of  cross-product                  √[10^2 + 7^2 + (-12)^2 ]      √293         √293

                   ________________________  =     ___________________ =  _____  =   _____

                             length of  s                               √  [1^2 + 2^2 + 2^2 ]            √9              3

So   d   = 293

a) 1/52 since there is one six of diamonds in a deck of cards.

b) 1/4, since there are 13 clubs out of 52 total cards.

c)3/13. 12 face cards in 51 total cards.

d)1/2 since there are 26 red cards