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 x^2 + x + 3 

___________    ≥   0

2x^2  + x - 6

Look at the graph of   x^2 + x + 3  here :

https://www.desmos.com/calculator/grdcjwwkrt

Note that it is positive for all x...so...the only thing that can make this function negative are x values that make the denominator negative   because 

 (positive) 

________   =   negative

(negative)

So  factor the bottom  function  and set it to  0   and we have

(2x - 3) ( x + 2)   = 0

Set  each factor to  0 and solve

2x - 3  = 0           x + 2  = 0

2x  =3                 x  = - 2

x  = 3/2 

So...we  have the  following values on the number line    x  =- 2  and  x  =3/2

And  the  values that  make  2x^2 + x - 6   negative  will come from these possible intervals :

(-infinity, -2)   or  ( -2, 3/2)  or  ( 3/2, infinity)

Let's pick a test point in the middle interval, say, 0  and put it into the function

2(0)^2 + 0  -  6   =  -2.....so...in the interval  ( -2, 3/2)....the  bottom  function is negative

You can check and  see that the other two intervals make the function positive

So.....the  x values that  make  the  given  function ≥ 0   are

(-infinity, -2)   U  ( 3/2, infinity)

See the graph that confirms this, here :

https://www.desmos.com/calculator/ckhao5mwig

6/(2 sqrt(3) - 3) -Simplify:

Multiply numerator and denominator of 6/(2 sqrt(3) - 3) by 2 sqrt(3) + 3:
(6 (2 sqrt(3) + 3))/((2 sqrt(3) - 3) (2 sqrt(3) + 3))

(2 sqrt(3) - 3) (2 sqrt(3) + 3) = -3×3 - 3×2 sqrt(3) + 2 sqrt(3)×3 + 2 sqrt(3)×2 sqrt(3) = -9 - 6 sqrt(3) + 6 sqrt(3) + 12 = 3:
(6 (2 sqrt(3) + 3))/3

6/3 = (3×2)/3 = 2:

=2 (2 sqrt(3) + 3)

There are at least 2 solutions:

a =1  and  b= 25, where 1 + 25 = 26. [This is likely the solution that you want]

a =5  and  b= 5,   where 5 + 5   = 10. [But these 2 values are NOT distinct]

[1/1 + 1/25] / [1 + 25] =
[ 1     + 0.04] / [  26     ] =
        1.04 / 26 x 100/100
104 / 2,600
= 1 / 25

nvm, whatever

\bump

Thank you Guest

How many different values can 

\(\lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 3x \rfloor +\lfloor 4x \rfloor \) 
take for \(0 \leq x \leq 1 \)

lowest common multiple \(\mathbf{ \text{lcm}(1,2,3,4) = 12 }\)

\(\begin{array}{|c|c|c|c|} \hline x & \lfloor 1\cdot x \rfloor & \lfloor 2\cdot x \rfloor & \lfloor 3\cdot x \rfloor & \lfloor 4\cdot x \rfloor & \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 3x \rfloor +\lfloor 4x \rfloor \\ \hline \frac{0}{12} & 0 & 0 & 0 & 0 & 0 \\ \hline \frac{1}{12} & 0 & 0 & 0 & 0 & 0 \\ \hline \frac{2}{12} & 0 & 0 & 0 & 0 & 0 \\ \hline \frac{3}{12} & 0 & 0 & 0 & 1 & 1 \\ \hline \frac{4}{12} & 0 & 0 & 1 & 1 & 2 \\ \hline \frac{5}{12} & 0 & 0 & 1 & 1 & 2 \\ \hline \frac{6}{12} & 0 & 1 & 1 & 2 & 4 \\ \hline \frac{7}{12} & 0 & 1 & 1 & 2 & 4 \\ \hline \frac{8}{12} & 0 & 1 & 2 & 2 & 5 \\ \hline \frac{9}{12} & 0 & 1 & 2 & 3 & 6 \\ \hline \frac{10}{12} & 0 & 1 & 2 & 3 & 6 \\ \hline \frac{11}{12} & 0 & 1 & 2 & 3 & 6 \\ \hline \frac{12}{12} & 1 & 2 & 3 & 4 & 10 \\ \hline \end{array}\)

There are 7 different values: \(\mathbf{0, 1, 2, 4, 5, 6, 10}\)

Hmmm, this question is very interesting, because it is related to the collatz conjecture, a famous unsolved problem in math.

A proper answer to this question was posted some months ago.

Ask Melody as to how to search for it.

Like so:

.

If there exists an integer  a  such that  f(a) = b ,   then  f(b)  is defined and

f(b) = 3b + 1  if  b  is odd      and      f(b) = b/2  if  b  is even .

We must have at least these defined values of the function:

f(14) = 7

f(7) = 22

f(22) = 11

f(11) = 34

f(34) = 17

f(17) = 52

f(52) = 26

f(26) = 13

f(13) = 40

f(40) = 20

f(20) = 10

f(10) = 5

f(5) = 16

f(16) = 8

f(8) = 4

f(4) = 2

f(2) = 1

f(1) = 2

There must be at least 18  integers in the domain of  f .

Let's see how Daniel might have gotten this answer.....

Let  a  > b   where  a and b are positive integers

Then  ....  .let  b^2  and a^2  be consecutive squares

So

a^2  -  b^2  =  59

( a + b) ( a - b)  = 59

Since   a, b are integers  and  59 is prime...its  only divisors  are 1  and 59

This implies that

a + b   =  59

a - b  = 1           add these

2a  = 60

a  = 30

And

a - b  =1

a - 1   = b

30 - 1  =  29

So

30^2  - 29^2  =

900  - 841  =

59

So  900  is the larger perfect square

I like yours better, hectictar....no extraneous lines need be drawn....!!!!

Also....

There are  180°  in triangle ACD, so..

m∠DAC  =  180° - 90° - 50°

m∠DAC =  40°

An inscribed angle is half the measure of its intercepted arc, so....

(1/2)m∠AOC  =  m∠ABC

(1/2)m∠AOC  =  70°

m∠AOC  =  140°

OA  and  OC  are radii of circle  O , so they are the same length, and triangle AOC  is isosceles. So...

m∠OAC  =  m∠OCA

m∠OAC + m∠OCA + 140°  =  180°

m∠OAC + m∠OAC + 140°  =  180°

2m∠OAC + 140°  =  180°

2m∠OAC  =  40°

m∠OAC  =  20°

m∠DAE  =  m∠DAC - m∠OAC

m∠DAE  =  40° - 20°

m∠DAE  =  20°

Where the two semi-circles intersect  in the middle of the square, call this point  E

Connect EC

Let the midpoint of  DC  be F

So FC  and FE are radii of the top semi-circle    = 4 cm

And FE  is perpendicular to FC

So  CFE is a right triangle  ....and the area of this triangle  = (1/2)FE * FC   =

(1/2) (4) (4)  = 8  cm ^2

And  FC , FE  and  arc EC will form  sector FEC  of this top semicircle

And the area of this sector  = (1/2) FE^2 ( pi/2)  = (1/2)  4^2  (pi/2)  = 4  pi cm^2

So...the area of the sector  minus  the area of the triangle will  equal 1/2 of the shaded area  =

[ 4pi  - 8]  cm^2

So...the shaded area  =  2 [ 4pi - 8 ]  = [ 8pi  - 16 ]  cm^2 

If arc CD   = 60°, the angle CAD is  1/2 of this  since it  intercepts this arc

So CAD  = 30°  = BAO

And since BAO  and ABO  =  30  + 60  = 90°

Then,in triangle BAO,  angle BAO  must =  90°  = angle  BOD

So....triangle  BAO  is a 30-60-90 right triangle

Now....draw  OC....and it will be  equal  to   OD  since both are radii

And if arc CD  = 60°....then angle central angle COD  = 60°

And since angle BOD   = 90 °

Then angle BOC  =  angle BOD  - angle COD  = 90 - 60  = 30°  

And angle ABO  = 60  so angle CBO  is supplemental  = 120°

So  angle   BCO  =  180  - angle BOC - angle CBO  =  180 - 30 - 120 = 30°

Thus triangle BOC  is isosceles with  angle BOC  = angle BCO

But BO  =  5....so.....BC  has the same measure  = 5

\(\sin(x +\frac{11\pi}{6})\,-\,\sin(x -\frac{11\pi}{6})\,=\,1\)

We can use the sum of angles formula for sine:

\(\sin(\alpha \pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta\\~\\ \ \\ {\color{BlueViolet} \sin(x +\frac{11\pi}{6})=\sin x\cos\frac{11\pi}{6}+\cos x\sin\frac{11\pi}{6}\quad=(\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)}\\~\\ {\color{Fuchsia} \sin(x -\frac{11\pi}{6})=\sin x\cos\frac{11\pi}{6}-\cos x\sin\frac{11\pi}{6}\quad=(\sin x)(\frac{\sqrt3}{2})-(\cos x)(-\frac12)}\)

So if...

\({\color{BlueViolet}\sin(x +\frac{11\pi}{6})}\,-\,{\color{Fuchsia}\sin(x -\frac{11\pi}{6})}\,=\,1\\~\\ [\,{\color{BlueViolet}(\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)}\,]\,-\,[\,{\color{Fuchsia}(\sin x)(\frac{\sqrt3}{2})-(\cos x)(-\frac12)}\,]\,=\,1\\~\\ (\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)-(\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)\,=\,1\\~\\ 2(\cos x)(-\frac12)\,=\,1\\~\\ - \cos x\,=\,1\\~\\ \cos x\,=\,-1\\~\\ x\,=\,\pi+2\pi n\quad \text{, where }\ n\ \text{ is an integer.}\)      then...

Extend AD through to intersect the circle at F

Now.....since angle ACB  =   50°...then minor arc AB has twice its measure  =100°

And since angle  CBA  =70°....then  angle BAD  = BDA  - CBA  = 90 - 70  = 20° = angle BAF

And angle BAF  =  intercepts minor arc BF...so  the measure of minor arc BF  is twice this = 40°

But arc ABE  =180°   =  minor arc AB  + minor arc BF  + minor arc FE

So

180  =  100 + 40  +  minor arc FE

180  =140 + minor arc FE

40  = minor arc FE

But angle FAE  intercepts minor arc FE, so it has  1/2 of its measure =  20°

But angle  FAE  = angle DAE...so  DAE  = 20°

Um....you see, i dont want a direct answer. At least a fraction or something is good. Can you at least show your steps or give me the solution, or hints?

(XY / PQ)²   =   area of UVWXY / area of LMNPQ

(XY / 10)²   =   198 / 88

(XY / 10)²   =   9 / 4

XY / 10   =   √[ 9 / 4 ]

XY / 10   =   3 / 2

XY   =   10 * 3 / 2

XY   =   30 / 2

XY   =   15

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We can use partial fractions, here

Note that   x^3 + x^2 -20x   can be factored  as   ( x^2 - 4x) ( x + 5)

x^2  - 2x + 10                               G                    H  

______________        =          _____         +   ________

( x^2 - 4x)(x + 5)                        x + 5               x^2  - 4x

Multiply through  by  ( x + 5) ( x^2  - 4x)

x^2   - 2x + 10  =  G(x^2 - 4x)  +  H (x + 5)    simplify

x^2  - 2x + 10  =  Gx^2 - 4Gx  + Hx + 5H      equate coefficients

1 = G

-2 = H  - 4G

10  = 5H

It's obvious that  G  = 1   and  H  = 2

So

H / G   =   2 / 1    = 2

diameter of circle  M  =  CB  =  36

radius of circle  M  =  36 / 2  =  18

area of semicircle  M  =  (1/2)pi( 18² )  =  162pi

radius of circle  O  =  OC + CB  =  32 + 36  =  68

diameter of circle  O  =  AB  =  2 * 68  =  136

area of semicircle  O  =  (1/2)pi( 68² )  =  2312pi

diameter of circle  K  =  AB - CB  =  136 - 36  =  100

radius of circle  K  =  100 / 2  =  50

area of semicircle  K  =  (1/2)pi( 50² )  =  1250pi

area of shaded region   =   area of semicircle O  -  area of semicircle K  -  area of semicircle M

area of shaded region   =   2312pi  -  1250pi  -  162pi

area of shaded region   =   900pi