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y = x (x^2 +1)^(-1)              take the  derivative and set to 0

y'  = (x^2 + 1)^(-1)  - 2x*x (x^2+ 1)^(-2)   =0

( x^2 + 1)(-2)  ( x^2 +  1 - 2x^2)  = 0

-x^2 + 1 = 0

x^2   -1   =  0

(x - 1) ( x + 1)  = 0

x -1  = 0                                                         

 =1   y =  1 / (1^2 + 1) = 1/2  = max  y                       

 x + 1 =  0

x =  -1    y =   -1 / ((-1)^2 + 1)  =  -1/2 = min y

The sum of max + min  y's   =  0

Very crafty, proyaop !!!

Continuing with proyaop's answer

s^2   = (3 -sqrt 5)   / 2  

So  Pn =  sqrt [ (3 -sqrt 5) / 2) ]  = s

But  P1 * P2     is  =  s * s  =   s^2

So....the given sum should just be  10(s*s) = 10s^2   =   10  (3 - sqrt 5)  /  2 =    5 ( 3 - sqrt 5)  = 15 - 5sqrt5

......Can you please plug it into latex and add the diagram for X?....

So basically we have a decagon inscribed in a unit circle, and the problem asks to find the perimeter.

Splitting the decagon up into 10 triangles and observing one of them, you will find that it is an isosceles triangle with vertex angle 360/10 = 36 degrees. Using the law of cosines, the side length of the decagon, s:

\(s^2=1^2+1^2-2*1*1\cos{36^o}\), if you know some trig, you will quickly find that cos(36 degrees) = (sqrt(5) + 1)/4, which can be substituted in to get \(s^2=2-{\sqrt{5}+1\over{2}}={3-\sqrt{5}\over{2}}\). 

Remember to square root both sides and simplify the radical to obtain s, and then multiply by 10 to get the perimeter: I don't want to directly spoil the answer for you Vxrtate- happy solving !

Floor 3: Room 3, 6, 9

Floor 5: Room 3, 6, 9

Floor 7: Room 3, 6, 9

9 total possible options, of which only 1 will be correct, so 1/9 (wait read below)

Also, doesn't it say that the hotel only has 5 floors, so there is no floor 7- and thus the answer could also be 1/6 if there were any typos in the question...

I'll  do  one.....the calculation of the  other two is similar

(The median to BC is AF , not CF)

Mid point of BC  = [ (-9 + 1)  /2 ,  (6 + -4) /2)  =  (-4, 1) =  F     (5,4) (-4,1)

Slope of a line through AF  =  ( 1 - 4)  / (-4 - 5)  =  -3/-9  = 1/3

Equation of line containing  median AF   using the  slope and (5,4)  =

y = (1/3) ( x - 5) + 4

y = (1/3)x - 5/3 + 4

y = (1/3) x + 7/3

If the mean  =  0......then the sum of all the oods must =  0  

The 500th positive odd   =  2 (500) - 1  =  999

The 500th negative odd  =  2 (-500) + 1   = -999  =  the least number

We can use  somethng called "Pick's Theorem"

Lattice point =  a point defined by the intersection of  grid  lines

Interior  lattice points   =  17

Boundary lattice points =   8

Area =  interior lattice points  + boundary lattice points / 2     - 1   =

17 + 8/2  -  1    =

14 + 4   - 1     =

20  

THX, Imcool.....very observant !!!!

Best to check the logic of some of these questions  before attempting an answer 

We want to find the intersection points of

(x -2)^2  + (y -7)^2   = 25       and  y = 5x -28

So

(x -2)^2  + (5x -28 - 7)^2   =   25

(x -2)^2  + (5(x -7))^2    = 25

(x -2)^2 + 25(x - 7)^2  = 25

x^2 - 4x + 4 + 25 (x^2 - 14x + 49)  = 25

x^2  - 4x + 4  + 25x^2 - 350x + 1225 - 25 =  0

26x^2 - 354x +  1204   =  0

x =  (354 + sqrt [ 354^2  - 4*26*1204 ])   / 52    =  7    and

x =   (354 - sqrt [ 354^2  - 4*26*1204 ])   / 52 =   86/13

So   y = 5(7) - 28 =  7

And

y = 5(86/13) - 28  =  66 /13

So  the points  are   (7,7)   and (86/13  , 66 / 13)

        W   60                     120   X

                                        -          

                        -      

          -                         

Z                                            120     Y

This is impossible  .....triangle WXZ's   interior angle sum would exceed 180°

Draw altitude CN  to AB =  h

We will have a right triangle CNB

Let BN = x

And angle NCB = 90 - 55 = 35

Using the Law of Sines

h /sin (55) = x / sin 35

h = xsin 55 / sin 35       (1)

Next, draw altitude DO  to AB  = h

We have another right triangle

Let AO = 96 - (44 + x) =  52 - x

Angle DAO = 180 - 110  = 70 

Angle ADO  = 90 - 70  = 20

Using the Law of Sines again

h / sin 70  = (52 - x)/ sin 20     

h =  (52 - x)sin 70 / sin 20    (2)

Equating (1) , (2)

xsin 55  / sin 35  =  (52 - x)sin (70) / sin 20

x (sin 55 / sin 35)  =  52(sin 70 / sin 20)  - x ( sin70 /sin20)

x [ sin 55 / sin 35 + sin 70 /sin20 ]  = 52 (sin70 /sin20)

x =  (52)(sin 70 / sin20) / ( sin55 / sin35 + sin 70/sin 20)  

h = (52)(sin 70 / sin20) / ( sin55 / sin35 + sin 70/sin 20)  * (sin 70 / sin20) ≈  94

Area =   (1/2) (94) (44 + 96)  = 47 * 140  ≈  6580 

This isn't always true, Imcool....see here :

However....it is true that the sum of the interior angles =  360

I believe that there is no answer. This is because opposite sides of quadrilaterals always add up to 180 degrees. 

\(110+55<180\)

\(\sqrt{\frac{3x + 4y}{6x + 5y + 4z}} + \sqrt{\frac{y + 2z}{5x + 6y + 4z}} + \sqrt{\frac{2z + 3x}{x + 2y + 3z}}\)

So, largest value of that.

\(u=3x+4y,v=y+2z,w=3x+2z\)

we have

\(\sqrt{\frac{u}{u+v+w}}+\sqrt{\frac{v}{u+v+w}}+\sqrt{\frac{w}{u+v+w}}\)

and also

\(\sqrt{\frac{u}{u+v+w}}+\sqrt{\frac{v}{u+v+w}}+\sqrt{\frac{w}{u+v+w}}\le\sqrt{3}\)

You can google why. I can't write it in latex since idk how.

Answer: \({\sqrt{3}}\)

Median is the sum of two bases/2.

One base is 12, the other base is 12+16=28.

So the median is \(\frac{28+12}{2}=20\).

Just as a side note, I'm not criticizing, but most of the time Vxrtate, you answer the question by rationalizing the denominator. In other words, you should put \(\frac{8\sqrt{3}}{3}\)unless it specifically states otherwise.

The answer is 9

In trapezoid PQRS, \(\overline{PQ} \parallel \overline{RS}\).  Let X be the intersection of diagonals \(\overline{PR},\overline{QS}\).  The area of triangle PQX is 2 and the area of triangle RSX is 2.  Find the area of trapezoid PQRS.

There are many solutions. The area could be 9, it also could be 12.5, ETC. Not enought info, I guess.

1/3.

Check out this similar question.

In triangle PQR, let X be the intersection of the angle bisector of \(\angle P\) with side \(\overline{QR}\), and let Y be the foot of the perpendicular from X to side \(\overline{PR}\).  If PQ = 8, QR = 5, and PR = 1, then compute the length of \(\overline{XY}\).

Ookie Dookie

No answer. Simply logic

Let me repeat the question, and you try to figure out whats wrong.

PQ = 8, QR = 5, PR = 1.

2 sides of a triangle has to add up to more than the third side, but 

\(5+1<8\)

so this triangle is not even possible in the first place.

Find all points (x,y) that are 5 units away from the point (2,7) and that lie on the line y = 5x - 28

If we plug it into desmos and draw a circle of \(\left(x-2\right)^{2}+\left(y-7\right)^{2}=25\), then we can see all the points of intersection, and there are 2.

\((7,7)\)

\((6.615,5.077)\) That is not the value, it is just an approximation.

\(y= 1/5\cdot f(5x + 5) - 5\)

\(y= 1/5\cdot f(5x + 5) - 5, x = -2/5\)

\(y= 1/5\cdot 4- 5\)

\(y = 6-5\)

\(y = 1\)

The point will be \((-2/5,1)\)