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btw, can you show me know your method of "System of equations" 

Let  the  point  be     (   x , 0)

Using  the  square of the  distance  formula

( x -  -2)^2    +   (4 - 0)^2  =   ( 0  - x)^2   +  (18  -  0 )^2      simplify

x^2  + 4x  + 4   +  16    = x^2  +  324

4x  + 20   =   324

4x   =   324  -  20

4x  =  304

x  =  76

The point  is  (76,  0  )

posted solution there; but probably wrong.

posted my solution there; but probably wrong.

Somehow,   we start  with   Bryan  and Harold  and  end up  with   Kaden  and  John   ?????

Ok, here is my answer; however, I think it is wrong..

Let a=b=0

\(2f(0)=2(f(0))^2\), which implied: \(2(f(0))^2-2f(0)=0\) ---> \(2f(0)(f(0)-1)=0\); thus \(f(0)=0\) or \(f(0)=1\)

Now, consider a=1 and b=0

\(2f(1)=(f(1))^2+(f(0))^2\) Suppose \(f(0)=1\), then: \(w^2-2w+1=0--->w=1\) (Where \(w=f(1)\))

But f(0)=f(1)=1 then... (Well, I think the original function is one-to-one function). (Not sure how to prove it.)

So, I ruled out f(0)=1, so f(0)=0, and then f(1)=2

Then, I let a=2 and b=0, a=3 and b=0, etc..

Realized the following pattern:
f(0)=0

f(1)=2

f(2)=4

f(4)=8

f(5)=10

So I concluded (again not sure how to prove it) that \(f(x)=2x\), where \(x=a^2+b^2\)

So let x=25

\(f(25)=2*25=50\)

Now, since this is a linear function, only 1 possible value, which is 50 (So \(n=1\))

and well, since it is only 1 possible value, then \(s=50\)

so: \(n*s=1*50=50\).

Pretty sure I did a mistake somewhere, especially these "claims" without proof.

                       A  =  15

        5sqrt 2

B =  45                                          C

Angle  C   = 180  -   45  -15   =    120

So....using the Law of Sines

AC   /sin 45   =  AB  /  sin 120

AC  / ( 1/ sqrt 2)  =  5 sqrt 2   /  ( sqrt (3)  / 2)

AC  =   (5 sqrt 2  )  ( 1 /sqrt 2)  / ( sqrt  (3)  / 2)

AC  =  5 * 2  /  sqrt 3

AC =   10  /sqrt 3    =     (10/3)sqrt (3)

N : H = 1 : 3

(x + 4) / (3x + 3) = 1/2

x = 5

Hannah had 3*5+3 = 18 apples 

                    A

         P  116

B                                       C

Since  APC   = 116

Then  BPC  =   180   -  116 =    64

And CP  =  BC

So  angle  PBC  = angle BPC   =  64

And  since  AB   = AC

Then  angle  ABC  =  angle  PBC  =  angle   ACB  = 64

And  angle  PCB    =   180  -  2 (64)  =   180  -  128  =  52

So 

ACP =   ACB  -  PCB   =    64   -   52   =    12°

Angle ACP = 12 degrees

Not a problem. 

THX,  civonamzuk for  spotting  my  earlier error   !!!!!

These are my answers:

a) 2.27

b) 137.62

Is that correct?

Check your answer again. Why didn't you use the quadratic formula to solve this problem?!?

36 - 4.144 = 31.856

4.144 * 31.856 = 132.011264

Let \(n\) be a positive integer greater than or equal to \(3\).
Let \(a,b\) be integers such that \(ab\) is invertible modulo \(n\) and
\((ab)^{-1} \equiv 2 \pmod {n}\).
Given \(a+b\) is invertible, what is the remainder when
\((a+b)^{-1} \left(a^{-1}+b^{-1} \right)\) is divided by \(n\)?

\(\begin{array}{|lrclcc|} \hline & && & \quad I. & \quad II. \\ & (ab)^{-1}=\dfrac{1}{ab} &\equiv& 2 \pmod{n} &| *a &| *b \\\\ \hline I. & \dfrac{a}{ab}&\equiv& 2a \pmod{n} \\ & \dfrac{1}{b} &\equiv& 2a \pmod{n} \\ & \mathbf{b^{-1}} &\equiv& \mathbf{2a \pmod{n}} \\\\ II. & \dfrac{b}{ab}&\equiv& 2b \pmod{n} \\ & \dfrac{1}{a} &\equiv& 2b \pmod{n} \\ & \mathbf{a^{-1}} &\equiv& \mathbf{2b \pmod{n}} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline (a+b)^{-1}(a^{-1}+b^{-1}) &\equiv& (a+b)^{-1}(2b+2a) \\ &\equiv& 2(a+b)^{-1}(b+a) \\ &\equiv& 2*\left(\dfrac{a+b}{a+b}\right) \\ &\equiv& 2*1 \\ \mathbf{ (a+b)^{-1}(a^{-1}+b^{-1}) } &\equiv& \mathbf{ 2 \pmod{n} }\\ \hline \end{array}\)

Let line AB and line CD be chords of a circle, that meet at the point Q inside the circle. If AQ = 16, BQ = 12, and CD = 36, then find the minimum length of CQ. 

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I didn't see the word integer

AQ * BQ = CQ * DQ

16 * 12 = CQ(36 - CQ)

CQ = 2(9 - √33)

N  / H   =  1   / 3    ⇒     N =  (1/3)H         (1)

And

( N + 4)  / ( H  + 3)   =     1   /  2            (2)

Sub   (1)  into (2)  for  N

[ (1/3)H  + 4  ]  / ( H + 3 )    =    1   / 2                   cross-multiply

2 [  (1/3) H  + 4 ]    =   H +  3

(2/3)H   +  8   =   H  +  3

8  - 3    =  H  -   (2/3) H

5  =   (1/3) H

H  =  5 (3/1)  =   15

In the end  Hannah  had    H + 3    =   18 apples

Intersecting Chord Theorem

AQ * BQ  =  CQ  *  DQ

16  *  12   =  CQ  ( 36   -  CQ)

192   =  36CQ   -  CQ^2        rearrange  as

CQ^2   -  36CQ    +  192     =   0         complete  the  square on CQ

CQ^2   - 36CQ    =  -192

CQ^2   -  36CQ   +  324    =   -192  +  324

(CQ   -  18)^2   =   132    take  the negative root

CQ -  18  =  -sqrt (132)

CQ  -  18  =  -2sqrt 33

CQ =  18  -  2sqrt (33)  ≈     6.51 =    min length of CQ

EDIT   TO   CORRECT   A  PREVIOUS   ERROR    !!!!!

Let  the  roots  =  x, y, z        in that order

And let  c  = 21

By Vieta

x + y + z  =  9     ⇒      y + z  =  9 - x          (1)

xy + xz + yz  = 11  ⇒   x ( y + z)  +  yz   = 11  ⇒   x ( 9 - x)  +  yz =   11     (2)      

xyz  = -21    ⇒  yz   =    -21   /  x          (3)

Sub   (3)  into (2)

x (9 - x)  -  21/x    =  11    

-x^2   +  9x    -  21/x   =  11                multiply through  by  -x

x^3  - 9x^2  +  21   =  -11x

x^3   - 9x^2   + 11x   +  21    =   0

By  inspection     x  =   -1  is a  root

So

y  +  z   =   9  -  - 1     =  10

And  

yz =    -21/ -1   = 21

To form an arithmetic  series,   y  =   3   and   z =   7

So.....the  series is

x     y     z

-1   3     7

https://web2.0calc.com/questions/find-a-monic-quartic-polynomial

textot has already answered this one. :))

=^._.^=

We  know  the oppositie and  adjacent sides to the  angle......the tangent is what we  need

tan x  =  40  / 70                   use the arctan to find x

arctan  (40 / 70)  =  x    ≈    29.7° 

Isosceles  triangle    with  base angles  equal

So

x  =    180   - 2 (68)     =  180  - 136   =   44°

h = -4.9t^2 + 18t + 1.2

Time to  max  ht =     -18  /  ( 2 * -4.9)    ≈   1.837   sec

Hi fellow guest, I replied to you as well.

Thank you guest, I've tried getting system of equations, but for some reason, I am always 1 equation behind the required amount. That is due to the constraint "non-negative". I would really like to know your solution.