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A circle of radius 5 with its center at $(0,0)$ is drawn on a Cartesian coordinate system.

How many lattice points (points with integer coordinates) lie within or on this circle?

A Calculation of the Number of Lattice Points within or on the circle:

Let \( \lfloor x \rfloor \) be the largest integer equal to or less than x.

Example:
\(\lfloor 3.53553390593 \rfloor = 3\)
\(\lfloor -3.53553390593 \rfloor = -4\)

Noted by Gauss:

Let r  radius of the circle = 5

Let \(x = r^2\)

\(\begin{array}{|rcll|} \hline A_2(x) &=& 1 + 4\lfloor \sqrt{x} \rfloor + 4 \lfloor \sqrt{\frac{x}{2}} \rfloor ^2 + 8 \sum \limits_{y_1= \lfloor \sqrt{\frac{x}{2}} \rfloor + 1 }^{\lfloor \sqrt{x} \rfloor} \lfloor \sqrt{x-y_1^2} \rfloor \qquad & | \quad x = r^2 = 5^2 \\\\ &=& 1 + 4\lfloor \sqrt{5^2} \rfloor + 4 \lfloor \sqrt{\frac{5^2}{2}} \rfloor ^2 + 8 \sum \limits_{y_1= \lfloor \sqrt{\frac{5^2}{2}} \rfloor + 1 }^{\lfloor \sqrt{5^2} \rfloor} \lfloor \sqrt{5^2-y_1^2} \rfloor \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 8 \sum \limits_{y_1= 3 + 1 }^{5} \lfloor \sqrt{5^2-y_1^2} \rfloor \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 8 \sum \limits_{y_1= 4 }^{5} \lfloor \sqrt{5^2-y_1^2} \rfloor \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 8 \cdot \left( \lfloor \sqrt{5^2-4^2} \rfloor +\lfloor \sqrt{5^2-5^2} \rfloor \right) \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 8 \cdot \left( 3 + 0 \right) \\\\ &=& 1 + 4 \cdot 5 + 4 \cdot 3 ^2 + 24 \\\\ &=& 1 + 20 + 36 + 24 \\ &\mathbf{=} & \mathbf{81} \\ \hline \end{array}\)

81 lattice points (points with integer coordinates) lie within or on this circle with radius 5.

Example:
\(r = 0 \ldots 20\)

Number of lattice points in circle:

\(\begin{array}{|r|r|r|} \hline r & \text{lattice points in circle} & \text{lattice points in sphere } \\ \hline 0 & 1 & 1 \\ 1 & 5 & 7 \\ 2 & 13 & 33 \\ 3 & 29 & 123 \\ 4 & 49 & 257 \\ {\color{red}5} & {\color{red}81} & 515 \\ 6 & 113 & 925 \\ 7 & 149 & 1419 \\ 8 & 197 & 2109 \\ 9 & 253 & 3071 \\ 10 & 317 & 4169 \\ 11 & 377 & 5575 \\ 12 & 441 & 7153 \\ 13 & 529 & 9171 \\ 14 & 613 & 11513 \\ 15 & 709 & 14147 \\ 16 & 797 & 17077 \\ 17 & 901 & 20479 \\ 18 & 1009 & 24405 \\ 19 & 1129 & 28671 \\ 20 & 1257 & 33401 \\ \hline \end{array} \)

The total number of   lattice ponts is given by

1  +  (4 * 5)  + 

4 *  [  floor √[ 5^2  - 1^2]  +  floor √ [5^2  - 2^2]  + floor √[5^2  - 3^2] + floor √[5^2 - 4^2]  ]  =

1  + 20  +

4 *  [  floor √24  +  floor  √21  +  floor √16  +  floor √9 ]  =

21  + 4 [  4 + 4 + 4  + 3 ]  =

21  +  4 [ 15]  =

81

Whatever  x  value we choose to plug in to  (g + f)(x)  has to be in the domain of both  g(x)  and  f(x) .

For instance, we can't find  g(1) + f(1)  because we can't find  g(1) .

The domain of  g + f  is all real  x  values such that   x ≥ -14  and  x ≤ -11

The domain of  g + f  is all real  x  values such that   -14 ≤ x ≤ -11

The minimum value in that domain is  -14  .

\($x_1,x_2$\)

\($|a+2|>2$\)

\($\frac 1{x_1}+\frac 1{x_2}<1$\)

The sum of the roots =   -a  

And since  the  product of the roots  = b  >  0.....this implies that both roots have the same sign    

Let  x₁  , x₂   be the positive roots

So  -x₁, -x₂  are the negative roots

The  sum of the negative roots  results in

-x₁ +   -x₂  =   -a

- [ x₁  + x₂  ] =  -a      ⇒  x₁  +  x₂  = a

Therefore......a  in this  case   >  0  so

l a + 2  l  >  2    is satisfied

Now....if both positive roots   are   >  0   but  ≤ 2   ....then.....

1 /x₁  +  1/x₂   ≥ 1         but this violates the condition that  1/x₁  + 1/x2  <  1

Therefore   x₁, x₂   must  be  >  2

Therefore

x₁  + x₂  =  -a

- [ x₁  + x₂  ]   =  a

But.....since   x1, x2  >  2   then their sum  must be  > 4

So....this implies that 

[x₁ + x₂]  >  4 

 - [ x₁  + x₂ ]  <  -4     ⇒    a  <  - 4

So.....in the case of two positive roots  .......

l a  + 2 l  >  2     is true

So......

l a  + 2  l  >  2   is proved for both positive and negative roots

Nice, EP....!!!!

To find these points..... first  find  the x  intersection of the graphs of

y  = (5/12)x  +  2     and    x^2  +  (y -2)^2  = 9

Manipulating the first....we have that  y  -  2  =  (5/12)x

Subbing this into the second, we have that

x^2  +  (5/12* x)^2  =  9      simplify

x^2  +  25x^2/ 144  = 9 

x^2  +  (25/144)x^2  = 9

169/144x^2  =  9

x^2  =  9*144/ 169       take the positive and negative roots

x  = ±√  [ 9 * 144 / 169 ]

x =  ± 36/13

So..using   y   =  (5/12)x  + 2...the y intersection points are

y  =  (5/12)(36/13)  +  2  =  (180/156) + 2  =  41/13

And 

y  =   (5/12)(-36/13)  +  2  =  (-180/156)  + 2  =  11/13

So....the points of intersection are

(36/13, 41/13)   and  (  -36/13, 11/13)

So  using these   points and the slope  (-12/5)   we have the  equations

y  =  (-12/5)( x - 36/13)  +41/13

So...when   x  = 0, the y intercept of this line  is  (-12/5)(-36/13)  + 41/13  = 49/5  = 9.8

And

y  =  (-12/5)( x +  36/13)  +  11/13

And.....when x  = 0, the y intercept of this line  is (-12/5)(36/13)  + 11/13  = -29/5  = -5.8 

Here's  a graph : 

https://www.desmos.com/calculator/ctru5i68ra

The other vertex  of the rectangle is  the circle's center  =   (5, -2)

Look  at the image here :  

The region common to both areas is just  the quarter circle  area  of the given circle

And this circle has a radius  of 3.....so....the quarter circle has an area  of

pi (r^2)  / 4  =   pi (3^2)  / 4  =    (9/4)pi   units ^2

x^2  +  y^2  ≤  36   (1)

y  =  x  - 4     (2)

Look at the graph, here  : https://www.desmos.com/calculator/leyhcfzzep

The integer pairs that make this true  are  :

(-1, -5)   (0, -4)  (1, -3)  (2, -2)  ( 3, -1)   ( 4, 0)   ( 5, 1)

\($\dfrac{30+19i}{4+9i}$\)       multiply top/bottm by  4  - 9i

[ 30  +  19i ] [ 4 - 9i ]  / (  4 + 9i) (4 - 9i) ]

[ 120 + 76i  - 270i   -  171i^2 ]  /  [ 16 -  81i^2  ]

[  120  +  171  -  194i ]  /  [ 16 + 81  ]

[ 291  -  194i  ] / [ 97 ]

(291/ 97)  -  (194/97) i

3 - 2i

\(\frac{1-i}{2+3i}\)     Multiply  top/bottom by   2 - 3i

[ 1  - i  ] [ 2 - 3i ]  /  [ ( 2 + 3i) ( 2 - 3i)  ]  

[ 2 - 2i - 3i + 3i^2  ] /  [ 4  - 9i^2 ]

[ 2  -  3  - 2i -  3i  ]  / [  4 + 9 ]

[ -1  - 5i  ] / [ 13 ] 

The question asks what is the minumum value in the domain of  (g+f)(x) :      -14

The domain   is   -14  -13   -12  -11         Cool?  

Start with the 4 "poles" i.e (5,0), (0,5), (-5,0), (0,-5) 

Each quadrant of the circle has 2 integer values (3,4) and (4,3) - Since (3,4,5) is a pythagorean triple

4 quadrants give 8 more points so 12 points in total

Thanks man. 

I want to say something like 'I was tired' or some other excuse, but I guess I am just a bit dumb. 

Thanks again though. 

Okay, I'll give this one a try also...

Let  t  be the number of 6 month periods in the term.

7000( 1 + 4% * t )   =   1400 + 7000( 1 + 4% * t/2 )

                                                         Divide through by  7000.

1 + 4% * t   =   0.2 + 1 + 4% * t/2

                                                         Subtract  1  from both sides.

4% * t   =   0.2 + 4% * t/2

                                                         Subtract  4% * t/2  from both sides.

4% * t  -  4% * t/2   =   0.2

t(4%  -   4% / 2)   =   0.2

t(4% - 2%)   =   0.2

t( 2% )   =   0.2

t   =   0.2 / ( 2% )   =   10

There were  10  periods of 6 months in the term. That is  5  years.

Thank you very much for the help! That's it. 

I'll enjoy it for a while, without wanting to abuse

- A capital of € 7000 was applied on a flat-rate basis at the half-yearly rate of 4% over a given period and produced an accumulated capital of more than € 1,400 to that which would be obtained if the period was reduced by half. Calculate the term of capital application.

0.5 ((2000 kg)  (9.81 m/s2) / 4)   = 2452 N - This is accurate.

[0.5 x 1,515 x 9.81] /4 = ~1,858 N - What your calculation should be. Don't square 9.81!! That is just the force of gravity on Earth=9.81 meters per second per second!!.

Here's my best guess, but I could easily be wrong.....

Let  c  be the starting capital,  and let  i  be the annual interest rate.

accumulated capital after  27  months   =   c(1 + i * 27/12)   =   6600

accumulated capital after  28  months   =   c(1 + i * 28/12)   =   6600 + 0.75% * c

Now we have two equations and two variables.

c(1 + i * 28/12)   =   c(1 + i * 27/12) + 0.75% * c         Divide through by  c .

1 + i * 28/12   =   1 + i * 27/12 + 0.75%                       Subtract  1  from both sides.

i * 28/12   =   i * 27/12 + 0.75%                                   Subtract  i * 27/12  from both sides.

i * 1/12   =   0.75%                                                      Multiply both sides by  12 .

i   =   9%

If I have interpreted the question correctly, this makes sense because it tells you that an additional month of interest adds 0.75% of the capital. That means that 0.75% is the annual rate / 12 . I might have not interepreted the question correctly.

Now....using this value of  i  , we can find  c .

c(1 + 9% * 27/12)   =   6600

c( 1.2025 )   =   6600                      Divide both sides of the equation by  1.2025 .

c   ≈   5488.57

OK, your problem is not stated very clearly! But, as far as I can understand you, this is what I think  your problem is:

If the initial Capital was C, then you have:

C x i% x 27 = 6,600..........................................(1)

C x i% x 28 = 6,600 + 0.0075C.........................(2), solve for i% and C

i =0.0075 pe month!!, and: 0.0075 x 12 x 100 =9% annual simple interest.

C=32,592.60 - Euros - The initial capital.

NOTE: I used SIMPLE INTEREST and not COMPOUND INTEREST, because your question says" applied 'simple interest'". Give us some feedback, so that we can help you solve the problem.

So x ≤ -14 and x e R ? 

Equation of a circle :  (x-h)^2  +  (y-k)^2  = r ^2        h, k being the center and r = radius

SO let's get to that form

x^2 +   8x       + y^2 + 4y    =  -c

Kinda 'complete the squares' now

(x+4)^2   + (y+2)^2  =  -c + 20            (if you complete the squares on the left side ...you will see we added 20)

so    -c + 20  = r^2 = 3^2 = 9     therefore   -c = -11    or  c= 11        center of circle (-4,-2)   radius = 3

It's a Financial Calculation problem.

A certain capital applied in simple interest at the annual rate during 27 months produced the accumulated capital of 6600 €. If the same capital was applied at the same rate but for 28 months, the cumulative capital reference would be increased by an amount corresponding to 0.75% of the initial capital. Calculate the starting capital and the annual rate to which it was applied.

I arrived at the initial formula, but then I do not know how to solve:

Equity = Initial Capital * (1 + rate * time), this is the calculation formula for accumulated capital.

6600 + 0,0075C₀ = C₀ * (1 + i * (28/12))

Little remark: I am Portuguese. 

What do you want to solve for? What do the instructions say? 

Since you have 5 protons + 5 neutrons it would be

\({10}_{B}\)

  then if you add an electron it would look like

\({10B}^{1-}\)

Pretty sure....

Hmmmmm.....   at exactly WHAT am I to be amazed?

The first one has a domain of

-14   -13   -12   -11   -10  -9 ......etc

The SECOND one has a domain of

-11  -12  -13  -14  -15  -16.....

The common numbers (in color) define the new domain after adding the functions....the minimum is   -14   that will satisfy both of the domain restrictions given....