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3 + 97 ==100

11 + 89 ==100

17 + 83 ==100

29 + 71 ==100

41 + 59 == 100

47 + 53 == 100

Hopefully, you can count the number of pairs.

Area  of large rectangle =  4 * 5  = 20

Area of larger right triangle at right =  (1/2) 4 * 5 =  10

Area of smaller right triangle at left  = (1/2) 3 * 4 =  6

Area of polygon   =

20  - 10  - 6   =     4

there it goes again.

Let the point  we are looking for  be  ( x  ,  (1/sqrt (2)) (x^2 -18) )

Using the square of the distance formula

D^2  = x^2  + [ (1/sqrt (2))(x^2 -18)]^2   

D^2 =  x^2  + (1/2) ( x^2 - 18)^2        take the derivative of  this and  set to 0

D^2'  =   2x  + 2x ( x^2 - 18)  = 0

2x ( x^2 - 18 + 1)  =  0

2x ( x^2 - 17)  = 0

The second factor set = 0  is what we want

x^2  - 17   =  0

x^2  = 17

x = sqrt (17)

y = -1/sqrt (2)

The shortest distance is

sqrt [  (sqrt (17))^2  + ( -1/sqrt (2))^2 ] =  sqrt [ 17 + 1/2 ]  =  sqrt (17.5) ≈   4.183 units

sin 45 - cos 30  = 

[sqrt 2]  / 2   -  [ sqrt 3 ]  / 2  =

[ sqrt (2)  - sqrt (3) ]  /  2

78  -  38 ==40

40 / 10 ==$4 - cost of each computer game

[942!  +  120!] = 2^116 * 3^58 * 5^28 * 7^19 * 11^10 * 13^9 * 17^7 * 19^6 * 23^5 * 29^4 * 31^3 * 37^3 * 41^2 * 43^2 * 47^2 * 53^2 * 59^2 * 61 ...........etc.

60 = 2^2 * 3 * 5

Since 5 is the largest prime factor of 60, it therefore follows that the exponent of 5 above, or 28, would be the largest n:

n ==28, so that: [942!  +  120!]  mod 60^28 ==0

[11111  +  22222  +  33333  +  55555  +  88888] ==211,109

211,109  x  24 ==5,066,616 -sum of all 120 permutations of 1, 2, 3, 5, 8.

What is the smallest distance?

Hello Guest!

The smallest distance is r (determined graphically with 4.179).

\(f(x) = \frac{1}{\sqrt{2}} (x^2 - 18)\\ g(x)= (r^2-x^2)^{0.5}\\ f'(x)=\frac{2x}{\sqrt 2}\\ g'(x)=0.5(r^2-x^2)^{-0,5}\cdot (-2x)\\ g'(x)=-\frac{x}{\sqrt{r^2-x^2}}\)

\(f'(x)=g'(x)\\ \sqrt 2\cdot x=-x(r^2-x^2)^{-0.5}\\ \sqrt{2}\cdot (r^2-x^2)^{0.5}=-1\)

\(-\sqrt{2}=\sqrt{r^2-x^2}\)

To be continued.

  !

3x - 5y = -11

4x + 2y = -12 

Solve with elimination:

12x - 20y = -44

12x + 6y = -36

Subtract:

-26y = -8

\(y = \frac{4}{13}\)

Substitute:

\(3x - 5(\frac{4}{13}) = -11\)

\(\rightarrow 3x - \frac{20}{13} = -11\)

\(\rightarrow x = -\frac{41}{13}\)

Solve with substitution:

2x = -8 - 5y 

\(x = -4 - \frac{5}{2}y\)

Substitute x into the second equation:

\(6(-4-\frac{5}{2}y) = 16+a-15y+4(-4-\frac{5}{2}y) +10y\)

\(\rightarrow -24-15y=16+a-15y-16-10y+10y\)

Simplify:

-24 = a

So, for the system to have infinite solutions, a = -24. 

That's incorrect.  

The largest x is 5/2.

Here's a hint: Note that the lengths of each side are the sum of 2 radii (biggest radii + middle radii; middle radii + smallest radii; biggest radii + smallest radii)

Now, use Heron's to solve for the area. 

Let the longer side of the triangle be \(x+3 \). This means that the shorter side of the triangle is length \(x\).

Now, let the total area of the triangle be \(a\). We have the equation \(4a + 9 = 89\), meaning \(a = 20\).

The area of the triangle is \(x (x+3) \div 2\), but we know that the area of each triangle is 20, so we have the equation: \({x(x+3) \over 2} = 20\), meaning \(x = 5\).

This means that the side opposite to \(\theta\) is 8, and the side adjacent to it is 5, so \(\tan \theta = {\text{opposite} \over \text{adjacent}} =\color{brown}\boxed {8 \over 5}\)

Nevermind, the answer is t = 0.65.

See https://web2.0calc.com/questions/quadratics_36

\(\tan \theta = \frac{8}{5}\)

Let the number ==N

N^5 ==1 / 3 * N^6

N ==3

3^12 / 3^9 ==27 :  1

GCD((121^2 + 233^2 + 345^2), (120^2 + 232^2 + 346^2))

Prime factorizations
187955 = 5×37591 (2 distinct prime factors)
187940 = 2^2×5×9397 (4 prime factors, 3 distinct)

GCD == 5

The gcd is 1.

Each ball has 3 choices so, it's 6^3=216.