geno3141

x = r·cos(theta)

y = r·sin(theta)

                                                               x² + (y + 5)²  =  25

                         ( r·cos(theta) )² + ( r·sin(theta) + 5 )²  =  25      (substitue)

  r²·cos²(theta) + r²·sin²(theta) + 10·r·sin(theta) + 25  =  25      (multiply out)

          r²·cos²(theta) + r²·sin²(theta) + 10·r·sin(theta)  =  0         (subtract 25)

          r²·[ cos²(theta) + sin²(theta) ] + 10·r·sin(theta)  =  0         (factor)

                                               r²·[ 1 ] + 10·r·sin(theta)  =  0         (replace sin² + cos² with 1)

                                                      r²·+ 10·r·sin(theta)  =  0

                                                    r·[ r + 10·sin(theta) ]  =  0         (factor)

Either  r = 0     or     r + 10·sin(theta)  =  0     

                                             --->     r  =  -10·sin(theta)

tan(2x)  =  [ 2tan(x) ] / [ 1 - tan²(x) ]

Problem:  tan(2x)  =  sin(x)

--->   [ 2·tan(x) ] / [ 1 - tan²(x) ]  =  sin(x)

                                    2·tan(x)  =  sin(x)· [ 1 - tan²(x) ]

                        2·sin(x) / cos(x)  =  sin(x)· [ 1 - tan²(x) ]

                                  2 / cos(x)  =  1 - tan2(x)       ( divide both sides by sin(x); now you must check for sin(x) = 0 )

                                  2 / cos(x)  =  1 - sin²(x) / cos²(x)          ( replace tan with sin/cos )

                                    2·cos(x)  =  cos²(x) - sin²(x)              ( multiply both sides by cos2(x ))

                                    2·cos(x)  =  cos²(x) - ( 1 - cos²(x) )    ( sin²(x)  =  1 - cos²(x) )

                                    2·cos(x)  =  cos²(x) - 1 + cos²(x)

                                               0  =  2·cos²(x) - 2cos(x) - 1 

Quadratic formula:  cos(x)  =  [ 2 + sqrt(12) ] / 4     or  cos(x)  =  [ 2 + sqrt(12) ] / 4

So:  possible answers:  x = sin^-1(0)     x  =  cos^-1( [ 2 + sqrt(12) ] / 4 )     x  =  cos^-1( [ 2 - sqrt(12) ] / 4 ) 

You'll need to check these to see which one or ones are correct solutions.

R_n  =  (2ⁿ - 1)  written in base 2.

p( R_n )  =  (2^{2n + 1} - 1)  written in base 2.

I'm assuming that you want the radius of the inscribed circle.

The only way that I can see to find the radius of the inscribed circle is to use coordinate geometry and a very nasty formula; so, if anyone else has a nicer solution, please post it!

A 3-4-5 triangle is a right triangle and the center of the inscribed circle is the point of intersection of the angle bisectors.

Let A = (0,3), B = (4,0), and C = (0,0).

The angle bisector of angle(C) is the line whose equation is y = x.

To find the angle bisector of angle(B), we can use this formula (where m = slope of the bisector,

                m₁ = slope of one of the sides, m₂ = slope of the other side):

      m =  [ m₁·m₂ + sqrt( 1 + m₁² )·sqrt(1 + m₂² ) - 1 ] / [ m₁ + m₂ ]

The slope of BC = 0; the slope of AB = (0 - 3) / (4 - 0) = -3/4

     m =  [ (0)·(-3/4) + sqrt( 1 + (0)² )·sqrt(1 + (-3/4)² ) - 1 ] / [ (0) + (-3/4) ]

         =  [ 0 + sqrt( 1 )·sqrt( 25/16 ) - 1 ] / [ -3/4 ]

         =  -1/3             <---  this is the slope of the angle bisector at vertex B

The equation of the angle bisector at B is:  y - 0  =  (-1/3)(x - 4)     --->     y  =  (-1/3)x + 4/3

The two lines:  y = x   and   y  =  (-1/3)x + 4/3   intersect at th point  (1,1) 

because:                              x  =  (-1/3)x + 4/3

                                     (4/3)x  =  4/3

                                             x  =  1

The distance from the point (1,1) to the x-axis is 1; as is the distance to the y-axis and also to line AB.

Thus, the radius of the incircle is 1.

If the coefficient of the first term is 1: to find the last term, divide the middle term by 2, and square the result.

22 / 2  =  11     --->     11²  =  121

x² + 22x + 121     --->     (x + 11)²

The person persons gets a bike; it doesn't matter which one.

For the second person to get a bike in the same row, there are 5 remaining in that row:  5/17

For the third person to get a bike in the same row, there are 4 reamining in that row:  4/16

Total probability:  1 x 5/17 x 4/16

x² + y²  =  4(x + y)     is an equation of a circle.

x² + y²  =  4(x + y)      --->                                   x² + y²  =  4x + 4y

                                             (x² - 4x      ) + (y² - 4y      )  =  0

Complete the squares:        (x² - 4x + 4) + (y² - 4y + 4)  =  8

Factor:                                                 (x - 2)² + (y - 2)²  =  8

This circle has its center at  (2, 2)  and has a radius of  sqrt(8).

The largest x-value occurs at the right end of the circle  --  the x-value of the center plus the radius  -- 2 + sqrt(8)

u₀  =  0

u_n  =  1 / ( 2 + u_n-1 )

We'll have to work our way up from u₀ to u₅.

u₁  =  1 / ( 2 + u₀ )     --->     u₁  =  1 / ( 2 + 0 )  =  1/2

u₂  =  1 / ( 2 + u₁ )     --->     u₂  =  1 / ( 2 + 1/2 )  =  2/5

u₃  =  1 / ( 2 + u₂ )     --->     u₃  =  1 / ( 2 + 2/5 )  =  5/12

u₄  =  1 / ( 2 + u₃ )     --->     u₄  =  1 / ( 2 + 5/12 )  =  12/29

u₅  =  1 / ( 2 + u₄ )     --->     u5  =  1 / ( 2 + 12/29 )  =  ............. 

Let  T  represent the time going from A to B.

The distance =  rate x time   --->   distance  =  40·T

Since the total time was 5.5 hrs, the time returning from B to A was  5.5 - T  hours.

The distance returning  --->   distance  =  70·(5.5 - T)

Since the distance going is the distance returning:  40·T  =  70·(5.5 - T)

Solving:                                                                      40T  =  385 - 70T

                                                                                 110T  =  385

                                                                                       T  =  3.5 hours

The distance going     =  40·3.5  =  140 km

The distance returning  =  70·2  =  140 km

You are adding two combinations:  _mC_k-1 + _mC_k

_{1)  m}C_k  =  m! / [ k! · (m - k)! ]

_{2)  m}C_k-1  =  m! / [ (k - 1)! · ( m - (k - 1) )! ]  =   m! / [ (k - 1)! · ( m - k +1 )! ] 

To add them, they must have common denominators.

Formula 1) needs the term (m - k + 1) in its denominator because (m - k + 1)! = (m - k + 1) ·(m - k)!

Formula 2) needs the term k in its denominator because k! = k ·(k - 1)!

So:  1)  m! / [ k! · (m - k)! ] · [ (m - k + 1) / (m - k + 1) ] =  [ m!·(m - k + 1) ] / [ k! · (m - k + 1 )! ]

       2)  m! / [ (k - 1)! · ( m - k +1 )! ]  · [ k / k ]  =  [ m! · k ] / [ k! · (m - k + 1)! ]

Adding the numerators of these two parts:  [ m!·(m - k + 1) ] + [ m! · k ]  =  m! · [ (m - k + 1) + k ]

                       =  m! · (m + 1)

The full fraction:  [ m! · (m + 1) ] / [ k! · (m - k + 1)! ]  =  (m + 1)! / [ k! · (m - k + 1)! ] 

Probably, to make sense of this, try some particular problems, such as ₈C₃ + ₈C₂