geno3141

Picture a right triangle with angle(x) being an acute angle;

then, the side opposite to angle(x) = 2ab 

  and the side adjacent to angle(x) = a² - b².

Finding the hypotenuse:  hyp²  =  (a² - b²)² + (2ab)²

                                                  =  [ a⁴ -2a²b² + b⁴ ] + [ 4a²b² ]

                                                  =  a⁴ + 2a²b² + b⁴ 

                                                  =  (a² + b²)² 

--->     hypotenuse  =  a² + b² 

sin(x)  =  opposite / hypotenuse  =  2ab / (a² + b²) 

x/y + y/x  =  -1     --->     xy( x/y + y/x )  =  xy( -1 )

                           --->                x² + y²  =  -xy

                           --->         x² + xy + y²  =  0

Since, when factored:  x³ - y³  =  (x - y)(x² + xy + y²)  and  x² + xy + y²  =  0

--->     x³ - y³  =  (x - y)( 0 )  =  0

If  a  is inversely proportional to  b  then there is some constant  k  such that:  a  =  k / b.

Since  a = -6  when  b = 12     --->     -6  =  k / 12     --->    k  =  -6 · 12  =  -72.

For this problem, the formula is:  a  =  -72 / b.

When  b = -2     --->     a  =  -72 / -2  =  36. 

Let  G  = number of Gello pens and  K  = number of Inko pens.

Number equation:  G + K  =  17

Value equation:     5G + 7K  =  109

G + K  =  17     --->     K  =  17 - G

Substituting:  5G + 7K  =  109     --->     5G + 7(17 - G)  =  109

                                                                5G + 119 - 7G  =  109

                                                                                 -2G  =  -10

                                                                                    G  =  5

Since  G + K  =  17  and  G  =  5,  K  =  12.

There is a formula to find the n^th term of a triangular number:  t_n  =  n(n + 1)/2

So, as Guest as shown, the 50^th term is:  50(50 + 1)/2  =  1275.

The conjugate of  a + b  is  a - b  and the conjugate of  a - b  is  a + b.

To simplify this problem, multiply both the numerator and the denominator of the fraction by

the conjugate of the denominator:  sqrt(5) + sqrt(2).

[ sqrt(5) + sqrt(2) ] / [ sqrt(5) - sqrt(2) ]  · [ sqrt(5) + sqrt(2) ] / [ sqrt(5) + sqrt(2) ]

=

[ sqrt(5) + sqrt(2) ] · [ sqrt(5) + sqrt(2) ]  /  [ sqrt(5) - sqrt(2) · [ sqrt(5) + sqrt(2) ]

=

[ 5 + 2·sqrt(5)·sqrt(2) + 2 ] / [ 5 - 2 ]

=

[ 7 + 2·sqrt(10) ] / 3

Try an example:  f(5)  =  11     f(4)  =  5     f(3)  =  3

A)  Does  f(5)  =  f(4) + f(3)?

B)  Does  f(5)  =  f(4) + 2·f(3)?

C)  Does  f(5)  =  2·f(4) + f(3)?

D)  Does  f(5)  =  2·f(4) + 2·f(3)?

26! / (13! · 13! )  +  26! / ( n! · (26 - n)! )  =  27! / ( 14! · 13! )

Rewrite as:     26! / (13! · 13! )  +  26! / (n! · (26 - n)! )  =  27 · 26! / ( 14 · 13! · 13! )

Divide by 26!:     1 / (13! · 13! )  +  1 / (n! · (26 - n)! )  =  27  / ( 14 · 13! · 13! ) 

1 / ( n! · (26 - n)! )  =  27  / ( 14 · 13! · 13! )  - 1 / (13! · 13! ) 

 Factor:                 =  1 / (13! · 13! )  ·  ( 27/14 - 1 )

 Simplify:              =  1 / (13! · 13! )  ·  ( 13/14 )

                             =  13 / ( 14! · 13! )

                             =  1 / ( 14! · 12! )

Therefore,  n! · (26 - n)!  =  14! · 12!

So,  n  is either  12  or  14 ...

If you are allowed to start with the identity:  tan( x/2 )  =  ( 1 - cos(x) ) / sin(x)  then: 

tan( x/2 )  =  ( 1 - cos(x) ) / sin(x)  =  

multiply both the numerator and denominator by  ( 1 + cos(x) )

=  ( 1 - cos(x) ) / sin(x)  ·  ( 1 + cos(x) ) / ( 1 + cos(x) )

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

=  [ 1 - cos²(x) ] / [ sin(x) · ( 1 + cos(x) ) ]

=  sin²(x) / [ sin(x) · ( 1 + cos(x) ) ]

=  sin(x) / ( 1 + cos(x) )

To determine vertical asymptotes, you must first factor both the numerator and denominator and cancel

out those factors which are found in both the numerator and denominator.

y  =  (x - 1) / (x² + 6x - 7)  =  (x - 1) / [ (x - 1)·(x + 7) ]  =  1 / (x + 7)

Because there is a factor of (x + 7) remaining in the denominator, there is one vertical asymptote,

at x = -7.

The factor (x - 1) does not give a vertical asymptote, instead, it creates a "hole" in the graph at x = 1.