CPhill

Call the number that Bertha has, N

Then Douglas has  N + 6

And Rhonda has (N + 6) + 12   =  N + 18

And we know that :

N + 18   = 2N     subtract N from  both sides

N = 18    this is Berha's number

And Douglas has 6 more = 24

cos θ = 20.8/33.6

Use the cosine inverse to find θ

arccos ( 20.8 / 33.6)   = θ   = about  51.75°

500 = 400(1.05)^x     divide both sides by 400

5/4   =  (1.05)^x

1.25  = (1.05)^x      take the log of both sides

log(1.25)  = log(1.05)^x     and we can write

log(1.25) = x * log(1.05)      divide both sides by log(1.05)

log(1.25) / log(1.05)  = x  =   about   4.5735

Nice graph, Melody......what did you use to create that???

The easiest way to see this is to imagine that something costs, lets say $100

A 22%  discount means  that the item now sells for 78%  of its original cost = $78

And.......another 15% discount means that the item now sells for 85% of the $78  = $66.30

So  the combined discount =  (100 - 66.30)%   =  33.7%

A simple method to figure this is :    1 - (1-.22)(1-.15)  = .337   = 33.7%

The interior angle is given by : (n-2) (180)/ n

And the exterior angle is given by :  360/n

And we have that

(n - 2) (180) / n  =  14* 360/ n      [multiply both sides by n  and simplify ]

180n -360  =  5040    add 360 to both sides

180n = 5400     divide both sides by 180

n = 30

OK, Solveit....although, technically.....Will beat me to it....!!!!!!

a(x - y)  = b(y - x)

ax - ay  = by - bx

ax + bx  = ay + by

x(a + b)  = y(a + b)

x / y  =  (a + b) / (a + b)    =  1

sec 2x = (sec^2 x)/ (2 - sec^2 x)

1 / cos2x  = (sec^2x) / (2 - sec^2x)  

Note, Shades....since    5/10   = 1/2    then it is also true that 10/5  = 2/1  .....thus....we can "flip" both fractions and write them as :

cos2x / 1  =  (2 - sec^2x) / (sec^2 x)  

cos2x =  2/sec^2x   -  sec^2x/ sec^2x

cos2x   = 2/ (1/cos^2x)  - 1

cos2x   = 2cos^2x  - 1        which is an identity   

Let's suppose that there could be.....call the functions

f(x)  = ax + b      and  g(x)  = cx + d

So ......   a and c must be reciprocals and so must b and d....so we have

(ax + b) ((1/a)x + (1/b)  = 

x^2 + (b/a)x + (a/b)x + 1      

And this = x^2 + 1

Which implies that

[ (b/a) + (a/b)]    = 0

Which implies that

[b^2 + a^2] / ab   = 0

Which implies that

b^2 + a^2   = 0

But this is only possible if a and b  = 0

But....if a and b = 0 

Then f(x)  =  0x +  0   = 0

And   g(x)   = (1/0)x + (1/0)

Thus :   f(x) * g(x)   = 0 times an undefined expression   ....which could never be any function!!!