CPhill

Draw AC

Now, because BQ  =(1/3) BA and BR =(1/3)BC, Q and R  divide BA and BC proportionally...

Thus.....BQ/BA  =BQ/BC...and by Euclid,  QR is parallel to AC

And <BQR = < BAC    and <QBR  = <ABC

So, by AA  congruency ΔQBR ~ ΔABC

And since AB=3QB.....then the area of ΔABC  will be 3^2  = 9 times that of ΔQBR

By similar reasoning, we can show that the area of ΔADC  = 9 times that of ΔVDU

And  area of ΔABC +  area of ΔADC  = 180....so....

9 * area of ΔQBR + 9 * area of ΔVDU  = 180

9 *  [ area of ΔQBR +  area of ΔVDU]  = 180     ....so...

[ area of ΔQBR +  area of ΔVDU]  = 180/9  = 20

So......the area of hexagon AQRCUV  =  180  -  20  =  160

Like so.....

You're working too hard here, EP......LOL!!!!!!!

It appears that the "Guest" took the "long way home".....LOL!!!!!

1.015^n * 4000 = 1.04^n * 3542  rearrange as

1.015^n / 1.04^n  = 3542 / 4000

[1.015/1.04]^n  = 3542 / 4000    log both sides

log [1.015/1.04]^n  =  log [ 3542 / 4000 ]

n * log [1.015/1.04]=  log [ 3542 / 4000 ]

n = log [ 3542 / 4000 ] / log [1.015/1.04]  ≈  4.99763

2^45  = 10^15   ???

45 log 2  = 15 log10  ???

45log2  = 15  ???

log 2 =  15/45  ???

log 2  = 1/3  ???

2 = 10^(1/3)  =   ???

2^3  = 10  ????      No......not equal....2^3   = 8

MaxWong

Wow. Any other methods? My classmates do not know calculus!!

I'm sure there is a geometric solution....but....I don't know how to go about it.....

The result  of this summation / subtraction is the same as the solution for the sllghtly more advanced equation found here :

http://web2.0calc.com/questions/how-to-do-this_699

p  = k / (q + 4)^2

And we know that

2  = k / (2 + 4)^2

2  = k / (6)^2

2 = k / 36

2(36)  = k  = 72  = the proportionality constant

So

p = 72 / ( -2 + 4)^2

p = 72 / (2)^2

p  = 72 / 4

p = 18     when q  = -2

Cost price + .20* Cost price  = $20.40

Cost price ( 1 + .20)  = $20.40     divide both sides by (1 + .20)  = (1.20)

Cost price  = $20.40 / 1.20  = $ 17

The answer is 42.....or something like that