CPhill

Note....    log_d e  *  log_e f =   log_d  f

Also .....     log e / log d   +  log f / log d  =   log_d (ef)

Also........ [ log a / log c ] [ log c / log a ]  =   1

Also ...........( log_b c *  log_c b )  =  1

And.....    log_d e + log_d f  =  log_d (ef)

So......

xyz  =   [  logb / loga   + log c /  log a ]  [  log c / log b  + log a / log b ] [ log a / log c + log b / log c ]  =

[ ( logb / loga)( log c / log b) + (logb / loga)( log a / log b) + ( log c /  log a)  (  log c / log b) +

(log c /  log a) (log c /  log a)]  [ log a / log c + log b / log c ]

[ log_a c  + 1  +  log c /  loga *  log c / log b  +  log_b c ]  [ log_c a + log_c b  ] =

[   log_a c  + 1   +  log_a c * log_b c   +   log_b c ]   [ log_c a + log_c b  ]  =

[ ( log_a c) ( log_c a) + (1)( log_c a) + ( log_c a * log_a c) * log_b c + (log_b c)(log_c a ) + ( log_a c)( log_c b) + (1)( log_c b) + log_a c *( log_b c *  log_c b )  +  ( log_b c)(log_c b )  ]

[  1 +  log_c a  + 1* log_b c  +  log_b a  + log_a b +  log_c b +   log_a c * 1  + 1  ]  =

[ 2  + log_a b +  log_a c +  log_b c  +  log_b a +   log_c a  +  log_c b ]  =

2   +  log_a (bc)  + log_b(ca)  + log_c (ab)

2  +  x  +  y  + z           (  subtract 2 from both sides)

xyz  - 2   =  x + y + z

 If $x + y =2$ and $xy = 23$, then what is $x^2 + y^2$?

x + y  = 2    square both sides

x^2 + 2xy + y^2  =  4

x^2 + 2(23) + y^2  = 4

x^2 + 46  +  y^2   =  4     subtract 46 from both sides

x^2  + y^2  =  -44

2mn - 16m - 7n  +   ???

2m (n - 8)  - 7 (n  - 8)

(2m  - 7)  ( n - 8)

2nm  - 16m -7n +  56

2x+3y=13  ⇒   3y   = 13 - 2x      (1)
12x-3y=-2 ⇒    3y  =   2 + 12x    (2)

Set (1)  = (2)

13 - 2x  =  2 + 12x     add 2x to both sides, subtract 2 from  both sides

11   = 14x         divide both sides by 14

11/ 14  = x

Putting this into (1), we have

3y  = 13 - 2 (11/ 14)

3y  = 13  - 11/7           get a common denominator    

3y  = 91/7 - 11/7

3y  = 80/7          divide both sides by 3

y  = 80/ 21

(a)        What is the probability that both cards are aces if the first card is put back into the deck (which is then shuffled) before the second card is drawn?  

P (ace on the first draw  = 4/52    = 1/13

P (ace on the second draw)  = 4/52   = 1/13

P(ace on both draws)  = (1/13) (1/13)  ≈  .0059

(b)       What is the probability that both cards are aces if the first card is not put back into the deck (drawing without replacement)?     

P (ace on the first draw  = 4/52    = 1/13

P (ace on the second draw)  = 3/51   = 1/17

P(ace on both draws)  = (1/13) (1/17)  ≈  .0045

Very nice, hectictar.....!!!!!

Call the amount invested at 4%  = .04 =   = x

So   the amount invested at   3.5%  =  .035  =  5000 - x

And     Amount * Interest Rate  = Total Interest   .....so...

Amount at 4%    +   Amount at 3.5%  =  190          and we have that

(x) (.04)   +  ( 5000 - x) (.035)  = 190      simplify

.04x  +  175  - .035x  =   190          combine like terms     

.005x +  175   = 190                   subtract 175 from both sides

.005x  =   15                divide both sides by  .005

x  = $3000   = amount   invested at  4%

5000 - x  =    5000  - 3000  = $2000  = amount invested at 3.5% 

Note that the sum of the first 21 integers  is   21 * 22 /2  =  231...this isn't large enough

And the sum of the first  22 integers  =   22 * 23 / 2  =   253

So    253 - 241  =  12 = omitted sum.....  but  the sum of two consecutive integers must be odd

And......the sum of the first 23 integers is  23 * 24 / 2  = 276

So.......276 - 241   =  35    = omitted sum

So....the   consecutive integers omitted must be  17  and 18

So....... the smallest value of n  is   23

Thanks, X^2....I neglected to spot that 1 - 4m^2  could be factored further as  (1 -2m) (1 + 2m)  !!!!!

7m - 28m^3           

The greatest common factor   between   7 and  28   = 7

The greatest common factor between m and m^3 =   m

So.... the greatest common factor  ( GCF )  = 7m

So we will have

7m ( a  -  b)

To  find out  what  " a '  will be....divide the first term of the given expression , 7m,  by the GCF

So      7m  /   7m    = 1

To find out what  "b" will be......divide the second term of the given expression, 28m^3,  by the GCF

So       28m^3  / 7m  =      4m^2

So......the complete factorization is

7m  ( 1  - 4m^2)