CPhill

( g  o  f)  

This is equal to   g [ f ( a) ] 

f (a) =  f ( .25)  =   sqrt (.25)  =  .5  =  1/2 =   b

So

g [ b ] = g [ 1/2 ]  =  1  / [ 1/2]  =   2

 Note that  triangle XOY  is similar to triangle  YPZ .....so....

 XO / OY  = YP /  PZ  →  XO * PZ  =  YO * YP

Also, triangle XOW  is similar to triangle WPZ ...so.... 

XO / OW  = WP / PZ  →  XO * PZ  = OW * WP

Which implies that

YO * YP  =  OW * WP

But YO  = YP + PO     and WP = OW + PO

Therefore....by substitution.....

[ YP + PO ] YP  =  OW [ OW + PO ]   simplify

YP^2 - OW^2  =  PO [ OW - YP ]     subtract the right side from both sides

[ YP + OW] [ YP - OW ] - PO [ OW -YP]  = 0      factor out a negative

[ YP + OW ] [ YP - OW] + PO [ YP - OW] = 0    factor out  [ YP - OW]  

[ YP - OW] [ YP + OW + PO ]  = 0

So, by the zero factor property,  either

[ YP + OW + PO ]  = 0

But this is impossible since  YP, OW and PO  are all > 0

Or

[ YP - OW]  = 0   

Which implies that  YP = OW    →  PY  = OW

"m" will be the slope.......

We can find it thusly.....

[ subtract the y coordinates in some order ]  /  [ subtract the x coordinates in the same order ]   

So we have

[ 5 - 2  ]  / [ 1 - 6]  =  [ 3] / [ -5]  =     -3/5       and this is the slope of AB

"b"  will be the y intercept........we need to get to this form  →  y  = mx + b

Using either one of the points and the slope, we can write

y  = -(3/5) ( x - 1) + 5    simplify

y = (-3/5)x +  3/5  + 5         →     "5"  can  be writen as  25/5 .....

So.....we have

y = (-3/5)x  +  3/5 + 25/5

y = (-3/5)x   +  28/5

So......  (-3/5)  = m    and   28/5  = b

Hope that helps  !!

No prob.....!!!!

log (1.5x + 14) + log (2)  =  0.5 log (x^4)

Other than a slight mistake, you were on the right track..... let's go from your last step

log (3x + 28)   =  0.5 * 4 log (x)

log (3x + 28 )  = 2 * log (x)

Let's write the "2"  back as an exponent

log (3x + 28)  =  log (x^2)

Since the we have a single log on each side....we can forget those and solve this :

3x + 28   = x^2    rearranging, we have that

x^2 - 3x  - 28  =0      this is factorable as

(x -7) ( x+4)  = 0

Setting both factors to 0 and solving for x, we have that x  = 7  or  x  = -4

Note that both solutions will solve the simplified form of the problem ,i.e.,  log (3x + 28)  =  log (x^2)    !!!!

Thanks, rarinstraw !!!!

Here's another way to compare fractions without having to worry about common denominators

Write  - (10 + 3/4)   as the improper fraction   -43/4

Now  let's compare these fractions

-43 / 4   ???   - 41 / 5

Multiply both sides by 5

(5) (-43)  / 4  ???  -41

Multiply both sides by 4

(5) (-43)   ???   (4) (-41)    

Simplify both sides

-215  ???  -164

Note that the left side is < the right side

This means that the original fraction on the left side (-43/4)   is smaller than the original fraction on the right side, (-41/5)

If S=P(1+rt), make t the subject 

Divide both sides by P

S / P  =  1 + rt       

Subtract 1 from both sides

S / P  - 1    =  rt     

Divide both sides by  r

S / (Pr)  - 1 / r   = t

(1/2)x+2/3=5/6     multiply through by the common denominator of 2,3 and 6  = 6    to clear the fractions

3x  + 4   =  5       subtract 4 from each side

3x  = 1              divide both sides by 3

x  = 1/3

As an alternative answer  to this.....

Note that each  element  will be summed  5  times

So we have  the sum  (1 + 2 + 3 + 4 + 5 + 6)   summed 5 times

And the sum of  the frist  6 elements  =  (6) (7)  / 2  =  21 

So......the total sum will be  

(5) * [ (6) * ( 7)  / 2 ]   =    (N -1) (N) ( N + 1)  / 2     which is the correct "formula"

As an addendum to my answer.....I believe  that frasinscotland's  answer will be true if "Point A"  is (5,5)

Point P will be located at  ( 50/5, 0)  = (10,0)

So AP  will have a length of   sqrt   [ (10 - 5)^2  + ( 5 - 0)^2 ]  =  sqrt ( 25 + 25)  = sqrt (50)

And this will be  equal  to  OA