CPhill

Thanks, geno....here's one more approach

56x^2  +  27  =  89x  - 8   rearrange as

56x^2  -  89x  +  35  = 0

The sum of the solutions  =  89/56

So  letting r be the other solution, this means that

5/7  +  r   =  89/56

r  =  89/56  -  5/7

r  =  89/56  -  40/56

r  =  49 / 56   =   7 / 8  

 x^3+2x^2-11x-12    =  0           we can write this as

x^3  + x^2  + x^2  - 11x  -  12 =  0     factor

x^2(x + 1)  +  ( x - 12) ( x + 1)  =  0

(x  + 1)  ( x^2  +  x  -   12)  = 0

(x + 1)  ( x + 4) ( x - 3)  =  0

Setting each factor to  0   and solving for x  gives the roots

x  = -4,  x  = -1    and  x  = 3

And their sum is    -2

We have that

255/512  =  (1/4) [  1 - (1/2)^n  ]  /  [  1 - 1/2]    simplify

255/512 =  (1/2) [ 1 - (1/2)^n ]  

255 / 256 =  1 - (1/2)^n

(1/2)^n  =   1 -  255/256

(1/2)^n  =  1/256

(1/2)^n  =  (1/2)^8

So....n  = 8

(r + 3)  / ( r - 2)   =  (r - 1)  / ( r + 1)      cross-multiply

(r + 3) ( r + 1)  =  (r - 2) (r - 1)     simplify

r^2 + 4r  +  3   =   r^2 - 3r  +  2         subtract r^2 from both sides

4r + 3   =    -3r  +  2           add 3r to both sides, subtract 3 from both sides

7r  =  - 1       divide both sides by 7

r  =  - 1/7

I actually think there is a more efficient way to do this........maybe someone else knows how

AB / CB   =  1/2  ⇒   2AB  = BC

AD/AB = 3   ⇒ 3AB  =  AD

BD/ AB  =  2  ⇒  2AB  = BD

So  this implies that  BD  =  BC

So....B  must  be the midpoint of  C and D  =   (2,2)

2AB  =  BC   ⇒  AB  =  BC/2

3AB  =  AD  ⇒  AB  =  AD/3

So logic dictates that A  must be the midpoint of BC  =  (1,3)

So....AB  =   sqrt [ (1^2 + 1^2) ]  =  sqrt (2)

And  AD  =  sqrt (3^2  + 3^2)  = sqrt (18)   =  3sqrt (2)  =  3AB

Proof 

AB   =  sqrt (2)

CB  =  sqrt [ 2^2  + (4 -2)^2]  =  sqrt (4 + 4)  =    sqrt (8)  = 2sqrt (2)

So  AB /CB  =  sqrt (2)/ 2sqrt (2)  =  1/2

And

AD  =  3sqrt(2)

BA  =  sqrt(2)

So  DA /BA  =   3sqrt(2)/ sqrt (3)  = 3

And

DB  = sqrt [ ( 4 - 2)^2  + 2^2 ]  =  sqrt (2^2 + 2^2 )  =  sqrt (8)  = 2sqrt(2)

BA  =  sqrt(2)

So DB / BA  =   2sqrt(2)/ sqrt (2)  =  2 

So  A  =  (1 ,3)

And   2x + y   =   2(1) + 3     =      5

We have that

a*4^2 + 5  = a*6^2  + 7   simplify

16a + 5  =  36a + 7  rearrange as

-2  =  20a   divide both sides by  20

-2/20  =  a   =  - 1/10

Not totally sure about this...but.....here's my best attempt...

abs [ 2n^2  + 23n   + 11 ]   factor

abs(2n  + 1) * abs(n + 11)

Note that this will  be a possible prime  if either factor  =  ±1 

But 2n + 1  will = 1  only when  n  = 0.....and the other factor will = abs ( 11 )  = 11

So....when  n = 0, the result will be prime, i.e, 11

And 2n + 1  will equal  - 1  when  n  = -1.....and the other factor will =  abs(-1  + 11) = 10......but this isn't prime

And n + 11  will equal  1  when n  =  -10

And the other factor will  be 2(-10) + 1  =  -19   which is prime for abs (2n + 1) = abs (2*-10 + 1) =

abs(-19)  =  19

And  n + 11  will =  - 1  when n  =  -12  ....so abs (-12 + 11)  = abs(-1)  =  1

And the other factor will be  abs (2(-12) + 1)  =  abs (-23)  =  23

So.....this will be prime  when n   = -12

So.....the  integers producing prime results  for  abs [ 2n^2  + 23n   + 11 ]  are 

n = 0 , n  = -10 and n = -12

EDITED ANSWER.....still don't know if it's correct, or not....!!!!!!

Hours * Rate per hour  = Total Amt

So

(t + 1) (3t - 3) =  (3t  - 5)(t + 2)  + 2    simplify

3t^2 + 3t  - 3t - 3  =  3t^2  - 5t + 6t - 10  + 2

-3   =  1t - 8         add 8  to both sides

5  =  1t

5  = t  

Check

(5 + 1)(3*5 - 3)  =  (3(5) - 5 ) ( 5 + 2)  + 2   ???

(6)(12)  = (10)(7)  + 2

72   =  70  + 2

So.....t  = 5  is correct

25.  We have this equation

y  =  497 (1/2)^x/38    where  y is the amount remaining after x  days have elapsed

[ This is the first answer ]

After 4 days have elapsed, the amt remaining is

497(1/2)^4/38  ≈  462.06  kg

26.    log ₄ 1024

We can use the change-of-base theorem, here

The answer is given by :

log 1024  /  log 4   =   

5