CPhill

(x^2+5x+6)^2+(px+q)(x^3+7x^2+3x)

x^4 + 10 x^3 + 37 x^2 + 60x + 36   +  px^4 + 7px^3 + 3px^2  + qx^3 + 7qx^2 + 3qx

( 1 + p)x^4  + (10 + 7p + q) x^3 + (37 + 3p + 7q)x^2 +( 60+ 3q)x + 36

If the resulting polynomial is of degree 2, then

1 + p   = 0    →  p  = -1

And

(10 + 7p + q)  = 0   

(10  - 7  + q)  = 0

3 + q  = 0

q  = -3

So

(x^2+5x+6)^2+(-1x-3)(x^3+7x^2+3x)  =  13 x^2 + 51 x + 36

So

p + q  =  ( -1 )  +  (- 3)   =  - 4

100x-200 > 50x-75    subtract 50 X from both sides, add 200 to both sides

50x  >  125    divide both sides by 50

x >   125 / 50    reduce fraction 

x >  5 / 2

6(a+3)=18 + 6a

6a + 18 =  18 + 6a

Since one side equals the other, we have infinite solutions for  "a" 

1/5(25−5a) = 4−a

5 - a  =  4 - a       add a to both sides

5  = 4

This is an absurd solution.....thus.....there are no solutions for "a"

9x =  5x+20      subtract 5x from both sides

4x  = 20            divide both sides by 4

x  = 5

I assume this is :

[6x−2] / 4 = x−2     multiply both sides by 4

6x - 2  = 4 (x - 2)

6x - 2  = 4x  - 8      add 2 to both sides, subtract 4x from both sides

 2x  = -6               divide both sides by 2

x  = -3

8/9(54x−36)+2=−3/4(−40+16x)+90x

48x - 32 + 2   =  30  -  12x  + 90x

48x - 30  =  30 + 78x           add 30 to both sides, subtract 78x from both sides

-30x  = 60          divide both sides by -30

x  = -2

A.) | 4 - 8 |   =  l  -4  l  = 4

B.) | -4 - (-8) |  =  l  -4 + 8 l    = l 4  l   = 4

2. Simplify

A.) (x+2)(x-3)    =   x(x -3)  + 2(x - 3)   =  x^2 - 3x  + 2x - 6   =   x^2  - x  -6

B.)  (x+2)^2   =   ( x + 2) ( x + 2)  =  x(x + 2)  + 2(x + 2)   =  x^2 + 2x + 2x + 4  = x^2 + 4x + 4

Let x be the number of grapes that Pierre grew in 2016.......then 2x  was the number that Alphonse grew that year

And let y be the number of grapes that Alphonse grew in 2017.....then  2y was the number that Pierre grew that year

So  we have that

2x  + y  = 49000    =     number of grapes that Alphonse grew over the two years

x +  2y  =  49000 + 7600  =  number of grapes that Pierre grew over the two years    

Simplify

2x  +  y  =  49000

x  +  2y   = 56600       

Multiply the 1st equation by  -2   and we have

-4x -  2y       =     -98000

  x +  2y       =      56600        add these equations

-3x  = -41400      divide both sides by -3

x  =  13800       

And Alphonse grew twice as many in 2016  = 27600

Arc length, S,  =  r * θ           wher r is the radius and θ  is  in radians

Area of a sector  =   (1/2) (r^2)*θ

So........

12  = r * θ      → θ  =   12 / r    (1)

And

108 =  (1/2) (r^2) * θ         subbing (1)  into this, we have that

108 = (1/2) (r^2) * (12 /r)        simplify

216  = 12 r         divide both sides by 12

18  = r  = the radius

And to find the angle subtended by the arc  is given by    θ  =   12 / r    →  θ  =   12 / 18  =  2 / 3

And this is in rads......so  to convert to degrees we have   .....     (2/3)  *  180° / pi  ≈  38.19°  

.99^x  = 0.5    take the log of both sides

log .99^x  =  log 0.5      and by a log property we can write

x *  log .99  =  log 0.5      divide both sides by log .99

x  =  log 0.5 / log .99  ≈  68.96  =   69 days

Here's another way of attacking this......

Note that  the sum of the  terms  with odd powers of r  can be represented by :

ar  + ar^3  + ar^5 + ar^7 +  ..... + ar^(2n - 1)  =  5   

So.....the common ratio between these terms is just r^2

And the sum of this series can be represented as

5  =   ar / ( 1 - r^2)   →   (1 - r^2)  =  ar / 5     (1)

And note that the sum of the  terms  with even powers of r  can be represented by :

a  +  ar^2  + ar^4 + ar^6 + ar^8 +  ..... + ar^(2n)  =  7

And the common ratio between these terms is just r^2

And the sum of this series can be represented by

7  = a / (1 - r^2)  →   (1 - r^2)  =  a / 7    (2)

Equating (1)  and (2)  we have that

ar / 5   =  a / 7              divide both sides by a

r / 5   =  1 / 7               multiply both sides by 5

r =  5 / 7

Note that 

r = 3/a  

And the sum of the series can be represented as

S = 4 / [ 1 - 3/a ]  =  4a / [ a - 3]

Note that if a = 4, then the sum  is   4 (4) / [ 4 - 3]  =  16/ 1  =  16   which is a perfect square 

The sum of the series  can be represented by :

a / ( 1 - r)  = 12

a = 12 ( 1- r )    (1)

So....note that the sum of the  terms  with odd powers of r  can be represented by :

ar  + ar^3  + ar^5 + ar^7 +  ..... + ar^(2n - 1)  =  5   (2)

And note that the sum of the  terms  with even powers of r  can be represented by :

a  +  ar^2  + ar^4 + ar^6 + ar^8 +  ..... + ar^(2n)  =  7   (3)

Multiply (2)  by r on both sides

ar^2  + ar^4  + ar^6 + ar^8 +  ..... + ar^(2n )  =  5r     (4)

Subtract   (4)  from (3)

a  = 7 - 5r    (5)

Sub (5)  into (1)

7 - 5r  = 12(1 - r)

7 - 5r  =  12 - 12r

7r = 5     →   r  =  5/7

Check

a = 7 - 5(5/7)  =  24/7

a / ( 1 - r)  =

(24/7) / ( 1- 5/7)  =

(24/7) / ( 2/7)  =

24 / 2   =

12