CPhill

x + y + 8z  =  10

x + 9y -4z  =  7

kx + 4y +2z = 3

Multiply the second equation  by 2

2x + 18y  - 8z  = 14

add to the first equation

3x + 19y  = 24        (1)

Multiply the last equation by 2

2kx + 8y + 4z  = 6

add to the second equation

(2k + 1)x + 17y = 13    (2)

Multiply (1)    by 17    and (2) by 19

51x + 323y  = 408

(38k + 19)x + 323y =  247

This system will  have no solution  when

38k + 19 = 51

38k  = 32

k = 32 / 38  =  16  /19

(x + 4)^2  = x^2 + (x + 2)^2 -  4x^3

x^2 + 8x + 16 =  x^2 + x^2 + 4x + 4  - 4x^3

4x^3  - x^2  + 4x + 12 = 0

The Rational Zeroes Theorem won't help us here, so a graph seems to be the way to go

Only one real solution  x ≈  -1.15

Second one

Let N be the center of the circle   and its radius =  r

Let M be the tangent point of the circle to OA

OM  =  3

ON = 5 - r

MN =  r

We can  form  a right triangle  ONM such that

OM^2  + MN^2  =  ON^2

3^2 + r^2  =  (5-r)^2

3^2 + r^2  = r^2  - 10r + 25

10r  =  25 - 9

10r  = 16

r = 16 / 10  =  8/5 =   1.6

First one

Law of Cosines

(x + 4)^2 = x^2 + (x + 2)^2  - 2(x * (x + 2))cos 120°          (cos 120°    = -1/2)

x^2 + 8x + 16  =  x^2 + x^2 + 4x + 4 + x (x + 2)

x^2 + 8x + 16  =  2x^2 + 4x + 4 + x^2 + 2x

x^2 + 8x + 16 =  3x^2 + 6x + 4

2x^2 - 2x - 12 =   0

x^2 - x  - 6 =  0

(x - 3) ( x + 2)  = 0

x  = 3

Simplify as

3x^2  - 11x - 7 =  0

Product of  roots    ab  =  -7/3

Sum of roots  = a + b  = 11/3        

1/ (a + 7)  +  1/ (b + 7)  =     [ a + b + 14 ] /  [ ab + 7 (a+b) + 49 ]   =

[ 11/3 + 14 ]  / [ -7/3 + 7(11/3) + 49 ] =

[ 53/3] / [ 70/3 + 49]  =

[53/11] / [  217/3]  =

[53/3][ 3/217]   =

53 / 217

\(\frac{xy}{x + y} = 0, \quad \frac{xz}{x + z} = 0, \quad \frac{yz}{y + z} = 1 \)

x  = 0 

yz  = y + z

zy - z  =  y

z ( y - 1)   = y

z =   y / (y - 1)

y             z

2             2

z =  2

Simplify as

x^2 -14x + y^2 -6y =  - 55   complete the square on x, y

x^2  -14x + 49  + y^2 -6y + 9 =  -55 + 49 + 9

(x - 7)^2  + (y - 3)^2  =  3

y = sqrt ( 3 - (x-7)^2 ) + 3

This is a circle  centered at (7,3)  with a rdius of sqrt 3

This is probably best approximated with a graph

The area we want is the area of triangle ABC   plus the area of  major sector BC

(1/2) (sqrt 3)^2 sin (131.81°)  + pi * (sqrt 3)^2 * ( 360 - 131.81) /  360    ≈ 7.09

Let O be the midpoint of  MB......this is also the  center of the the semi-circle on the right

MO = AB / 4 =  10/4 =  2.5

MC = 5 - r      where  5 is the radius of the large semi-circle and r is the radius of the circle

CO = 2.5 + r

We have a right triangle such that

MC^2 + MO^2 = CO^2

(5 - r)^2   + 2.5^2 = (2.5 + r)^2

r^2 -10r + 25 +  6.25  + r^2 + 5r + 6.25

-10r + 25  = 5r

15r  = 25

r = 25/15  =  5/3  

AO = 1  = CO

AD  =   AO + CO  - CD  =  1 + 1  - 1/5  =   2 -1/5 = 9/5

DC = 1/5

AC perpendicular to BF  ....so  chord BF is bisected by AC

So.....BD = DF

Intersecting Chord Theorem

AD * DC =  BD * DF

AD * DC  = BD * BD

AD * DC = BD^2

(9/5) (1/5) = BD^2

9/25 = BD^2

BD = 3/5

DE = (1/2)(BD)   = (1/2)(3/5)  = 3/10

Triangle AED is right

Pythagorean Theorem

AE^2 = AD^2 + DE^2

AE^2  = (9/5)^2 + (3/10)^2

AE^2  = 81/25 + 9/100

AE^2  =  [324 + 9 ] / 100

AE^2  = 333/100

AE = sqrt (333) / 10 ≈ 1.82

Let the time to complete the job with 5 workers = T days

So......the total man-days for the job  is  5T

If one additional worker is  hired, the man-days for the job is 6 (T - 8)

So....since the man-days will be the same for the job.....

5T  = 6(T - 8)

5T = 6T - 48

48 = T 

So   the number of man-days to complete the job is   5(48)  = 240

So.....to complete the job 28 days earlier

(number of  workers needed) * ( 48 - 28)  =  240

(number of workers needed)  * (20)   = 240

number of workers needed  = 240 / 20     =    12

So.....(12  - 5) =  7  additional workers should be hired to  complete the job 28 days earlier