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(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

The period is the distance between any repetition of the function. In the sinusoidal curve, this is usually the distance between every two times it passes the midline.

It's easy to see that from 0-2 the function curves up, from 2-4 the function curves down, then back up from 4-6 etc. 

Therefore the period is 4.

From the start, we know the porportion of the amount of water in the beaker, to the total amount of solution in the beaker.

The amount of acid drained away to the amount of water drained away is in porportion to their original amounts.

We know \(\frac{150}{200+800}\)of the total amount is drained away, so \(200*\frac{150}{200+800}=30\).

So there is \(200-30=170 \)mL left.

Adding 150mL water gives 320mL water.

We know that the total amount of solution didn't change, so there is \(1000-320=680\)mL acid.

Similarly, now \(320*\frac{150}{320+680}=48\) mL drained away.

Subtracting this, and adding the 150 mL of water, we have 422mL of water now.

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  

\({(3+\sqrt{5})}^{2}*{(2-\sqrt{5})}^{7}\)

Simplify the left side first:

\({(3+\sqrt{5})}^{2}=14+6\sqrt{5}\)

Split some terms apart to find the right side.

\((2-\sqrt{5})^{2}=9-4\sqrt{5}\)

\({(2-\sqrt{5})}^{4}={(9-4\sqrt{5})}^{2}=161-72\sqrt{5}\)

\({(2-\sqrt{5})}^{7}={(2-\sqrt{5})}^{4}*{(2-\sqrt{5})}^{2}*{(2-\sqrt{5})}\).

\(=(161-72\sqrt{5})(9-4\sqrt{5})(2-\sqrt{5})=12238-5473\sqrt{5}\)

With a little help from calculators, this is 

\((14+6\sqrt{5})*(12238-5473\sqrt{5})=7142-3194\sqrt{5}\).

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

My bad, you are right. The second one is also correct

x^2 - 8x + 8 + c

The x coordinate of the vertex is   8/ (2*1)  = 4

So....we  need y to be 1  when x  = 4

1  = (4)^2 - 8(4) + 8 + c

1 = 16 - 32 + 8 + c

1 = -8 + c

c  =  9

Here's the graph :

Note that  if c is larger than 9, the graph is  shited above y =1....so 1 will not be in the range

Simplify as

sqrt [ -3x^2 - x + 6  ]

-3x^2 - x + 6 =  0

3x^2 + x - 6   = 0

x  =   [ -1 - sqrt 73 ] / 6   ≈ -1.59     and  x = [ 1 +sqrt 73  ] / 6 ≈ 1.26

The root will be real when x =  [  (-1- sqrt 73)/6  , (-1 +sqrt 73)/6 ]  = the domain

Second one

Let  O be the center of the circle

Let  M be the mid-point  of UV

Let d be the distance from O to M

We can form a right triangle such that

d^2 + (UV/2)^2  = r^2

d^2 + 19^2  =  r^2      (1)

Let the midpoint of YZ = N

Let the distance from O to N =  d + 4

We can  form another right triangle such that

(d + 4)^2 + (YZ/2)^2  = r^2 

(d + 4)^2 + 11^2  = r^2      (2)

Equate (1) , (2)

d^2 + 19^2   =  (d + 4)^2  + 11^2

d^2 + 361  = d^2 + 8d + 16 + 121

8d =  361  - 16 - 121

8d =  224

d =  224/8  = 28

28^2 +19^2   =r^2

1145 =  r^2

And we can form a third right triangle such that

(d + 2)^2 + (WX / 2)^2  = r^2

(28 + 2)^2  + WX^2 / 4  = 1145

30^2  + WX^ 2 /4 = 1145

900 + WX^2/4 = 1145

WX^2 / 4  = 1145 - 900

WX^2 / 4  = 245

WX^2   =  245 * 4 = 980

WX  = sqrt 980   ≈ 31.3

It's correct

Here's the graph :

x^2 + (k - 9)x + 16

We need to have the discriminant =   0 

(k - 9)^2  - 4 * 16  = 0

(k -9)^2  - 64 =  0

(k -9)^2   = 64      take both roots

k - 9  =  8                    k -9  = -8

k  =17                        k = 1

Sum of k's   =  17 + 1  =  18

Let x be the total distance

Let B  = the distance he has driven

So

(1/2)x    = 12           multiply through by 2

x  = 24

I dont think thats correct

2x^2 - 1x - 28  = -31 + jx

2x^2 - (1 + j )x + 3  = 0

We will have one root when  the discriminant =  0 

(1 + j)^2  - 4(2)(3)   =  0

j^2 + 2j + 1  - 24  =  0

j^2 + 2j - 23  = 0

j = [ -2 - sqrt [ 2^2 + 4*23 ] ] / 2

j =  [ -2 - sqrt [96] ] / 2

j = [ -2 - 4sqrt 6 ]  /2

j = - 1 - 2sqrt 6          and     j  =  -1 + 2sqrt 6

\(\frac{-21-\sqrt{201}}{10}\)

[-21 + sqrt (201) / 10   is also  a  root

product of the roots   = [  (-21)^2   - 201 ] / 100  =   240/100  =  2.4

product of roots =  v / 5

v / 5  = 2.4

v = 5 * 2.4    = 12