CPhill

cos (x/2)  =  -√ [ (1 + cosx) / 2 ]

cos (x/2)  = -√ [ (1 -1/5) / 2 ]   =  -√ [ 4/10]  =  -2 / √10  =    -2√10/10  =   -√10 / 5

See the following image :

Let E  = (2,4)    F  = (6,4)     G = (4,0)     H = (0,0)

FH  =  sqrt  ( 6^2 + 4^2)  = sqrt ( 36 + 16)  = sqrt (52)   

Midpoint of EF  =  (4,4)   = M

Midpoint of  EH = (1,2)  = N

Equation of line  through MG  is  x  =  4

Equation of line through FH is  y = (4/6)x  = (2/3)x

The x coordinate of P  = 4

The y coordinate of P  is  y = (2/3)(4)  = 8/3

So P =  (4, 8/3)

Similarly

The slope  of the line through GN  is ( 2-0) / (1 - 4)   = -2/3

And the equation of of this line is   y = (-2/3)(x - 4)  = (-2/3)x + 8/3

The find the x coordinate of Q

(-2/3)x + 8/3  = (2/3)x

8/3 = (4/3)x

x = 2

And the y coordinate of Q  is    (2/3)(2)  = 4/3

So Q  = ( 2, 4/3)

And PQ  =  sqrt  [( 4-2)^2  + (8/3 - 4/3)^2]  =  sqrt  [ 2^2 + (4/3)^2]  = sqrt [ 4 + 16/9]  =

sqrt [36 + 16] / 3  =   sqrt (52)/3

So

PQ                  sqrt(52) / 3                 1

___      =         __________   =      ____

FH                   sqrt (52)                     3

Mmmmm...I don't see how they got that  ....but.....I'm not an expert on this !!!!

See the graph  here : https://www.desmos.com/calculator/g6co3qmnlt

We can imagine these points as being in the complex plane

The line  y  =  (-2/3)x   goes through both  points  .....so...they are  separated by 180° 

Note  that if we multiply  (3 - 2i)  by i^3   we get   3i^3 - 2i^4  = -3i - 2  =  -2 - 3i   which is a 90° rotation clockwise  and puts the product in Q3

Maybe someone else on here knows why 90° is correct

We might use Calculus to solve this but a graph seems way easier

See here :  https://www.desmos.com/calculator/ttbkffmhmb

This is minimized when x  =  3

And the minimum value  =  10

OK, jenny   !!!

1 + 1/2  =  3/2

So

(3/2) / 6  =  x /18          cross-multiply

18 (3/2) =  6x

27  = 6x       divide both sides by 6

27/6  = x  =   9/2

3 -2i   lies  in  Quadrant IV

Multiplying this by i^2  gives us

(3 - 2i) * i^2   =   3i^2 - 2i^3   =  - 3 - 2 (-i)  =   -3 + 2i

This  is a 180°  rotation counter-clockwise

The  new  product  lies in Quadrant II

t₄  = -5

t₉ = 10

We have this

t₉ =  t₄  + d ( 9- 4)

10  = -5  + d (5)      add -5 to both sides

15  =  d (5)            divide both sides by 5

3  =  d

To find t₁   we have

t₄  = t₁  + d ( 4 - 1)

-5  = t₁ + 3 ( 3)

-5 = t₁ +  9          subtract  9 from both sides

-14  =  t₁

a.     

210  =  14 + 7n    subtract 14 from both sides

196  = 7n             divide both sides by 7

28  = n

However....since "between" is stated, we need to subtract 1 from this  =  27  multiples of 7 between 14 and 210

b.  similarly

348  = -51 + 7n        add 51  to both sides

399 = 7n          divide both sides by 7

57 = n

Suntract 1 from this

56  multiples of 7 between - 51  and 348

I think you mean "Integers"

How about

-4  and  3    ???

(-4) (3)  = -12

-4 + 3  =  -1