✔2-4 correct points

✔5-7 correct points 

✔plotting all points

y = 29,22 + 0,89(25)
= 51,47
OR
y = 51,38       [calculator use]
ANSWER ONLY FULL MARKS

y = - 1 x + 7
        2
m_BC = m_AD  = - ¹/₂
y - 5 = - 1 (x + 1)
             2
y = - 1 x + 9
        2       2

- 1 x + 9 = 0
  2       2
x = 9
D(9; 0)

OR

m_AD   = m_BC
0 - 5 = - 1
x + 1      2
- 10 = -x - 1
x = 9
D(9; 0)

m_FD = 2 - 0
         10 - 9
= 2
m_BC  × m_FD  = 2 × - 1
                                 2
= -1
FD ⊥ BC     [m_BC × m_FD   = -1]

AD =√(-1 - 9)²  + (5 - 0)²
= √125
= 5√ 5

✔ subt. into correct formula
✔  AD = 5√ 5

FD = √(9 -10)²  + (0 - 2)²
=  √5
A of ABCD = b ×  h
= 5 √5 ×√5
= 25

✔ subst. into correct formula
✔  FD = √5
✔  subst into correct formula
✔  answer

m_AB  = m_DC
=   6 - 5    
   2 - (-1)
= 1
   3
∴ Inclination of DC = 18, 43º
Inclination of AD = 180º - tan^-1 [½]
= 153, 43º
ADC = 153, 43º -18, 43º
= 135º
∴ ABC = 135º   [opp∠s of a parm.]

✔ subst. into correct formula
✔ m_AB  = 1
                3
✔  θ_DC = 18, 43º
✔  θ_AC = 153, 43º
✔ AD C = 135º 
✔  ABC = 135º 

x_S  = 0 + 4               y_S   = 0 - 6
            2                               2
= 2                                   = -3
S(2; - 3)
Answer only full marks

✔ subst. into correct formula
✔  both coordinates

SP = √(2 - 4)²  + (- 3 + 6)²
= √13

✔ subst. into correct formula
✔  answer

✔ correct subt. centre
✔  r ² = 13.

m_rad  = - 3 + 6
              2 - 4
= - 3
     2
m_tan = 2           [radius ⊥ tan] 
           3
y + 6 = 2 (x - 4)
            3
= 2 x - 26
   3       3

✔ subst. into correct form.
✔ m_rad  = - 3
                   2
✔ m_tan = 2         
                3
✔ subst. into correct form

(0 - 2)²  + (y + 3)²   = 13
(y + 3)²   = 9
y + 3 = ±3
y_T = -6
T (0; - 6)

OR 

(0 - 2)²  + (y + 3)²   = 13
y ² + 6 y = 0
y(y + 6) = 0
y_T = -6
T (0; - 6)

OR 

Draw horizontal line y = -3 with M on OT
OM = MT         [⊥ from centre bisect the chord]

OM = 3
∴ MT = 3
∴ OT = 6
∴ T (0; - 6)

✔ x = 0
✔ (y + 3)²   = 9
✔ y + 3 = ±3
✔ y_T = -6
✔ subs.x = 0 in eqn of circle
✔ standard form
✔ factors
✔ y_T = -6
✔ S/R
✔ length of MT
✔ length of OT
✔ answer

At U , y = - 26
                   3
∴TU = 26 - 6
           3
= 8
   3
Area ΔOTP = 1/2 × 6 × 4
Area ΔPTU   1/2 × 8/3 × 4
= 9
   4

sin 2x = √15
                8 

cos2x = 2 cos² x - 1
2 cos² x = cos2x + 1
cos x = √cos2x + 1
                      2
= √ 7/8  + 1 
         2
= √15  
      4
 OR

cos2x = 2 cos² x -1
7 = 2 cos² x -1
8
15 = cos² x
16
∴ cos x = √15  
                   4

✔ diagram 
✔  identity of cos2x 
✔  cos x subject of formula 
✔ cos2x = 7
                 8
✔  answer
✔  identity 
✔  cos2x = 7
                  8
✔ cos² x = 15
                  16
✔ answer

sin(180º - θ ).sin(540º - θ ).cos(θ - 90º)
            tan(- θ ).sin² (360º - θ )      
= (sinθ)(sinθ ).(sinθ ).
   (- tanθ )(- sinθ )²
=         sinθ      
        - sinθ
         cosθ
= -cosθ

✔ sin θ
✔  sin θ
✔  sin θ
✔  (- tanθ )
✔  (- sinθ )²
✔  sinθ = tan θ
     cosθ
✔  (- cosθ )

LHS = sin 5x. cos3x - cos5x.sin 3x -1
                          tan 2x
= sin(5x - 3x) -1
        tan 2x
=        sin 2x          -1
          sin 2x
         cos 2x
= cos 2x -1
= 1- 2 sin ² x -1
= -2 sin ² x
∴ LHS = RHS

✔ sin (5x - 3x)
✔ tan 2x =  sin 2x
                  cos 2x

✔ simplification
✔ identity  
      1- 2 sin ² x

tan 2x = 0
2x = 0º  or 180º
x = 0º  or 90º

OR

tan 2x is undefined
2x = 90º or 270º
x = 45º or 135º

f
🗸 both intercepts 
🗸asymptote
🗸shape  

g
🗸intercepts
🗸min& max values 
🗸shape

🗸critical values 
🗸notation

🗸 Area rule formula 
🗸 correct subst. 
🗸answer

AC ² = AB² + BC² - 2.AB.BC.cosB
= 17² + 17² - 2.17.17.cos105º

= 727,5974081
∴AC = 26,97

🗸 cosine rule formula
🗸 correct subst into cosine rule
🗸 AC = 26,97

sinACD = sinD
    AD         AC
sinACD = sin75º
  13          26,97
sinACD = 13 × sin75º
                         26,97
= 0,4655927231
ACD = 27,75º

🗸 sine rule formula
🗸 correct subst. into sine rule
🗸 answer 

Converse opp ∠ s of a cyclic quad

OR

int.opp.∠ s of a quad suppl

🗸 reason

OR

🗸 reason

T₂   = S = 90º       [corresp.∠ s  RS ΙΙ Q]
∴ T is the midpt of SP         [ line from centre ⊥ to chord]
QP = 10 andTP = 8
QT² = (10)²  -  (8)2                     [Pyth.Theorem]
∴ QT = 6
QU = QP = 10                [radii]
∴ TU = 4

🗸      S/R
🗸      S/R
🗸      subst. into Pyth. 
🗸       QT
🗸       S/R
🗸      TU

Q₂  = S₃        [∠  s opp =sides] 
Q₂ =  180º -150º                  [sum of ∠ sin a Δ] 
                  2
= 15º

🗸 S/R
🗸 S
🗸 answer

O₁  = P₁ + A and  O₂ = P₂ + B  [ext ∠ of a Δ] 
AO = OP = OB                           [radii] 
P₁  = A  and  P₂   = B                 [∠ s opp =sides] 
∴ O₁  = 2P₁  and O₂   = 2P₂
∴O₁  + O₂  = 2P₁  + 2P₂
AOB = 2 (P₁ + P₂)
= 2APB

RUT = 90º        [∠ in a semi - circle] 
X₁   = 90º       [line from centre to midpoint] 
∴RU ΙΙ SY [corresp.∠ s =] 

OR

RUT = 90º        [∠ in a semi - circle] 
X₁   = 90º       [line from centre to midpoint]
∴ RUT + X₂ = 90º + 90º = 180º
∴RU ΙΙ SY [co-int.∠ s supp] 

🗸S         🗸 R
🗸S         🗸 R
🗸 R
🗸S        
🗸 R
🗸S        
🗸 R
🗸 R

R₂  = y                 [tan chord theorem] 
= O₁                      [alt ∠s, RU ΙΙ SY]
∴T₁ = 1 O₁     [∠ at centre = twice ∠at circumf.]
          2     
= 1 y
   2

O₄  = y = O₁        [vert opp.∠ s] 
TUV = y            [tan from same point =in length] 
∴ TOUVis a cyclicquad            [converse same segment]

OR 

VTO = 90º                      [tan ⊥ radius] 

VUO = 90º                      [tan ⊥ radius] 

∴VTO + VUˆ O = 180º

∴TOUVisa cyclicquad   [converseoppÐsof cyclicquad.supp]

🗸S         🗸 R
🗸S         🗸 R
🗸 R
🗸S         🗸 R
🗸S         🗸 R
🗸 R

BA = BC    [prop theorem; EF ΙΙ  AC]
EA    FC
= CA   [prop theorem; ED ΙΙ BC]   
   DA
∴BC = CA
  FC    DA