AJ – Amavila Ft. Eltonnick & Lizwi

AJ - Amavila ft. Eltonnick & Lizwi

AJ – Amavila Ft. Eltonnick & Lizwi MP3 DOWNLOAD

AJ – Amavila Ft. Eltonnick & Lizwi. Talented South African singer AJ dropped a new song titled Amavila Featuring Eltonnick and Lizwi.

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46 thoughts on “AJ – Amavila Ft. Eltonnick & Lizwi

  1. (1a)
    Given A={2,4,6,8,…}
    B={3,6,9,12,…}
    C={1,2,3,6}
    U= {1,2,3,4,5,6,7,8,9,10}

    A’ = {1,3,5,7,9}
    B’ = {1,2,4,5,7,8,10}
    C’ = {4,5,7,8,9,10}
    A’nB’nC’ = {5, 7}

    (1b)
    Cost of each premiere ticket = $18.50
    At bulk purchase, cost of each = $80.00/50 = $16.00

    Amount saved = $18.50 – $16.00
    =$2.50

    =================

    (2ai)
    P = (rk/Q – ms)⅔
    P^3/2 = rk/Q – ms
    rk/Q = P^3/2 + ms
    Q= rk/P^3/2 + ms

    (2aii)
    When P =3, m=15, s=0.2, k=4 and r=10
    Q = rk/p^3/2 + ms = 10(4)/(3)^3/2 + (15)(0.2)
    = 40/8.196 = 4.88(1dp)

    (2b)
    x + 2y/5 = x – 2y
    Divide both sides by y
    X/y + 2/5 = x/y – 2
    Cross multiply
    5(x/y) – 10 = x/y + 2
    5(x/y) – x/y = 2 + 10
    4x/y = 12
    X/y = 3
    X : y = 3 : 1

    =====================
    (3a)
    Diagram
    CBD = CDB (base angles an scales D)
    BCD+CBD+CDB=180° (Sum of < in a D)
    2CDB+BCD=180°
    2CDB+108°=180°
    2CDB=180°-108°=72°
    CDB=72/2=36°
    BDE=90°(Angle in semi circle)
    CDE=CDB+BDE
    =36°+90
    =126

    (3b)
    (Cosx)² – Sinx given
    (Sinx)² + Cosx
    Using Pythagoras theory thrid side of triangle
    y²= 1²+√3
    y²= 1+ 3=4
    y=√4=2
    (Cosx)² – sinx/(sinx)² + cosx

    (1/2)² – √3/2/
    (√3/2)² + 1/2 = 1/4 – √3/2 = 1-2√3/4
    3/4+1/2 = 3+2/4
    =1-2√3/4 * 4/5
    =1-2√3/5

    =====================

    (4a)
    Given: r : l = 2 : 5 (ie l = 5/2r)
    Total surface area of cone =πr² + url
    224π = π(r² + r(5/2r))
    224 = r² + 5/2r²
    224 = 7/2r²
    7r² = 448
    r² = 448/7 = 64
    r = root 64 = 8.0cm

    (4b)
    L = 5/2r = 5/2 × 8 = 20cm
    Using Pythagoras theorem
    L² = r² + h²
    h² = l² – r²
    h² = 20² – 8²
    h² = (20 + 8)(20 – 8)
    h² = 28 × 12
    h = root28×12
    h = 18.33cm

    Volume of cone = 1/3πr²h
    = 1/3 × 22 × 7 × 8² × 18.33
    =1229cm³

    =======================

    (5a)
    Total income = 32+m+25+40+28+45
    =170+m
    PR(²)=m/170+m = 0.15/1
    M=0.15(170+m)
    M=25.5+0.15m
    0.85m/0.85=25.5/0.85
    M=30

    (5b)
    Total outcome = 170 + 30 = 200

    (5c)
    PR(even numbers) = 30+40+50/200
    =115/200 = 23/40

    ==========================

    === SECTION B ====
    ================================

    (7a) Click here to view image

    Using Pythagoras theorem, l²=48² + 14²
    l²=2304 + 196
    l²=2500
    l=√2500
    l=50m
    Area of Cone(Curved) =πrl
    Area of hemisphere=2πr²
    Total area of structure =πrl + 2πr²
    =πr(l + 2r)
    =22/7 * 14 [50 + 2(14)]
    =22/7 * 14 * 78
    =3432cm²
    ~3430cm² (3 S.F)

    (b)
    let the percentage of Musa be x
    Let the percentage of sesay be y
    x + y=100 ——————-1
    (x – 5)=2(y – 5)
    x – 5=2y – 10
    x – 2y=-5 ——————-2

    Equ (1) minus equ (2)
    y – (-2y)=100 – (-5)
    3y=105
    y=105/3
    y=35
    Sesay's present age is 35years

    ====================================
    (8a)
    Let Ms Maureen's Income = Nx
    1/4x = shopping mall
    1/3x = at an open market

    Hence shopping mall and open market = 1/4x + 1/3x
    = 3x + 4x/12 = 7/12x

    Hence the remaining amount
    = X-7/12x = 12x-7x/12 =5x/12

    Then 2/5(5x/12) = mechanic workshop
    = 2x/12 = x/6
    Amount left = N225,000
    Total expenses
    = 7/12x + X/6 + 225000
    = Nx

    7x+2x+2,700,000/12 =Nx
    9x + 2,700,000 = 12x
    2,700,000 = 12x – 9x
    2,700,000/3 = 3x/3
    X = N900,000

    (ii) Amount spent on open market = 1/3X
    = 1/3 × 900,000
    = N300,000

    (8b)
    T3 = a + 2d = 4m – 2n
    T9 = a + 8d = 2m – 8n
    -6d = 4m – 2m – 2n + 8n
    -6d = 2m + 6n
    -6d/-6 = 2m+6n/-6
    d = -m/3 – n
    d = -1/3m – n

    ====================================

    (9a)
    Draw the triangle

    (9b)
    (i)Using cosine formulae
    q² = x² + y² – 2xycosQ
    q² = 9² + 5² – 2×9×5cos90°
    q² = 81 + 25 – 90 × 0
    q² = 106
    q = square root 106
    q = 10.30 = 10km/h
    Distance = 10 × 2 = 20km

    (ii)
    Using sine formula
    y/sin Y = q/sin Q
    5/sin Y = 10.30/sin 90°
    Sin Y = 5 × sin90°/10.30
    Sin Y = 5 × 1/10.30
    Sin Y = 0.4854
    Y = sin‐¹(0.4854), Y = 29.04

    Bearing of cyclist X from y
    = 90° + 19.96°
    = 109.96° = 110°

    (9c)
    Speed = 20/4, average speed = 5km/h

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