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  • Neco 2017 Mathematics Obj And Theory Answers – May/June Expo@ Blackloaded
  • Neco 2017 Mathematics Obj And Theory Answers – May/June Expo @ blackloaded

    That for the Subscribers

    You always enjoy our Ans early morning

    VERIFIED
    Verified-Maths-Obj
    1CDAAEDEECE
    11EEBABAABCC
    21BBCEDABECD
    31BCCCDDCADE
    41BDCBDCCECD
    51BCDCBEDCDB

    1)
    tabulate
    Number-
    5932
    *6141
    num
    3679
    *3113
    den
    Log
    3.7732
    3.7882
    7.5612
    3.5657
    3.4931
    7.0588
    anti log
    7.5612
    7.0588/0.5024/3=0.1675
    anti log
    =1471*10
    CLICK HERE FOR THE IMAGE
    2)
    a = 3, b = 20, C = -7
    Sum is
    (α/1 + 1/β + β/1 + 1/ α) Note α means Alpha
    α^2 β + β^2 + α + β / α β
    = -7/3 (-20/3) + (-20/3) = 140/9 – 20/3 / -7/3 (Note / means
    Divided)
    = 80/9 x -3/7 = -80/21
    Product:
    -7/3 + 2 + 1/-7/3 = -7/3 + 2/1 – 3/7
    = -49+42-9/21
    = -16/21
    x^2 – (sum)x + product= 0
    x^2 – 80/21x + (-16/21) = 0
    Final Answer : 21x^2 – 80x – 16 = 0
    CLICK HERE FOR THE IMAGE
    3a)
    P=100I/RT=100*30,000/4*3
    P=N250,000
    3b)
    #7000 T0 FRANCS
    8FRANS= #1
    ?FRANCS =7000
    CROSS MULTIPLY
    =#56,000 FRANCS
    SPENDING 49,400 FRANCS
    56,000-49,400 = 6600
    CONVERTING 6,600 FRANCS TO # AT 10 FRANCS TO 1#
    10 FRANCS—– 1#
    6600 FRANCS—?#
    =6,600/10 X1
    4i)
    difference in latitude=36degree + 36 degree =72degree
    distance travelled=@/360*2pir
    =72/360*2*22/77*64
    =20275200/2520
    =8045.71
    =8050km(35.f)
    4ii)
    given that speed =800km/hr
    distance travelled =804571km
    time=distance travelled/speed
    =8045.71/800
    =10.057
    =10hrs(to the nearest hour)
    CLICK HERE FOR THE IMAGE
    5)
    tabulate
    Mark- 21-25,26-30,31-35,36-40,41-45,46-50
    F- 6,8,12,4,6,4=40
    X- 23,28,33,38,43,48
    FX- 138,224,396,152,258,192=1360
    Class Boundaries- 20.5-20.5, 25.5-30.5,35.5-40.50,40.5-45.5
    5i)
    MEAN
    (X)=EFX/EF=1360/40=34
    MODE
    L1+[Fm-Fa/2fm-fa-fb]c
    =30.5+[12-8/2(12)-8-4]5
    =30.5+[4/12]5
    =30.5+1.67
    =32.17
    CLICK HERE FOR THE IMAGE
    SECTION B
    ANSWER 5 QUESTIONS
    6i)
    7+8+x*x=47
    15+2x=47
    2x=47-15
    2x/2 =32/2
    x=16
    6ii)
    Tram only
    7+8+3+y =30
    18+y =30
    y=30-18=12
    Tram only =12
    6iii)
    at least two
    7+16+3+8
    =34//
    6iv)
    16+7+12+16+8+3+9=95
    62+9=95
    9=95-62
    9=33// or car only =33
    CLICK HERE FOR THE IMAGE
    7a)
    a+2d=11 —-eq (1)
    -(a+8d)=29 —-eq (2)
    -6d/-6d = -18/-6
    d=3
    sub for d=3 in —-eq (1)
    a+2(3)=11
    a+6=11
    a=11-6=5
    7b)
    L = ar^n-1
    729/3 = 3/3 x 3^n-1
    243 = 3^n-1
    3^5 = 3^-1
    5 = n-1
    N = 5+1
    = 6
    Sn = a(r^-1)/r-1 = 3(3^6 – 1)/3-1
    Sn = 3( 729 -1)/2
    S6 = 3 x 728/2
    S6 = 1092
    7c)
    3x^3/3 + x | 2,1
    (2^3 + 2) – (1^3+1)
    (8+2) – (1+1)
    = 8
    CLICK HERE FOR THE IMAGE
    8ai)
    y=(4x+9)^3
    let u =4x+9, y=U^3
    du/dx=4, dy/du=3U^2
    dy/dx=dy/du * du/dx = 4*3u^2=12u^2
    recall= U=4x+9
    dy/dx=12(4x+9)^2
    8aii)
    y=(3x-2)^3 (x^2 + 4)^2
    u=(3x-2)^3 V=(x^2+4)^2
    du/dx=3.3(3x-2)^3-1 , dv/dx=2x.2(x^2+4)^2-1
    =4x(x^2+4)
    in other words
    dy/dx=Vdu/dx + udv/dx
    =(x^2+4)^2 9 (3x-2)^2+(3x-2)^3 4x(3x-2)
    =(x^2+4) (3x-2)^2 (9x^2+36+12x^2-8x)
    =(x^2+4) (3x-2)^2 (21x^2-8x+36)
    8b)
    y-y1=m(x-x1)
    y-6=2(x-2)
    y-6=2x-4
    y=2x-4+6=2x+2
    CLICK HERE FOR THE IMAGE
    9)
    CLICK HERE FOR THE IMAGE
    10)
    CLICK HERE FOR THE IMAGE
    11a)
    Tabulate
    X- |-1| |0| |1| |2| |3| |4|
    6 |6| |6| |6| |6| |6| |6|
    +X |-1| |0| |1| |2| |3| |4|
    -X^2 |-1| |0| |-1| |-4| |-9| |16|
    Y- |4| |6| |6| |4| |0| |-6|
    11b)
    CLICK HERE FOR THE GRAPH
    1ci)
    gradient at x/x
    dy/dx =B-A/B-C
    =7-4/2-1=3/1=3
    11cii)
    roots of equation from the graph
    x=-2.2 and x=3
    11ciii)
    minimum value of y
    =6.4.
    12a)
    from economics
    3x+2y-12=90
    3x+2y=90+12
    3x+2y=102 —–(1)
    sum of angle in a circle=360degree
    5x+y-9+3x+2y-12+3y
    +4x+21=360
    5x+3x+4x+y+2y+3y-9-12+21=360
    12x+6y=360 —–(2)
    solving equation 1and2
    simultaneously
    13x+2y=102
    12x+6y=360
    using elimination method
    3x+2y=102*4
    12x+8y=408
    12x+6y=360
    2y/2=48/2
    y=24
    substitute y=24 into
    equation (1) for x
    3x + 2y =102
    3x+2*24=102
    3x+48=102
    3x=102-48
    3x/3=54/3
    x=18
    x+y=18+24
    x+y=42
    12b)
    candidate that registerd for literature
    =4x+21/360 * 1200
    =4*18+21/360 * 1200
    =72+21/360 * 1200
    93/360*1200 =111600/360
    =310
    candidate that registered for government
    =3y/360 * 1200
    =3*24/360 * 1200
    =72/360 * 1200 =86400/360
    =240
    310 – 240 = 70
    the candidate that registered for literature are 70, more than candidate that registered for government

             COMPLETED
    ================

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