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Expo on Solved Past Question for MTH 111

2. Resolve (3X-10) / (X²-7X+10)


(3X-10) / (X²-7X+10) =(3X-10) / (X²-5x-2x+10)
(3X-10) /[ x(x-5)-2(x-5) ] = (3X-10)/[X-2][X-5](3X-10)/[X-2][X-5] = A/(X-2) + B(X-5)
(3X-10)/[X-2][X-5] =[ A(X-5)+B(X-2) ]/[X-2][X-5][X-2][X-5] will cancel out
(3X-10) = A(X-5)+B(X-2)
If x-5 = 0
When x=5
(3[5]-10) = A(5-5)+B(5-2)
15-10=A(0) + B(3)
5=3B
B = 5/3
When x =2
(3[2]-10) = A(2-5)+B(2-2)
6-10=-3A
-4=-3A
A = 4/3
Therefore,
[ A(X-5)+B(X-2) ]/[X-2][X-5] = [ 4/3(X-5)+ 5/3(X-2) ]/[X-2][X-5]Or [ 4(X-5)+ 5(X-2) ]/3[X-2][X-5]

B. (a) In a class of 40 students, 32 are good in mathematics, 4 are good in physics and 4 do not take either mathematics or physics. How many are good in mathematics and physics as well.
Solution


40 = 32-x +x +24-x
40=56-x
X=56-40
X = 16 ( are good in both maths and physics

  1. Without using table find the value of Sin 75
    Solution

    Sin 75 = sin(45⁰+30⁰)
    Sin45Cos30 + Sin30Cos45
    1√2 × √3/2 + 1/2 × 1/√2
    √3/2√2 + 1/2√2
    (√3+1)/2√2 (final answer NO!)
    Using rationalization method,
    (√3+1)/2√2) ×( 2√2×2√2)
    [√3×2√2 + 1×2√2]/[2√2 × 2√2][2√6+2√2]/ 8
    [√6+√2]/4 (final answer)
    a. If tanA = ¾ (0<a<90) and SinB = 5/13 (90<b<180). Evaluate (I) CosA + secB. (II) TanB
    Solution

    TanA = ¾
    TanA = 0.75
    A = tan–¹(0.75)
    A = 36.86⁰
    When SinB = 5/13
    Sin B = 0.385
    B = Sin–¹(0.385)
    180-22.64⁰ = 157.36⁰
    B = 157.36⁰
    Evaluate CosA + secB.
    Cos(36.86) + sec(157.36)
    Cos(36.86) + 1/Cos(157.36)
    0.8+(0.923) = 0.8+0.923 = 1.723
    Tan B = Tan (157.36)
    = 0.417
    Or
    When tanA = ¾
    Using Pythagorean theorem
    |Hyp|² = |opp|² + |Adj|²
    |Hyp|² = |3|² + |4|²
    9+16 = 25

√25 = 5

When SinB = 5/13
Using Pythagorean theorem,

|Hyp|² = |opp|² + |Adj|²
|13|² = |5|² + |Adj|²
|Adj|² = 169-25
√144 = 12
Evaluate
(I) CosA + secB.
⅘+cos–¹
⅘+13/12
(48 + 65)/60= 113/60
(II) TanB
Tan B = 5/12
= 0.417

C. Show that (1+sinA)² + Cos²A = 2(1+sinA)
Solution

(1+sinA)² + Cos²A = 2(1+sinA)
Proving
(1+sinA)(1+SinA) + Cos²A
1 + SinA + SinA + Sin²A + Cos²A
Recall, Sin²A + Cos²A = 1
1 + 2SinA + (Sin²A + Cos²A)

1 + 2SinA + 1 = 2 + 2SinA
2(1 + SinA)

  1. Expand (2+3x)⁵ in ascending power of x using binomial expansion. Hence solve (2.3)⁵ Solution

Pascal triangle
  1. 1(2⁵3x⁰) + 5(2⁴3x¹) + 10(2³3x²) + 10(2²3x³) + 5(2¹3x⁴) +1(2⁰3x⁵)
    32+240x+10×8×9×x² + 10×4×27×x³ + 5×2×81×x⁴ + 243x⁵
    32 + 240x + 720x² + 1080x³ + 810x⁴ + 243x⁵
    Hence solve (2.3)⁵ = (2+0.3)⁵
    1(2⁵0.3⁰) + 5(2⁴0.3¹) + 10(2³0.3²) + 10(2²0.3³) + 5(2¹0.3⁴) +1(2⁰0.3⁵)
    32 + 5×16×0.3 + 10×8×0.09 + 10×4×0.027 + 5×2×0.0081 + 0.00243
    32+24+7.2+1.08+0.081+0.00243
    = 64.36343
    64.4(approximately to 1 decimal point)

B. In how many ways can three men and two women be chosen from six men and 4 women?
Solution


3 men out of 9 can be selected in 9C3, 2 women out of 4 women can be selected in 4C2
9C3 × 4C2. = 9!/(9-3)!3! × 4!/(4-2)!2!
9!/6!3! × 4!/2!2! =( 9×8×7×6!)/6!3! × (4×3×2!)/2!2!
504/6 × 12/2 = 84×6
= 504ways

Oluwamuyide Peter

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