Examloaded home
Home
jamb questions
Jamb
waec questions
Waec
neco questions
Neco
nabteb questions
Nabteb
Welccome To Examloaded.Com>>>>  Home of JAMB, WAEC and NECO Qestions and Answers Expo Site

PAYMENTS 4 WAEC & JAMB 2024 EXPO ENDS ON 29TH MARCH

   2024 WAEC answers link  2024 WAEC ANSWERS LINK:
ExamAnswer.Net

This is Our Only Contact | Click On The Phone Number

    +2348162563540

To Chat Us on WhatsApp

   
Score 280+ And Above in Your JAMB 2024
       
  2024 WAEC QUESTIONS AND ANSWERS | 2024 WAEC EXPO

(SCORE A,B,C IN YOUR RESULT)
 

2024 WAEC UPDATES
--2024 WAEC TIMETABLE      

--2024 WAEC SPECIMENS/PRACTICALS  


« | »

2024/2025 Waec SSCE Further Mathematics Expo Questions and Answers

Answers Loading….. ….. Keep Refreshing this Page. FURTHER MATHS ANSWERS (1) 5^4^(3x/4 -1) + 5^3(x-1)/5^(3x – 2) 5^3x/4 – 4 + 5^3x – 3/5^3x – 2 5^3x – 4 + 5^3x – 3/5^3x – 2 (5^3x ÷ 5^4) + 5^3x ÷ 5^3÷5^3x – 2 5^3x/625 + 5^3x/125÷5^3x- 2 5^3x + 5(5^3x) ÷ 5^3x-3 5^3x + 5(5^3x)/625 ÷ 5^3x/25 Let 5^3x = y y + 5(y)/625 ÷ y/25 y + 5y/625 × 25/y =6y/625 × 25/y =6/25 1(b) F(x+2)-6x^2+5x-8 f(5 X+2=5z X=5-2 X=3 6x^2 +5x-8 =6(3)^2 + 5(3)-8 =54+15-8 =6! (2a) Using y2 – y1/x2 – x1 Where y2 = 7, y1 = -5, X2 = -2, and X1 = 7 7 – -5/-2 – 7 =7+5/-9 =12/-9 =4/-3 Coordinate points : -4/3(3 : 2) =-12/3 : -8/3 = -4 : -2⅔ X = (-4, 2⅔) (2b) 2/1-√2 – 2/2+√2 =2(2+√2)-2(1-√2)/(1-√2)(2+√2) =4+2√2 – 2+2√2/2+√2-2√2 – 2 =2 + 4√2/-√2 =(2+4√2)(-√2)/-√2(-√2) = -2√2 – 4(2)/2 = -8 – 2√2/2 = -4 – √2 (3) Sn = A/2[2n+(n-1)d] Where Sn = 165 a = -3, d = 2 165 = A/2[2(-3)+(n-1)2] 165 = n[-6+2n-2]/2 165×2 = n[2n – 8] 330 = 2n² – 8n 2n² – 8n – 330 = 0 n²-4n-165 = 0 Using -b±√b²-4ac/2a 4±√-4²-4(1)(-165)/2(1) 4±√16 + 660/2 4±√676 = 4±26/2 4+26/2 = 30/2 = 15 terms (4) Draw the right angled triangle Using Pythagoras theorem Third side = √(p+q)² – (p-q)² =√(p+q+p-q)(p+q-p+q) Difference of two squares. =√(2p)(2q) =√4pq Adjacent side = 2√pq Tanx = opp/adj = p – q/2√pq 1 – tan²X = 1-(p-q)²/4pq =(4pq)-(p²-2pq+q²)/4pq = -p²+6pq-q²/4pq = -(p² – 6pq + q²)/4pq (5) Draw the diagram Using cosine law Cos∅ = 16²+10²-14²/2(16)(10) Cos∅ = 256 + 100 – 196/320 Cos∅ = 160/320 Cos∅ = 0.5 ∅ = cos-¹(0.5) ∅ = 60° Angle between 10N and 16N = 180 – ∅ (sum of angles on a straight line) = 180 – 60 =120° (6) Draw the diagram Taking moment about the pivot, (T × 25)=(50×10)+(20×45) 25T = 500 + 900 25T = 1400 T = 1400/25 T = 56N (7) In a tabular form Under class interval: 1-5, 6-10, 11-15, 16-20, 21-25, 26-30 Under class mark (X): 3, 8, 13, 18, 23, 28 Under X-Xbar: -10, -5, 0, 5, 10, 15 Under frequency: 18, 12, 25, 15, 20, 10 Ef = 100 Under f(X – XbarA): -180, -60, 0, 75, 200, 150 f(X – XbarA) = 185 Where xA = 13 Mean = xA + Ef(X – Xbar)/Ef =13 + 185/100 =13 + 1.85 =14.85years PLS NOTE THAT XBAR LOOKS LIKE X WITH MINUS SIGN ON TOP. (10a) given 4x² – px +1 = 0 For real roots: b² – 4ac >0 (-p) ² – 4(4) (1) > 0 p² – 16 > 0 p² >16 p > ± 4 (10bi) Given: (1 +3x)⁶ Using pascal’s triangle: 1, 6, 15, 20, 15, 6, 1 (1)⁶(3x)º + 6(1)⁵ (3x)¹ + 15 (1)⁴ (3x)₂ + 20(1)³ (3x)³ + (15) (1)²(3x)⁴+6(1)¹(3x)⁵ + 1(1)º(3x)⁶ 1 + 6(3x) + 15 (9x²) + 20 (27x³) + 15 (81x⁴) + 6(243 x⁵) 729x⁶ 1 + 18x + 135x² + 540x³ + 1215x⁴ 1458x⁵ + 729x⁶ (ii) (1.03)⁶ = (1 + 3(0.01)] Therefore (1.03)⁶ = 1 + 18(0.01) + 135 (0.01)² + 540(0.01)³ + 1215(0.01)⁴ + 1458 (0.01)⁵ + 729 (0.01)⁶ + 1 + 0.18 + 0.0135 + 0.005 + 0.00001215 + 0.0000001458 + 0.000000000729 = 1.1940523 = 1.194 (4s.f) (12) prob (pass) = 60% = 60/100 =3/5 Prob (fail) 1-3/5 = 2/5 (a) Prob (atleast two failed) = 1 – prob (ome pass) = 1 – 10Ci (3/5)¹ (2/5)⁹ 1 – (10!/9!) (3/5) (2/5)⁹ = 1 – 10 (3/5) (2/5)⁹ = 1 – 10 (3/5) (0.000262144)) = 1 – 0.001572864 = 0.9984 (12b) prob (exastly half passed) = 10C5 (3/5)⁵ (2/5)⁵ = 10!/5!5! (6/25)⁵ = 252 (6/25)⁵ = 252 x 0.0007962624 = 0.2007 (12c) prob (at most two failed) = prob (zero/fail) + prob (one/fail) +prob (two/fail) = 10C (2/5)º (3/5)10 + 10C (2/5)¹ (3/5)⁹ + 10 C2(2/5)² (3/5)⁸ = (3/5)10 + 10 (2/5) (3/5)⁹ + 45 (2/5)²(3/5)⁸ = 0.060466176 + 0.040310784 + 0.120932352 = 0.1673 (14a) Draw the diagram (14b) From the diagram a = v – 20/4 2.5 = v – 20/4 V – 20 = 10 V = 10 + 20 = 30m/s acceleration/retardation = 3/4 2.5/30/T-12 = 3/4 2.5(T – 12)/30 = 3/4 (T – 12) = 30×3/2.5×4 = 90/10 = 9 T – 12 = 9 T = 9 + 12 = 21 t = T – 12 t = 21 – 12 t = 9secs (14c) Total distance of the journey = Area of BCDI + Area of AFEO + Area of DFI = 1/2(12+8)10 + 1/2(9×10) + 1/2(21+21)20 = (20/2)10 + 90/2 +(42/2)20 =10(10) + 45 + 21(20) =100 + 45 + 420 = 565m          
Download Nulled WordPress Themes
Download Best WordPress Themes Free Download
Download WordPress Themes
Download WordPress Themes
lynda course free download
download karbonn firmware
Download WordPress Themes Free
download udemy paid course for free

Tags: , , , , , , , , , , , , , , , , , , ,

Categories: WAEC Syllabus Past Questions and Answers

0 Responses
Leave a Reply


Your Comment

« | »