I STILL REMAIN ENGR NASER DAHER

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2020 mathematics answer

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MATHS-OBJ 1CBCDACDCCD 11AADBDACBBC 21BDDABDADAD 31CDACCCCCDA 41BBBCDCACDB Completed. ====== FOR TOMORROW'S PAPER SEND 500MTN TO 09066669882 AND YOUR NAME ============ THEORY – ANSWERS Always subscribe with us to get your accurate questions and answers A1 GUARANTEED 09066669882 ...... (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)2/3 p3/2 = (rk/Q -ms)2/8 * 3/2 p3/2 = rk/Q = ms p3/2 + ms = rk/Q q(p3/2 + ms ) = rk Q =rk/p3/2 + ms ... 2aii) Where : r = 10, k = 4, s = 0.2 p = 3, m = 15 ... Q = 10 * 4/3^3/2 + (15 * 0.2) Q = 40/5.2 + 3 Q = 40/8.2 Q = 4.88 ... (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=...
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STUDENTS REVISIONS SWEETS SIXTEEN/QUES & ANSWERS

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*STUDENTS REVISIONS* SWEETS SIXTEEN/QUES & ANSWERS* *PLEASE THIS CONTENT IS NOT FOR SALE* POWERED BY MR WHY HURRY SOLUTION TEAM 09066669882 CHAPTER 1 - THE LETTER 1. How old was Aliya when she read the book ‘Sweet Sixteen’? *(a)12 years* (b) 14 years (c)16 years 2. The book ‘Sweet Sixteen’ made Aliya realize that at 16 years, one is no longer a child but a_____ (a)Teenager  (b) Mature child *(c)Young adult* 3. At the time Aliya was born, her father was a______ (a)Public relation officer (b) Imam *(c)Journalist* 4. Aliya’s father is_____ (a)Mr Ahmed (b) Mr Ben *(c)Mr Bello* 5. Aliya turned 12 when she was in _____ (a)JSS2 *(b) JSS1* (c)SS1 6. The closest to any form of celebration in Aliya’s house was _____ (a)Elaborate parties and elaborate dinners (b) Elaborate prayers and organized parties *(c)Elaborate dinners and elaborate prayers* 7. On one of Aliya’s birthday cards from her dad, he wrote “…only _____ is greater than you” (a)Man (...
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2018 WAEC GCE MATH powered by Engr NASER DAHER

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WAEC GCE MATHEMATICS THEORY AND OBJECTIVES (2018 Our work is to DELIVER COMFIRMED AND VERIFIED WAEC JAMB NECO NABTEB BECE ANSWERS _OBJ LOADING_   (1a) S = ½n[a + L] Where S = 130 S = n/2[a+a+(n-1)d] 2S = n[2a+(n-1)d] 2(130) = 10[2a + (10 - 1)d] 260 = 10[2a + 9d] 260 = 20a + 90d .....(I) a + 4d = 3a...... (II) 4d = 3a - a 4d = 2a 2d = a...... (III) Substitute a = 2d into Eqn (I) 260 = 20(2d) + 90d 260 = 40d + 90d 260 = 130d 260/130 = 130d/130 d = 2 Hence common difference d = 2 Engr NASER DAHER 09030302775 (1b) First term a = 2d = 2(2) = 4 (1c) L = a+(n-1)d 28 = 4+(ń - 1)2 28 = 4+2n - 2 2n = 28 - 2 2n = 26 2n/2 = 26/2 n = 13 (4a) (2y+x) + (6y-2x+1) + 4y = 28 ...(i) 6y-2x + 1 = 4y... (II) 2y+ x +6y - 2x + 1 +4 = 28 12y + x +6 -2x + 1 + 4 = 28 12y - x + 1 = 28 12y - x = 29... (III) 6y-2x + 1 = 4y, 6y-2x - 4y = 1 2y - 2x = -1... (iv) 24y - 2x = 54 2y - 2x = 1 22y/2w = 55/22 y = 2.5 12y - x = 27 12 (2.5) - x = 27 30 x 27 x=3 (b) 2y + x = 2 (2.5) + 3 = 5+3 = 8cm...
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