2022/2023 WAEC/NECO Answers: NECO GCE 2022 - MATHS ANSWERS



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ID 152

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Saturday 7th Dec. 2019

English Language Paper II (Essay) 10.00 am 11.45 am
English Language Paper III (Objective) 11.45 am 12.45 am
English Language Paper IV (Test of Orals) 1.00 pm 1.45 pm

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Oral-Obj
1-10 EDDBDBCBEC
11-20 AABAEECCAA
21-30 ABCCCEDCEE
31-40 ECDEDCDDCB
41-50 BEDAAAADCA
51-60 ABBEDABACA
=============


English OBJ
1-10: DEEDBEBABB
11-20: AABDACEACC
21-30: DDACBCCDAD
31-40: BCBADCCADA
41-50: CDDCCDCDAA
51-60: DBABBBBCDA
61-70: DCDDADDCCE
71-80: CBEDECABEC
81-90: BCACBBAEAB
91-100: ECEEACDBAB
This is actually the NECO GCE 2019 - MATHS ANSWERSAnswers as questions sent to our subscribers on WHATSAPP, if you are looking for 2022 WAEC/NECO Answers midnight before the exam, try contacting us.

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(9a)
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(9c)
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(8a)
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(8b)
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(3a)
Given : R(3,5) and S(-2, -6)
equation to line through them is :
y-5/x-3 = -6-5/-2-3
y-5/x-3 = -11/-5
y-5/x-3 = 11/5
5(y-5)= 11(x-3)
5y-25 = 11x-33
5y-11x = 25-33
5y-11x = -8 or 11x-5y = 8

(3b)
RS= √(X1X2)^2 (y1y2)^2
√(-2-3)^2 + (-6-5)^2
√(-5)^2 + (-11)^2
√25 + 121
√146
= 12.08

(3)
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(4a)
Given: curve; y x^2 - 3x
gradient ; dy/dx = 2x - 3
At (2-2), gradient = 2(2) - 3
4-3 = 1

(4b)
Given; y= 1+x^2/1-x^-2
dy/dx = (1-x^2)(2x) - (1+x^2)(-2x)/(1-x^2)^2
= (1-x^2)(2x) - (1+x^2)(2x)/(1-x^2)^2
= 2x(1-x^2 + 1+x^2)/(1-x^2)^2
= 2x(2)/(1-x^2)^2
= 4x/(1-x^2)^2

(4)
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(1)
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(2)
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(7)
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(6)
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(10)
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(12)
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(5)
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