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03/02/2021

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20/12/2019

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18/04/2018

[4/18, 10:28 AM] ‪+234 806 731 1339‬: NO2) Given that y = 2pxˆ² – p² x – 14

AT (3, 10)

10 = 2p(3)² – p² (3) – 14

10 = 18p – 3p² – 14

3p² – 18p + 24 = 0

p² – 6p + 8 = 0

using factor method,

p² – 2p -4p + 8 = 0

p(p-2) – 4(p-2) = 0

(p-4)(p-2) = 0

p-4 = 0 or p-4 = 0

p= 4 or p =2
[4/18, 10:28 AM] ‪+234 806 731 1339‬: *4A ii*
RSQ=(x+90)°
RSQ=37.5°+90°
RSQ=127.5°
[4/18, 10:28 AM] ‪+234 806 731 1339‬: *MATHEMATICS ANSWERS*

(1)
On February 28th 2012, value = (100-30/100) * #900,00.00

= 70/100 * #900,00

= #630,000.00

On february 28th 2013, value = (100-22/1000 * #630,00

= 78/100 8 #630,000

= #491,400

On february 28th 2014, value = 78/100 8 #491,400

=383,292

On february 28th 2015, value = 78/100 * #383,292

= #298,967.76

=============================

(2)
Given that y = 2pxˆ² – p² x – 14

AT (3, 10)

10 = 2p(3)² – p² (3) – 14

10 = 18p – 3p² – 14

3p² – 18p + 24 = 0

p² – 6p + 8 = 0

using factor method,

p² – 2p -4p + 8 = 0

p(p-2) – 4(p-2) = 0

(p-4)(p-2) = 0

p-4 = 0 or p-4 = 0

p= 4 or p =2

2b)
The lines must be solved simultenously

3y – 2x = 21 ——- (1)

4y + 5x = 5 ——-(2)

using elimination method,

(4) 3y – 2x = 21

(30 4y + 5x = 5

12y – 8y = 84 ——— (3)

12y + 15x = 15 ——-(4)

equ (4) minus equ(3)

23x = -69

x = -69/23

x = -3

Put this into equation (1)

3y -2(-3) = 21

3y = 6 = 21

3y = 21 -6

3y = 15

y =15/3

y = 5

coordinates of Q is (-3, 5)
[4/18, 10:28 AM] ‪+234 806 731 1339‬: 2b) The lines must be solved simultenously

3y – 2x = 21 ——- (1)

4y + 5x = 5 ——-(2)

using elimination method,

(4) 3y – 2x = 21

(30 4y + 5x = 5

12y – 8y = 84 ——— (3)

12y + 15x = 15 ——-(4)

equ (4) minus equ(3)

23x = -69

x = -69/23

x = -3

Put this into equation (1)

3y -2(-3) = 21

3y = 6 = 21

3y = 21 -6

3y = 15

y =15/3

y = 5

coordinates of Q is (-3, 5)
[4/18, 10:28 AM] ‪+234 806 731 1339‬: (3a)
The diagonal = 10.2m and 9.3cm
Using Pythagoras theory
Ac² = 10.2² + 9-3²
Ac² = 104.04 + 86.49
Ac² = 190.53
Ac² = √190.53
Ac² = 13.80

(3b)
DRAW THE DIAGRAM
Using Pythagoras theory
5² = 3² + x²
x² = 5² - 3²
X²= 25 - 9
X² = √16
X= 4cm
CosX = adjacent/hyp
= 4/5
Tan X = opp/adj. = 3/4
5cos x - 4tan x
5(4/5)- 4(3/4)
20/5 - 12/4
4-3= 1
[4/18, 10:28 AM] ‪+234 806 731 1339‬: Nigeria🇳🇬👇🏽👇🏽

(4a) Rate = 2/100 * N0.02 per month Rate per annum = 0.02 * 12 = 0.24 per annum

(4b) Draw the Diagram =======================

*4A ii*
RSQ=(x+90)°
RSQ=37.5°+90°
RSQ=127.5°
[4/18, 10:28 AM] ‪+234 806 731 1339‬: 3a)
The diagonal = 10.2m and 9.3cm
Using Pythagoras theory
Ac² = 10.2² + 9-3²
Ac² = 104.04 + 86.49
Ac² = 190.53
Ac² = √190.53
Ac² = 13.80
3b)
DRAW THE DIAGRAM
Using Pythagoras theory
5² = 3² + x²
x² = 5² - 3²
X²= 25 - 9
X² = √16
X= 4cm
CosX = adjacent/hyp
= 4/5
Tan X = opp/adj. = 3/4
5cos x - 4tan x
5(4/5)- 4(3/4)
20/5 - 12/4
4-3= 1

4ai)
sum of angle in a D =180degree
xdegree + 90degree + 180degree - (3x+15)=180degree
xdegree + 90degree + 180degree - 3x+15=180degree
-2x=180degree - 255
+2x/2=+75/2
x=37.5
4aii)

18/04/2018

2)
Given that y = 2pxˆ² – p² x – 14
AT (3, 10)
10 = 2p(3)² – p² (3) – 14
10 = 18p – 3p² – 14
3p² – 18p + 24 = 0
p² – 6p + 8 = 0
using factor method,
p² – 2p -4p + 8 = 0
p(p-2) – 4(p-2) = 0
(p-4)(p-2) = 0
p-4 = 0 or p-4 = 0
p= 4 or p =2
2b)
The lines must be solved simultenously
3y – 2x = 21 ——- (1)
4y + 5x = 5 ——-(2)
using elimination method,
(4) 3y – 2x = 21
(30 4y + 5x = 5
12y – 8y = 84 ——— (3)
12y + 15x = 15 ——-(4)
equ (4) minus equ(3)
23x = -69
x = -69/23
x = -3
Put this into equation (1)
3y -2(-3) = 21
3y = 6 = 21
3y = 21 -6
3y = 15
y =15/3
y = 5
coordinates of Q is (-3, 5)

18/04/2018

Core Maths 2018 WASSCE

6a A regular polygon contains 35 diagonals .
(I) what type of polygon is that
(II) what is the sum of all the interior angles of the polygon
(III) what is the measure of each interior angle
(IV) what is the measure of an exterior angle of the polygon

b. Simplify the equation
3²ⁿ - 3ⁿ - 2 = 0

7a . Probability that Evans scores more than 50% in an exam is twice that of Michael's. If the probability that Michael scores the said % is ⅓ .
(I) what is the probability that Evans scores below 50% .
(II) Both of them score more than 50%
(III) at least one of them scores 50%

(b) the gradient of a line to the points ( 2,4 ) and ( x , 8 ) is 2 . Find the value of x

8. Graph of relation
( Full question )

9 . Solve the simultaneous equations

2x - y + 3z = 5
y + z = 4
2x + 3y - z = 7

9b. Statistics, Finding mean . Table already provided.

10 ......

18/04/2018

2018 WAEC MATHEMATICS THEORY AND OBJ ANSWERS ANSWERS from wapgist. com.ng

2018 waec Maths obj Answer
Loading.....
2018 Waec Maths Theory Answer

(1)
On February 28th 2012, value = (100-30/100) * #900,00.00
= 70/100 * #900,00
= #630,000.00
On february 28th 2013, value = (100-22/1000 * #630,00
= 78/100 8 #630,000
= #491,400
On february 28th 2014, value = 78/100 8 #491,400
=383,292
On february 28th 2015, value = 78/100 * #383,292
= #298,967.76
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

(2)
Given that y = 2pxˆ² – p² x – 14
AT (3, 10)
10 = 2p(3)² – p² (3) – 14
10 = 18p – 3p² – 14
3p² – 18p + 24 = 0
p² – 6p + 8 = 0
using factor method,
p² – 2p -4p + 8 = 0
p(p-2) – 4(p-2) = 0
(p-4)(p-2) = 0
p-4 = 0 or p-4 = 0
p= 4 or p =2

(2b)
The lines must be solved simultenously
3y – 2x = 21 ——- (1)
4y + 5x = 5 ——-(2)
using elimination method,
(4) 3y – 2x = 21
(30 4y + 5x = 5
12y – 8y = 84 ——— (3)
12y + 15x = 15 ——-(4)
equ (4) minus equ(3)
23x = -69
x = -69/23
x = -3
Put this into equation (1)
3y -2(-3) = 21
3y = 6 = 21
3y = 21 -6
3y = 15
y =15/3
y = 5
coordinates of Q is (-3, 5)
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

(3)
DRAW THE DIAGRAM.
Using pythagoras theorem,
lˆ² = 5.1ˆ² + 4.65ˆ²
lˆ² = 26.01 + 21.6225
lˆ² = 47.6325
l = √47.6325
l = 6.9cm (1d.p)
perimeter of phombus = 41
4 * 6.9
= 27.6cm

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12/04/2018

1-10: CCACDADDCA
11-20:BDACBADACD
21-30:DBDAACCDCB
31-40:DABCCAADAB
41-50:DCBADBCBAA
51-60:ABCDBBCABB
61-70:CCADDABBCB
71-80:DCDDDCACAD
OBJ COMPLETED✅✅✅✅✅✅✅

12/04/2018

ENGLISH LANGUAGE OBJ

1-10:
11-20:
21-30:
31-40:
41-50:
51-60:
61-70:
71-80: DCCBDCACAD
Still solving on Nigeria English obj
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