WAEC SSCE - Further Mathematics - 2022

Tambaya 1 Rahoto

Consider the following statement:

x: All wrestlers are strong

y: Some wresters are not weightlifters.

Which of the following is a valid conclusion?

All strong wrestlers are weightlifters

Some strong wrestlers are not weightlifters

Some weak wrestlers are weightlifters

All weightlifters are wrestlers

Logical Reasoning

Duba Amsa

Tambaya 2 Rahoto

Find the coefficient of x $^{3}$ y $^{2}$ in the binomial expansion of (x-2y) $^{5}$

-80

10

40

80

Bayanin Amsa

x $^{3}$ y $^{2}$ in (x-2y) $^{5}$

n = 5, r = 3, p = x, q = -2y

5C $_{3}$ * x $^{3}$ $^{2}$

5C $_{3}$ = $\frac{5!}{[5 - 3]! 3!}$

$\frac{5 * 4 * 3!}{2! 3!}$ → $\frac{5 * 4}{2}$

5C $_{3}$ = 10

: 5C $_{3}$ * x $^{3}$ -2y $^{2}$ = 10 * x $^{3}$ 4y $^{2}$

40x $^{3}$ y $^{2}$
the coefficient is 40

Karanta bayanin darasi akan Binomial Theorem (WAEC)

Binomial Theorem

Duba Amsa

Tambaya 3 Rahoto

If f(x-1) = x $^{3}$ + 3x $^{2}$ + 4x - 5, find f(2)

61

25

20

13

Bayanin Amsa

x - 1 = 2
x = 3
f(2) = (3) $^{3}$ + 3(3) $^{2}$ + 4(3) - 5
f(2) = 27 + 27 + 12 - 5

= 61

Karanta bayanin darasi akan Functions (WAEC)

Functions

Duba Amsa

Tambaya 4 Rahoto

Find the coefficient of x $^{2}$ in the binomial expansion of $(x + \frac{2}{x^{2}})^{5}$

10

40

32

80

Bayanin Amsa

$(x + \frac{2}{x^{2}})^{5}$

n = 5, r = 4, p = x and q = $\frac{2}{x^{2}}$

5C $_{4}$ x $^{4}$ ( $\frac{2}{x^{2}}$ )1 = 5C $_{4}$ $\frac{2 x^{4}}{x^{2}}$

5C $_{4}$ 2x $^{2}$ = $\frac{5!}{[5 - 4]! 4!}$ * 2x $^{2}$

$\frac{5 * 4!}{4!} * 2 x^{2}$ = 5 * 2x $^{2}$ = 10x $^{2}$

The coefficient is 10.

Karanta bayanin darasi akan Binomial Theorem (WAEC)

Binomial Theorem

Duba Amsa

Tambaya 5 Rahoto

If α and β are roots of x $^{2}$ + mx - n = 0, where m and n are constants, form the

equation

whose

roots

are

1α

and

1β

.

mnx

^{2}

- n

^{2}

x - m = 0

mx

^{2}

- nx + 1 = 0

nx

^{2}

- mx + 1 = 0

nx

^{2}

- mx - 1 = 0

Bayanin Amsa

x $^{2}$ + mx - n = 0

a = 1, b = m, c = -n

α + β = $\frac{- b}{a}$ = $\frac{- m}{1}$ = -m

αβ = $\frac{c}{a}$ = $\frac{- n}{1}$ = -n

the roots are = $\frac{1}{α}$ and $\frac{1}{β}$

sum of the roots = $\frac{1}{α}$ + $\frac{1}{β}$

$\frac{1}{α}$ + $\frac{1}{β}$ $\frac{α + β}{α β}$

α + β = -m
αβ = -n

$\frac{α + β}{α β}$

product of the roots = $\frac{1}{α}$ * $\frac{1}{β}$

$\frac{1}{α}$ + $\frac{1}{β}$ = $\frac{1}{α β}$ → $\frac{1}{- n}$

x $^{2}$ - (sum of roots)x + (product of roots)
x $^{2}$ - ( m/n )x + ( 1/-n ) = 0
multiply through by n
nx $^{2}$ - mx - 1 = 0

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Polynomial Functions

Duba Amsa

Tambaya 6 Rahoto

Given

| \begin{matrix} 2 & - 3 \\ 1 & 4 \end{matrix} | | \begin{matrix} - 6 \\ k \end{matrix} | | \begin{matrix} 3 \\ - 26 \end{matrix} | = 15

Solve for k.

-8

-5

-4

-3

Bayanin Amsa

$| \begin{matrix} 2 & - 3 \\ 1 & 4 \end{matrix} | | \begin{matrix} - 6 \\ k \end{matrix} | | \begin{matrix} 3 \\ - 26 \end{matrix} | = 15$

$| \begin{matrix} 2 [- 6] & - 3 k \\ 1 [- 6] & + 4 k \end{matrix} | = | \begin{matrix} 3 \\ - 26 \end{matrix} |$

$| \begin{matrix} - 12 & - 3 k \\ - 6 & + 4 k \end{matrix} | = | \begin{matrix} 3 \\ - 26 \end{matrix} |$

-12 - 3k = 3

-3k = 3 + 12

k = $\frac{15}{- 3}$

k = -5

Karanta bayanin darasi akan Surds (WAEC)

Surds

Duba Amsa

Tambaya 7 Rahoto

If $x^{2} + y^{2} + - 2 x - 6 y + 5 = 0$ , evaluate dy/dx when x=3 and y=2.

2

-2

-4

4

Bayanin Amsa

$x^{2} + y^{2} + - 2 x - 6 y + 5 = 0$

When differentiated:

$x^{2} + y^{2} + - 2 x - 6 y + 5 = 0$

where x=3 and y=2

2[3] + 2[2] - 8 = 0

6 + 4 - 8 = 2

Duba Amsa

Tambaya 8 Rahoto

Evaluate $\int_{- 1}^{0}$ (x + 1)(x - 2) dx

7/6

5/6

-5/6

-7/6

Bayanin Amsa

$\int_{- 1}^{0}$ (x + 1)(x - 2) dx

= $\int_{- 1}^{0}$ $x^{2} - x - 2$ dx

Integrated $x^{2} - x - 2$ = $\frac{x^{3}}{3} - \frac{x^{2}}{2} - 2$

=

(0

-

0

-

0)

-

(

-13

-

12

+

2)

=

0

-

(

76

)

=

-76

Duba Amsa

Tambaya 9 Rahoto

A linear transformation T is defined by T: (x,y) → (3x - y, x + 4y). Find the image of (2, -1) under T.

(7, -2)

(5, -2)

(-2, 7)

(-7, 2)

Matrices And Linear Transformation

Duba Amsa

Tambaya 10 Rahoto

Evaluate $1_{0}^{\int} x^{2} (x^{3} + 2)^{3}$

$\frac{56}{12}$

$\frac{65}{12}$

12

65

Bayanin Amsa

$1_{0}^{\int} x^{2} (x^{3} + 2)^{3}$ dx

let $u = x^{3} + 2, d u = 3 x^{2} d x$

when x = 1, u = 3

when x = 0, u = 2

dx = $\frac{d u}{3 x^{2}}$

$3_{2}^{\int}$ $\frac{x^{2} [u]^{3}}{3 x^{2}}$

$3_{2}^{\int}$ $\frac{u^{3}}{3}$ du

= $\frac{u^{4}}{3 * 4}$ $_{2}$ 3

$\frac{1}{12} [u^{4}]$ $_{2}$ 3

$\frac{1}{12} [3^{4} - 2^{4}]$

$\frac{1}{12} [81 - 16]$

$\frac{65}{12}$

Karanta bayanin darasi akan Integration (WAEC)

Integration

Duba Amsa

Tambaya 11 Rahoto

The probability that a student will graduate from college is 0.4. If 3 students are selected from the college, what is the probability that at least one student will graduate?

0.06

0.22

0.78

0.80

Probability

Duba Amsa

Tambaya 12 Rahoto

The equation of a circle is given as 2x $^{2}$ + 2y $^{2}$ - x - 3y - 41 = 0. Find the coordinates of its centre.

(

\frac{- 1}{4}

,

\frac{3}{4}

)

(

\frac{1}{4}

,

\frac{3}{4}

(

\frac{- 1}{2}

,

\frac{3}{2}

)

. (

\frac{- 1}{2}

,

\frac{- 3}{2}

)

Bayanin Amsa

2x $^{2}$ + 2y $^{2}$ - x - 3y - 41

standard equation of circle
(x-a) $^{2}$ + (x-b) $^{2}$ = r $^{2}$
General form of equation of a circle.
x $^{2}$ + y $^{2}$ + 2gx + 2fy + c = 0
a = -g, b = -f., r2 = g2 + f2 - c
the centre of the circle is (a,b)
comparing the equation with the general form of equation of circle.
2x $^{2}$ + 2y $^{2}$ - x - 3y - 41

= x $^{2}$ + y $^{2}$ + 2gx + 2fy + c
2x $^{2}$ + 2y $^{2}$ - x - 3y - 41 = 0
divide through by 2

g = $\frac{- 1}{4}$ ; 2g = $\frac{- 1}{2}$

f = $\frac{- 3}{4}$ ; 2f = $\frac{- 3}{2}$

a = -g → - $\frac{- 1}{4}$ ; = $\frac{1}{4}$

b = -f → - (\frac{-3}{4}\) = (\frac{3}{4}\)

therefore the centre is ( $\frac{1}{4}$ , $\frac{3}{4}$ )

Karanta bayanin darasi akan Co-ordinate Geometry (WAEC)

Co-ordinate Geometry

Duba Amsa

Tambaya 13 Rahoto

A particle is acted upon by forces F = (10N, 060º), P = (15N, 120º) and Q = (12N, 200º). Express the force that will keep the particle in equilibrium in the form xi + yj, where x and y are scalars.

17.55i + 13.78j

17.55j - 13.78i

-17.55i + 13.78j

-17.55i - 13.78j

Statics

Duba Amsa

Tambaya 14 Rahoto

If g(x) = √(1-x $^{2}$ ), find the domain of g(x)

x < -1 or x > 1

x ≤ -1 or x ≥1

-1 ≤ x ≤ 1

-1 < x < 1

Bayanin Amsa

1 - x $^{2}$ ≥ 0
-x $^{2}$ ≥ -1
x $^{2}$ ≤ 1
√x $^{2}$ ≤ 1
|x| ≤ 1
-1 ≤ x ≤ 1

Karanta bayanin darasi akan Surds (WAEC)

Surds

Duba Amsa

Tambaya 15 Rahoto

The gradient of a function at any point (x,y) 2x - 6. If the function passes through (1,2), find the function.

x

^{2}

- 6x - 5

x

^{2}

- 6x + 5

x

^{2}

- 6x - 3

x

^{2}

- 6x + 7

Bayanin Amsa

dy/dx = 2x - 6

y = ∫ 2x - 6

y = $\frac{2 x^{2}}{2} - 6 + c$ y = x $^{2}$ - 6x + c
passes through (1,2)
2 = 1 $^{2}$ - 6(1) + c
2 = 1 - 6 + c
c = 7
y = x $^{2}$ - 6x + c

y = x $^{2}$ - 6x + 7

Karanta bayanin darasi akan Functions (WAEC)

Functions

Duba Amsa

Tambaya 16 Rahoto

A binary operation ∆ is defined on the set of real numbers R, by x∆y = $\sqrt{x + y - \frac{x y}{4}}$ , where x, yER. Find the value of 4∆3

16

8

4

2

Bayanin Amsa

x∆y = $\sqrt{x + y - \frac{x y}{4}}$

4∆3 = $\sqrt{4 + 3 - \frac{4 * 3}{4}}$

= $\sqrt{4 + 3 - 3}$

= $\sqrt{4}$

= 2.

Karanta bayanin darasi akan Binary Operations (WAEC)

Binary Operations

Duba Amsa

Tambaya 17 Rahoto

A straight line makes intercepts of -3 and 2 on the x and y axes respectively. Find the equation of the line.

2x + 3y + 6 = 0

3x - 2y - 6 = 0

-3x 2y - 6 = 0

-2x + 3y - 6 = 0

Co-ordinate Geometry

Duba Amsa

Tambaya 18 Rahoto

Evaluate $4 p_{2} + 4 C_{2} - 4 p_{3}$

18

6

-6

-18

Bayanin Amsa

$4 p_{2} + 4 C_{2} - 4 p_{3}$

$n p_{r} = \frac{n!}{[n - r]!} a n d n C_{r} = \frac{n!}{[n - r]! r!}$

= $\frac{4!}{[4 - 2]!} + \frac{4!}{[4 - 2]! 2!} - \frac{4!}{[4 - 3]!} = \frac{4!}{2!} + \frac{4!}{2! 2!} - \frac{4!}{1!}$

= $\frac{4 * 3 * 2!}{2!} + \frac{4 * 3 * 2!}{2! 2!} - \frac{4 * 3 * 2 * 1}{1!}$

12 + 6 - 24 = -6

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Permutation And Combinations (WAEC)

Sets

Permutation And Combinations

Duba Amsa

Tambaya 19 Rahoto

If 36, p, $\frac{9}{4}$ and q are consecutive terms of an exponential sequence (G.P), find the sum of p and q.

9/16

81/16

9

\frac{9}{16}

Bayanin Amsa

GP : 36, P, $\frac{q}{4}$ , q, ... p + q = ?

Recall,

common

ratio,

r

=

TnTn-1

=

T2T1

=

T3T2

=

T4T3

∴

P36

=

94

÷

p

;

p

^{2}

=

94

x

36

;

p

^{2}

=

81

p

=

9

∴

r

=

T2T1

=

936

=

14

Also

r

=

T4T3

=

q

÷

94

∴ $\frac{1}{4}$ = q ÷ $\frac{9}{4}$ ;

$\frac{9}{4}$ = 4q

16q

=

9

,

q

=

916

∴

p

+

q

=

9

+

916

=

9

916

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Polynomial Functions

Duba Amsa

Tambaya 20 Rahoto

The functions f:x → 2x $^{2}$ + 3x -7 and g:x →5x $^{2}$ + 7x - 6 are defined on the set of real numbers, R. Find the values of x for which 3f(x) = g(x).

x = -3 or -5

x = -3 or 5

x = 3 or -5

x = 3 or 5

Functions

Duba Amsa

Tambaya 21 Rahoto

Solve: $3^{2 x - 2} - 28 (3^{x - 2}) + 3 = 0$

x = -2 or x = 1

x = 0 or x = -3

x = 2 or x = 1

x = 0 or x = 3

Bayanin Amsa

$3^{2 x - 2} - 28 (3^{x - 2}) + 3 = 0$

$\frac{3^{2 x}}{3^{2}} - \frac{{28.3}^{x}}{3^{2}} + 3 = 0$

$\frac{3^{2 x}}{9} - \frac{{28.3}^{x}}{9} + 3 = 0$

let p = 3 $^{x}$

$\frac{p^{2}}{9} - \frac{28 p}{9} + 3 = 0$

multiply through by 9
p $^{2}$ - 28p + 27 = 0
p $^{2}$ - p - 27p + 27 = 0
p (p - 1) - 27(p - 1) = 0
(p-1)(p-27) = 0
p = 1 or 27
when p = 1
p = 3 $^{x}$
3 $^{x}$ = 1
3 $^{x}$ = 3 $^{0}$
x = 0
when p = 27
3x = 27
3x = 33
x = 3

Karanta bayanin darasi akan Differentiation (WAEC)

Differentiation

Duba Amsa

Tambaya 22 Rahoto

Given that P = (-4, -5) and Q = (2,3), express →PQ in the form (k,θ). where k is the magnitude and θ the bearing.

(10 units, 053º)

(9 units, 049º)

(10 units, 037º)

(9 units, 027º)

Vectors

Duba Amsa

Tambaya 23 Rahoto

Find the range of values of x for which 2x $^{2}$ + 7x - 15 ≥ 0.

x ≤ -5 or x ≥

\frac{3}{2}

x ≥ -5 or x ≤

\frac{3}{2}

-5 ≤ x ≤

\frac{3}{5}

\frac{3}{5}

≤ x ≤ -5

Bayanin Amsa

2x $^{2}$ + 7x - 15 ≥ 0

2x $^{2}$ -3x + 10x - 15 ≥ 0
x(2x - 3) + 5(2x - 3) ≥ 0
(x+5)(2x-3) ≥ 0
the points on x-axis where the graph ≥ 0

x ≤ -5 or x ≥ $\frac{3}{2}$

Duba Amsa

Tambaya 24 Rahoto

Given that $\frac{8 x + m}{x^{2} - 3 x - 4} \equiv \frac{5}{x + 1} + \frac{3}{x - 4}$

23

17

-17

18

Bayanin Amsa

$\frac{8 x + m}{x^{2} - 3 x - 4} \equiv \frac{5}{x + 1} + \frac{3}{x - 4}$

$\frac{8 x + m}{x^{2} - 3 x - 4}$ ≡ $\frac{5 (x - 1) + 3 (x + 4)}{x^{2} - 3 x - 4}$

multiplying both sides by x2-3x-4
8x+m ≡ 5(x-4)+3(x+1)
8x + m ≡ 5x - 20 + 3x + 3
8x - 5x - 3x + m = -20 + 3
m = -17

Karanta bayanin darasi akan Binomial Theorem (WAEC)

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Binomial Theorem

Polynomial Functions

Duba Amsa

Tambaya 25 Rahoto

Solve: 4sin $^{2}$ θ + 1 = 2, where 0º < θ < 180º

60º 0r 120º

30º 0r 150º

30º 0r 120º

60º 0r 150º

Bayanin Amsa

4sin $^{2}$ θ + 1 = 2

4sin $^{2}$ θ = 2 - 1

4sin $^{2}$ θ = 1

$\sqrt{s} i n^{2} θ$ = $\sqrt{\frac{1}{4}}$

sinθ = $\frac{1}{2}$

θ = $s i n^{- 1} \frac{1}{2}$

θ = 30º 0r 150º

Karanta bayanin darasi akan Trigonometry (WAEC)

Trigonometry

Duba Amsa

Tambaya 26 Rahoto

If $l o g_{10} (3 x + 1) + l o g_{10} 4 = l o g_{10} (9 x + 2)$ , find the value of x

\frac{1}{3}

1

2

3

Bayanin Amsa

Explanation

$l o g_{10} (3 x + 1) + l o g_{10} 4 = l o g_{10} (9 x + 2)$

$l o g_{10} 4 (3 x + 1) = l o g_{10} (9 x + 2)$

4(3x+1) = 9x + 2

12x -4 = 9x + 2

12x - 9x = 2 + 4

3x = 6

x = 2

Duba Amsa

Tambaya 27 Rahoto

A particle of mass 3kg moving along a straight line under the action of a F N, covers a line distance, d, at time, t, such that d = t $^{2}$ + 3t. Find the magnitude of F at time t.

0N

2N

3(2t + 3)N

6N

Bayanin Amsa

F = m * a

d = t $^{2}$ + 3t.

a = $\frac{d^{2} d}{d t^{2}}$

$\frac{d [d]}{d t}$ = 2t + 3

$\frac{d^{2} d}{d t^{2}}$ = 2m/s $^{2}$

a = 2m/s $^{2}$

F = m * a

F = 3 × 2 = 6N

Duba Amsa

Tambaya 28 Rahoto

If →PQ = -2i + 5j and →RQ = -i - 7j, find →PR

-3i + 12j

-3i - 12j

-i + 12j

i - 12j

Vectors

Duba Amsa

Tambaya 29 Rahoto

The table shows the distribution of the distance (in km) covered by 40 hunters while hunting.

Distance(km)	3	4	5	6	7	8
Frequency	5	4	x	9	2x	1

If a hunter is selected at random, find the probability that the hunter covered at least 6km.

\frac{3}{5}

\frac{2}{5}

\frac{3}{8}

\frac{9}{40}

Probability

Duba Amsa

Tambaya 30 Rahoto

Given that P = {x: x is a multiple of 5}, Q = {x: x is a multiple of 3} and R = {x: x is an odd number} are subsets of μ = {x: 20 ≤ x ≤ 35}, (P⋃Q)∩R.

{20, 21, 25, 30, 33}

{21, 25, 27, 33, 35}

{20, 21, 25, 27, 33, 35}

{21, 25, 27, 30, 33, 35}

Sets

Duba Amsa

Tambaya 31 Rahoto

The length of the line joining points (x,4) and (-x,3) is 7 units. Find the value of x.

4√3

2√6

3√2

2√3

Surds

Duba Amsa

Tambaya 32 Rahoto

If Un = kn $^{2}$ + pn, U $_{1}$ = -1, U $_{5}$ = 15, find the values of k and p.

k = -1, p = 2

k = -1, p = -2

k = 1, p = -2

k = 1, p = 2

Bayanin Amsa

Un = kn $^{2}$ + pn,

U $_{1}$ = -1,

U $_{5}$ = 15,

when n = 1
U1 = k(1) $^{2}$ + p(1) = -1
k + p = -1 --------eqn1
when n = 5
U $_{5}$ = k(5) $^{2}$ + p(5) = 15
25k + 5p = 15 --------eqn2
multiply eqn1 by 5 and eqn2 by 1
5k + 5p = -5 -------eqn3
25k + 5p = 15 -------eqn4
eqn4 - eqn3
20k = 20
k = 1
sub for k in eqn1
1 + p = -1
p = -1 -1 = -2

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Polynomial Functions

Duba Amsa

Tambaya 33 Rahoto

The mean heights of three groups of students consisting of 20, 16 and 14 students each are 1.67m, 1.50m and 1.40m respectively. Find the mean height of all the students.

1.63m

1.54m

1.52m

1.42m

Statistics

Duba Amsa

Tambaya 34 Rahoto

In how many ways can six persons be paired?

5

10

15

20

Bayanin Amsa

6C $_{2} = \frac{6!}{[6 - 2]! [2!]}$

$\frac{6 * 5 * 4!}{4! * 2!}$

= $\frac{6 * 5}{2}$

= 15

Karanta bayanin darasi akan Permutation And Combinations (WAEC)

Permutation And Combinations

Duba Amsa

Tambaya 35 Rahoto

The first, second and third terms of an exponential sequence (G.P) are (x - 4), (x + 2), and (3x + 1) respectively. Find the values of x.

\frac{- 1}{2}, 8

\frac{1}{2}, - 8

\frac{- 1}{2}, - 8

\frac{1}{2}, 8

Bayanin Amsa

U1 = x - 4
U2 = x + 2
U3 = 3x + 1

$\frac{u_{2}}{u_{1}} = \frac{u_{3}}{u_{2}}$

$\frac{x + 2}{x - 4} = \frac{3 x + 1}{x + 2}$

(x+2)(x+2) = (x-4)(3x+1)
x $^{2}$ + 4x + 4 = 3x2 - 11x - 4
collecting like terms
2x $^{2}$ - 15x - 8 =0
2x $^{2}$ + x - 16x - 8 = 0
x(2x + 1) - 8(2x + 1) = 0
(x-8)(2x+1) = 0

x = ( $\frac{- 1}{2}, 8$ )

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Polynomial Functions

Duba Amsa

Tambaya 36 Rahoto

The table shows the distribution of the distance (in km) covered by 40 hunters while hunting.

What is the mode of the distribution?

Distance(km)	3	4	5	6	7	8
Frequency	5	4	x	9	2x	1

5

6

7

8

Statistics

Duba Amsa

Tambaya 37 Rahoto

A particle moving with a velocity of 5m/s accelerates at 2m/s $^{2}$ . Find the distance it covers in 4 seconds.

16m

26m

36m

46m

Bayanin Amsa

from the equation of motion

u = 5m/s, a = 2m/s $^{2}$ , t = 4s

s = ut + $\frac{1}{2} a t^{2}$

s = 5*4 + $\frac{1}{2} 2 * 4^{2}$

s = 20 + 16

s = 36m

Duba Amsa

Tambaya 38 Rahoto

Differentiate $\frac{5 x^{3} + x^{2}}{x}$ , x ≠ 0 with respect to x.

10x + 1

10x + 2

x(15x + 1)

x(15x + 2)

Bayanin Amsa

$\frac{5 x^{3} + x^{2}}{x}$ → $\frac{5 x^{3}}{x} + \frac{x^{2}}{x}$

5x $^{2}$ + x

Then dy/dx = 10x + 1

Duba Amsa

Tambaya 39 Rahoto

Simplify $\frac{9 * 3^{n + 1} - 3^{n + 2}}{3^{n + 1} - 3^{n}}$

3

9

27

81

Bayanin Amsa

$\frac{9 * 3^{n + 1} - 3^{n + 2}}{3^{n + 1} - 3^{n}}$

= $\frac{3^{n} * 3^{n} * 3^{1} * - 3^{2} * 3^{2}}{3^{n} * 3^{1} - 3^{n}}$

= $\frac{3^{n} (3^{2} * 3^{1})}{3^{n} (3^{1} - 1)}$

= $\frac{27 - 9}{3 - 1}$

= $\frac{18}{2}$

= 9

Karanta bayanin darasi akan Functions (WAEC)

Functions

Duba Amsa

Tambaya 40 Rahoto

Find correct to the nearest degree, the acute angle formed by the lines y = 2x + 5 and 2y = x - 6

76

^{\circ}

53

^{\circ}

37

^{\circ}

14

^{\circ}

Bayanin Amsa

tanθ

=

m1 - m21 + m1m2

y = 2x + 5
m1 = 2
2y = x - 6

y

=

12

x

-

3

m2

=

12

tanθ

=

2 -

\frac{1}{2}

1+2(

\frac{1}{2}

)

tanθ = $\frac{3}{2}$ ÷ (1+1)

tanθ = $\frac{3}{2}$ ÷ 2

tanθ

=

34

θ = $t a n^{- 1} (\frac{3}{4})$

θ = 36.87º
θ = 37º

Karanta bayanin darasi akan Co-ordinate Geometry (WAEC)

Co-ordinate Geometry

Duba Amsa

Tambaya 41 Rahoto

Express $\frac{4 π}{2}$ radians in degrees.

288º

200º

144º

120º

Bayanin Amsa

$\frac{4 π}{2}$ = $\frac{4}{5} * 180$

= 144º

Karanta bayanin darasi akan Trigonometry (WAEC)

Trigonometry

Duba Amsa

Tambaya 42 Rahoto

( $\frac{3 \sqrt 6}{\sqrt 5} + \frac{\sqrt 54}{3 \sqrt 5}$ ) $^{- 1}$

\frac{5 \sqrt 3}{6}

\frac{3 \sqrt 15}{6}

\frac{5 \sqrt 6}{12}

\frac{5 \sqrt 3}{12}

Bayanin Amsa

( $\frac{3 \sqrt 6}{\sqrt 5} + \frac{\sqrt 54}{3 \sqrt 5}$ ) $^{- 1}$

= $\frac{\sqrt 5 (3 \sqrt 5)}{3 \sqrt 6 + 3 \sqrt 6}$

= $\frac{3 * 5}{6 \sqrt 6} = \frac{5}{2 \sqrt 6}$

= $\frac{5 * 2 \sqrt 6}{2 \sqrt 6 + 2 \sqrt 6} = \frac{10 \sqrt 6}{4 * 6}$

= $\frac{5 \sqrt 6}{12}$

Karanta bayanin darasi akan Surds (WAEC)

Surds

Duba Amsa

Tambaya 43 Rahoto

Which of the following is the semi-interquartile range of a distribution?

Mode - Median

Highest score - Lowest score

1/2 (Upper Quartile - Median)

1/2 (Upper Quartile - Lower Quartile)

Statistics

Duba Amsa

Tambaya 44 Rahoto

A body of mass 18kg moving with velocity 4ms-1 collides with another body of mass 6kg moving in the opposite direction with velocity 10ms-1. If they stick together after the collision, find their common velocity.

\frac{1}{2}

m/s

\frac{1}{3}

m/s

2m/s

3m/s

Bayanin Amsa

m1u1 + m2u2 = (m1 + m2)v

m1 = 18kg, m2 = 6kg, u1 = 4ms-1, u2 = -10m/s

18(4) + 6(-10) = (18+6)v

72 - 60 = 24v
12 = 24v
v = $\frac{1}{2}$ m/s

Karanta bayanin darasi akan Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

Duba Amsa

Tambaya 45 Rahoto

Two functions f and g are defined on the set of real numbers, R, by

f:x → x $^{2}$ + 2 and g:x → $\frac{1}{x + 2}$ .Find the domain of (g∘f) $^{- 1}$

Amsa

f:x → x $^{2}$ + 2 and g:x → $\frac{1}{x + 2}$ .

(g∘f) $^{- 1}$

(g∘f) $^{x^{2} + 2}$

g $^{x^{2} + 2}$ = $\frac{1}{(x^{2} + 2) + 2}$

(g∘f) = $\frac{1}{(x^{2} + 4}$

let y = $\frac{1}{(x^{2} + 4}$

y $((x^{2} + 4)$ = 1

$y x^{2} + 4 y = 1$

x $^{2}$ = $\frac{1 - 4 y}{y}$

x = $\sqrt{\frac{1 - 4 y}{y}}$

(g∘f) $^{- 1}$ = $\sqrt{\frac{1 - 4 x}{x}}$

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Karanta bayanin darasi akan Functions (WAEC)

Bayanin Amsa

f:x → x $^{2}$ + 2 and g:x → $\frac{1}{x + 2}$ .

(g∘f) $^{- 1}$

(g∘f) $^{x^{2} + 2}$

g $^{x^{2} + 2}$ = $\frac{1}{(x^{2} + 2) + 2}$

(g∘f) = $\frac{1}{(x^{2} + 4}$

let y = $\frac{1}{(x^{2} + 4}$

y $((x^{2} + 4)$ = 1

$y x^{2} + 4 y = 1$

x $^{2}$ = $\frac{1 - 4 y}{y}$

x = $\sqrt{\frac{1 - 4 y}{y}}$

(g∘f) $^{- 1}$ = $\sqrt{\frac{1 - 4 x}{x}}$

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Polynomial Functions (WAEC)

Karanta bayanin darasi akan Functions (WAEC)

Sets

Polynomial Functions

Functions

Duba Amsa

Tambaya 46 Rahoto

A solid rectangular block has a base that measures 3x cm by 2x cm. The height of the block is ycm and its volume is 72cm $^{3}$ .

i. Express y in terms of x.

ii. An expression for the total surface area of the block in terms of x only;

iii. the value of x for which the total surface area has a stationary value.

Amsa

The volume of a solid rectangular block is given by the formula V = lwh, where l, w, and h are the length, width, and height of the block, respectively. In this problem, we are given that the base of the block has dimensions 3x cm by 2x cm, so we have l = 3x cm and w = 2x cm. The height of the block is y cm, so h = y cm. We are also given that the volume of the block is 72 cm³, so we have:

V = lwh

72 = (3x)(2x)(y)

72 = 6x^2y

Solving for y, we get:

y = 72/6x^2

y = 12/x^2

Therefore, the height of the block is 12/x^2 cm.

b.

To find the total surface area of the solid rectangular block, we need to consider the six faces of the block: the top face, bottom face, front face, back face, left face, and right face.

Given:
Base length = 3x cm
Base width = 2x cm
Height = y cm
Volume = 72 cm^3

The volume of a rectangular block is given by the formula:

Volume = Base Area * Height

Therefore, we can write the equation:

72 cm^3 = (3x cm * 2x cm) * y cm

Simplifying this equation, we have:

72 = 6x^2 * y

Now, let's express the total surface area of the block in terms of x only.

The total surface area of the block can be calculated by adding the areas of all six faces:

Total Surface Area = 2 * (Base Area) + (Front Face Area) + (Back Face Area) + (Left Face Area) + (Right Face Area)

The base area is given by:

Base Area = Length * Width = (3x cm) * (2x cm) = 6x^2 cm^2

The front face and back face both have the same dimensions, so their areas are equal:

Front Face Area = Back Face Area = Length * Height = (3x cm) * (y cm) = 3xy cm^2

Similarly, the left face and right face both have the same dimensions, so their areas are equal:

Left Face Area = Right Face Area = Width * Height = (2x cm) * (y cm) = 2xy cm^2

Now, let's substitute these values into the equation for the total surface area:

Total Surface Area = 2 * (6x^2 cm^2) + 2 * (3xy cm^2) + 2 * (2xy cm^2)

Simplifying further, we have:

Total Surface Area = 12x^2 cm^2 + 6xy cm^2 + 4xy cm^2

Finally, we can express the total surface area of the block in terms of x only as:

Total Surface Area = 12x^2 cm^2 + 10xy cm^2

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Trigonometry (WAEC)

Bayanin Amsa

The volume of a solid rectangular block is given by the formula V = lwh, where l, w, and h are the length, width, and height of the block, respectively. In this problem, we are given that the base of the block has dimensions 3x cm by 2x cm, so we have l = 3x cm and w = 2x cm. The height of the block is y cm, so h = y cm. We are also given that the volume of the block is 72 cm³, so we have:

V = lwh

72 = (3x)(2x)(y)

72 = 6x^2y

Solving for y, we get:

y = 72/6x^2

y = 12/x^2

Therefore, the height of the block is 12/x^2 cm.

b.

To find the total surface area of the solid rectangular block, we need to consider the six faces of the block: the top face, bottom face, front face, back face, left face, and right face.

Given:
Base length = 3x cm
Base width = 2x cm
Height = y cm
Volume = 72 cm^3

The volume of a rectangular block is given by the formula:

Volume = Base Area * Height

Therefore, we can write the equation:

72 cm^3 = (3x cm * 2x cm) * y cm

Simplifying this equation, we have:

72 = 6x^2 * y

Now, let's express the total surface area of the block in terms of x only.

The total surface area of the block can be calculated by adding the areas of all six faces:

Total Surface Area = 2 * (Base Area) + (Front Face Area) + (Back Face Area) + (Left Face Area) + (Right Face Area)

The base area is given by:

Base Area = Length * Width = (3x cm) * (2x cm) = 6x^2 cm^2

The front face and back face both have the same dimensions, so their areas are equal:

Front Face Area = Back Face Area = Length * Height = (3x cm) * (y cm) = 3xy cm^2

Similarly, the left face and right face both have the same dimensions, so their areas are equal:

Left Face Area = Right Face Area = Width * Height = (2x cm) * (y cm) = 2xy cm^2

Now, let's substitute these values into the equation for the total surface area:

Total Surface Area = 2 * (6x^2 cm^2) + 2 * (3xy cm^2) + 2 * (2xy cm^2)

Simplifying further, we have:

Total Surface Area = 12x^2 cm^2 + 6xy cm^2 + 4xy cm^2

Finally, we can express the total surface area of the block in terms of x only as:

Total Surface Area = 12x^2 cm^2 + 10xy cm^2

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Trigonometry (WAEC)

Sets

Trigonometry

Duba Amsa

Tambaya 47 Rahoto

Evaluate: lim $_{x \to - 2}$ $\frac{x^{3} + 8}{x + 2}$ .

Amsa

$\frac{x^{3} + 8}{x + 2}$

x $^{3}$ + 8 = x $^{3}$ + 23
recall that
x $^{3}$ + y $^{3}$ = (x+y)(x $^{2}$ - xy + y $^{2}$ )
x = x, y = 2
x $^{3}$ + 8 = (x+2)(x $^{3}$ - 2(x) + 22)

$\frac{x^{3} + 8}{x + 2} = \frac{(x + 2) (x^{3} - 2 (x) + 22)}{x + 2}$ - 2x + 4

x = -2
(-2) $^{2}$ - 2(2) + 4 = 4 - 4 + 4

= 4

Karanta bayanin darasi akan Differentiation (WAEC)

Bayanin Amsa

$\frac{x^{3} + 8}{x + 2}$

x $^{3}$ + 8 = x $^{3}$ + 23
recall that
x $^{3}$ + y $^{3}$ = (x+y)(x $^{2}$ - xy + y $^{2}$ )
x = x, y = 2
x $^{3}$ + 8 = (x+2)(x $^{3}$ - 2(x) + 22)

$\frac{x^{3} + 8}{x + 2} = \frac{(x + 2) (x^{3} - 2 (x) + 22)}{x + 2}$ - 2x + 4

x = -2
(-2) $^{2}$ - 2(2) + 4 = 4 - 4 + 4

= 4

Karanta bayanin darasi akan Differentiation (WAEC)

Differentiation

Duba Amsa

Tambaya 48 Rahoto

Given that p = (8N,030º) and q = (9N, 150º), find, in component from, the unit vector along(p - q).

Amsa

p = 8cos30i + 8sin30j
q = 9cos150i + 9sin150j
p = 6.9282i + 4j
q = -7.7942i + 4.5j
pq = (-7.7942 - 6.9282)i + (4.5 - 4)j
pq = -14.7224i + 0.5j
|pq| = √((-14.7224)2 + (0.5)2)
|pq| = √(216.7491 + 0.25)
|pq| = √216.9991
|pq| = 14.7309

the unit vector in the direction of p - q = $\frac{- 14.7224 i + 0.5 j}{14.7309}$

-0.9994i + 0.0339j ≌ -i + 0.0339j

Karanta bayanin darasi akan Vectors (WAEC)

Bayanin Amsa

p = 8cos30i + 8sin30j
q = 9cos150i + 9sin150j
p = 6.9282i + 4j
q = -7.7942i + 4.5j
pq = (-7.7942 - 6.9282)i + (4.5 - 4)j
pq = -14.7224i + 0.5j
|pq| = √((-14.7224)2 + (0.5)2)
|pq| = √(216.7491 + 0.25)
|pq| = √216.9991
|pq| = 14.7309

the unit vector in the direction of p - q = $\frac{- 14.7224 i + 0.5 j}{14.7309}$

-0.9994i + 0.0339j ≌ -i + 0.0339j

Karanta bayanin darasi akan Vectors (WAEC)

Vectors

Duba Amsa

Tambaya 49 Rahoto

The vectors 6i + 8j and 8i - 6j are parallel to ?OP and ?OQ respectively. If the magnitude of ?OP and ?OQ are 80 units and 120 units respectively, express: ?OP and ?OQ in terms of i and j;

ii. |?PQ|, in the form c?k, where c and k are constants.

Amsa

Vectors

Duba Amsa

Tambaya 50 Rahoto

The probability that Abiola will be late to the office on a given day is 2/5. In a given working week of six days, find, correct to four significant figures, the probability that he will:

(a) only be late for 3 days.

(b) not be late in the week:

(c) be late throughout the six days.

Amsa

he will be late to office = 2/5
he will not be late to office is = 1 - 2/5 = 3/5

If he will be late for 3 days only, he will also not be late for 3 days

$\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$

p = 0.0138

(b) not be late to office is = 1 - 2/5 = 3/5

$\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$

p = 0.0467

(c) be late to office = 2/5

$\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$

p = 0.0041

Karanta bayanin darasi akan Probability (WAEC)

Bayanin Amsa

he will be late to office = 2/5
he will not be late to office is = 1 - 2/5 = 3/5

If he will be late for 3 days only, he will also not be late for 3 days

$\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$

p = 0.0138

(b) not be late to office is = 1 - 2/5 = 3/5

$\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$ * $\frac{3}{5}$

p = 0.0467

(c) be late to office = 2/5

$\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$ * $\frac{2}{5}$

p = 0.0041

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 51 Rahoto

Solve 3cos2x - sinx = 0 for 0º≤x≤360º

Amsa

3cos2x - sinx = 0
cos 2A = 1 – 2 sin $^{2}$ A
3(1 - 2sin $^{2}$ x) - sinx = 0
3 - 6sin $^{2}$ x -sinx = 0
let sinx = p
3 - 6p $^{2}$ -p = 0
-6p2 - p +3 = 0

factors of p in the eqn:

p = 0.7953 or 0.6287
when p = 0.7953
p = sinx
x = sin $^{- 1}$ p
x = sin $^{- 1}$ (0.7953)
x = 52.68º

when p = 0.6287
p = sinx
x = sin $^{- 1}$ p
x = sin $^{- 1}$ (0.6287)
x = 38.95º

Karanta bayanin darasi akan Trigonometry (WAEC)

Karanta bayanin darasi akan Co-ordinate Geometry (WAEC)

Bayanin Amsa

3cos2x - sinx = 0
cos 2A = 1 – 2 sin $^{2}$ A
3(1 - 2sin $^{2}$ x) - sinx = 0
3 - 6sin $^{2}$ x -sinx = 0
let sinx = p
3 - 6p $^{2}$ -p = 0
-6p2 - p +3 = 0

factors of p in the eqn:

p = 0.7953 or 0.6287
when p = 0.7953
p = sinx
x = sin $^{- 1}$ p
x = sin $^{- 1}$ (0.7953)
x = 52.68º

when p = 0.6287
p = sinx
x = sin $^{- 1}$ p
x = sin $^{- 1}$ (0.6287)
x = 38.95º

Karanta bayanin darasi akan Trigonometry (WAEC)

Karanta bayanin darasi akan Co-ordinate Geometry (WAEC)

Trigonometry

Co-ordinate Geometry

Duba Amsa

Tambaya 52 Rahoto

A particle initially at rest moves in a straight line with an acceleration of (10t - 4t $^{2}$ )m/s $^{2}$
Find the:
a. velocity of the particle after t seconds;

ii. average acceleration of the particle during the 4th second.

b. A load of mass 120kg is placed on a lift. Calculate the reaction between the floor of the lift and the load when the lift moves upwards at a constant velocity. [Take g = 10m/s $^{2}$ ]

ii. with an acceleration of 3m/s $^{2}$ . [Take g = 10m/s $^{2}$ ]

Amsa

a. a = (10t - 4t $^{2}$ )m/s $^{2}$ , t = t seconds, u = 0
from first equation of motion,
v = u + at
v = 0 + (10t - 4t $^{2}$ )t
v = (10t $^{2}$ - 4t $^{3}$ )m/s $^{2}$

ii. average acceleration of the particle during the 4th second = the average acceleration between the t = 3 and t = 4
Δa4 = 10(t4-t3) - 4(t4-t3) $^{2}$
Δa4 = 10(4-1) - 4(4-3) $^{2}$
Δa4 = 10(1) - 4(12)
Δa4 = (10-4)m/s $^{2}$
Δa4 = 6ms-2

b. R = m(g+a)
When the lift moves with a constant velocity, a = 0
R = mg
R = 120 × 10
R = 1200N

ii. when the lift moves upward with an acceleration of 3m/s $^{2}$
R = m(g+a)
R = 120(10+3)
R = 120 × 13
R = 1560N

Bayanin Amsa

a. a = (10t - 4t $^{2}$ )m/s $^{2}$ , t = t seconds, u = 0
from first equation of motion,
v = u + at
v = 0 + (10t - 4t $^{2}$ )t
v = (10t $^{2}$ - 4t $^{3}$ )m/s $^{2}$

ii. average acceleration of the particle during the 4th second = the average acceleration between the t = 3 and t = 4
Δa4 = 10(t4-t3) - 4(t4-t3) $^{2}$
Δa4 = 10(4-1) - 4(4-3) $^{2}$
Δa4 = 10(1) - 4(12)
Δa4 = (10-4)m/s $^{2}$
Δa4 = 6ms-2

b. R = m(g+a)
When the lift moves with a constant velocity, a = 0
R = mg
R = 120 × 10
R = 1200N

ii. when the lift moves upward with an acceleration of 3m/s $^{2}$
R = m(g+a)
R = 120(10+3)
R = 120 × 13
R = 1560N

Duba Amsa

Tambaya 53 Rahoto

A basket contains 12 fruits: orange, apple and avocado pear, all of the same size. The number of oranges, apples and avocado pear forms three consecutive integers.

Two fruits are drawn one after the other without replacement. Calculate the probability that:

i. the first is an orange and the second is an avocado pear.

ii.both are of same fruit;

iii. at least one is an apple

Amsa

They form 3 consecutive integers
let the number of orange be x, apple = x+1 and avocado pear = x+2
x + x+1 + x+2 = 12
3x + 3 = 12
3x = 9
x = 3
apple = 4
avocado pear = 5
probability that orange is drawn = $\frac{3}{12}$ = $\frac{1}{4}$

probability that apple is drawn = $\frac{4}{12}$ = $\frac{1}{3}$

probability that avocado pear is drawn = $\frac{5}{12}$

probability that the first is orange and the second is an avocado pear = $\frac{3}{12}$ * $\frac{5}{11}$ = $\frac{5}{44}$

ii.

both are of same fruit, they are both Orange or apple or Avocado pear

p(O∩O) + p(A∩A) + p(P∩P) = $\frac{1}{4} * \frac{2}{11} + \frac{1}{3} * \frac{3}{11} + \frac{5}{12} * \frac{4}{11}$

= $\frac{1}{22} + \frac{1}{11} + \frac{5}{33}$

= $\frac{19}{66}$

iii.

at least one is an apple = p(A∩O) or p(O∩A) or p(A∩P) or p(P∩A) or p(A∩A)
p(A∩O) + p(O∩A) + p(A∩P) + p(P∩A) + p(A∩A) =

= $\frac{4}{12} * \frac{3}{11} + \frac{3}{11} * \frac{4}{12} + \frac{4}{12} * \frac{5}{11} + \frac{5}{12} * \frac{4}{11} + \frac{4}{12} * \frac{3}{11}$

= $\frac{12}{132}$ + $\frac{12}{132}$ + $\frac{20}{132}$ + $\frac{20}{132}$ + $\frac{12}{132}$

= $\frac{76}{132}$

= $\frac{19}{33}$

Karanta bayanin darasi akan Probability (WAEC)

Bayanin Amsa

They form 3 consecutive integers
let the number of orange be x, apple = x+1 and avocado pear = x+2
x + x+1 + x+2 = 12
3x + 3 = 12
3x = 9
x = 3
apple = 4
avocado pear = 5
probability that orange is drawn = $\frac{3}{12}$ = $\frac{1}{4}$

probability that apple is drawn = $\frac{4}{12}$ = $\frac{1}{3}$

probability that avocado pear is drawn = $\frac{5}{12}$

probability that the first is orange and the second is an avocado pear = $\frac{3}{12}$ * $\frac{5}{11}$ = $\frac{5}{44}$

ii.

both are of same fruit, they are both Orange or apple or Avocado pear

p(O∩O) + p(A∩A) + p(P∩P) = $\frac{1}{4} * \frac{2}{11} + \frac{1}{3} * \frac{3}{11} + \frac{5}{12} * \frac{4}{11}$

= $\frac{1}{22} + \frac{1}{11} + \frac{5}{33}$

= $\frac{19}{66}$

iii.

at least one is an apple = p(A∩O) or p(O∩A) or p(A∩P) or p(P∩A) or p(A∩A)
p(A∩O) + p(O∩A) + p(A∩P) + p(P∩A) + p(A∩A) =

= $\frac{4}{12} * \frac{3}{11} + \frac{3}{11} * \frac{4}{12} + \frac{4}{12} * \frac{5}{11} + \frac{5}{12} * \frac{4}{11} + \frac{4}{12} * \frac{3}{11}$

= $\frac{12}{132}$ + $\frac{12}{132}$ + $\frac{20}{132}$ + $\frac{20}{132}$ + $\frac{12}{132}$

= $\frac{76}{132}$

= $\frac{19}{33}$

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 54 Rahoto

A body of mass of 18kg is suspended by an inextensible string from a rigid support and is pulled by a horizontal force F until the angle of inclination of the string to the vertical is 35º. If the system is in equilibrium, calculate the:

i. value of F

ii. tension in the string

Amsa

If the system is in equilibrium, then the weight of the body = the force pulling it
and the force = the vertical component of force F
since Force F is inclined to vertical then the vertical component = Fcos35º
the weight of the body = mg
Fcos35º = 18(10)
0.8192F = 180

F = $\frac{180}{0.8192}$

F = 219.7N

tension in the string

T = mg
T = 18kg × 10
T = 180N

Karanta bayanin darasi akan Statics (WAEC)

Bayanin Amsa

If the system is in equilibrium, then the weight of the body = the force pulling it
and the force = the vertical component of force F
since Force F is inclined to vertical then the vertical component = Fcos35º
the weight of the body = mg
Fcos35º = 18(10)
0.8192F = 180

F = $\frac{180}{0.8192}$

F = 219.7N

tension in the string

T = mg
T = 18kg × 10
T = 180N

Karanta bayanin darasi akan Statics (WAEC)

Statics

Duba Amsa

Tambaya 55 Rahoto

A binary operation * is defined on the set T = {-2,-1,1,2} by p*q = p $^{2}$ + 2pq - q $^{2}$ , where p,q ∊ T.
Copy and complete the table.

*	-2	-1	1	2
-2		7		-8
-1		2	-2
1	-7			1
2		-1

Amsa

p*q = p $^{2}$ + 2pq - q $^{2}$

when p = -2 and q = -2

p*q = -2 $^{2}$ + 2(-2)(-2) - (-2) $^{2}$
p*q = 4 + 8 - 4 = 8

when p = -2 and q = -1
p*q = -2 $^{2}$ + 2(-2)(-1) - (-1) $^{2}$
p*q = 4 + 4 - 1 = 7

when p = -2 and q = 1
p*q = -2 $^{2}$ + 2(-2)(1) - (1) $^{2}$
p*q = 4 - 4 -1 = -1

when p = -1 and q = -2
p*q = -1 $^{2}$ + 2(-1)(-2) - (-2) $^{2}$
p*q = 1 + 4 - 4 = 1

when p = -1 and q = -1
p*q = -1 $^{2}$ + 2(-1)(-1) - (-1) $^{2}$
p*q = 1 + 2 -1 = 2

when p = -1 and q = 2
p*q = -1 $^{2}$ + 2(-1)(2) - (2) $^{2}$
p*q = 1 - 4 - 4 = -7

when p = 1 and q = -2
p*q = 1 $^{2}$ + 2(1)(-2) - (-2) $^{2}$
p*q = 1 - 4 - 4 = -7

when p = 1 and q = -1
p*q = 1 $^{2}$ + 2(1)(-1) - (-1) $^{2}$
p*q = 1 - 2 - 1 = -2

when p = 1 and q = 1
p*q = 1 $^{2}$ + 2(1)(1) - (1) $^{2}$
p*q = 1 + 2 - 1 = 2

when p = 1 and q = 2
p*q = 1 $^{2}$ + 2(1)(2) - (2) $^{2}$
p*q = 1 + 4 - 4 = 1

when p = 2 and q = -2
p*q = 2 $^{2}$ + 2(2)(-2) - (-2) $^{2}$
p*q = 4 - 8 - 4 = -8

when p = 2 and q = -1
p*q = 2 $^{2}$ + 2(2)(-1) - (-1) $^{2}$
p*q = 4 - 4 - 1 = -1

when p = 2 and q = 1
p*q = 2 $^{2}$ + 2(2)(1) - (1) $^{2}$
p*q = 4 + 4 - 1 = 7
when p = 2 and q = 2
p*q = 2 $^{2}$ + 2(2)(2) - (2) $^{2}$
p*q = 4 + 8 - 4 = 8

*	-2	-1	1	2
-2	8	7	-1	-8
-1	1	2	-2	-7
1	-7	-2	2	1
2	-8	-1	7	8