JAMB UTME - Mathematics - 1984

Pergunta 1 Relatório

A trader in country where their currency 'MONI'(M) is in base five bought 1035 oranges at M145 each. If he sold the oranges at M245 each, what will be his gain?

1035

10305

1025

20025

30325

Detalhes da Resposta

Total cost of 1035 oranges at ₦145 each
= 1035 x 145
= 20025
Total selling price at ₦245 each
= (103)5 x 245
= 30325
Hence his gain = 30325 - 20025
= 10305

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Pergunta 2 Relatório

Using $ \triangle $ XYZ in the figure, find XYZ.

29

^{\circ}

31

^{\circ}

31

^{\circ}

20'

31

^{\circ}

18'

Detalhes da Resposta

\frac{\sin y}{3} = \frac{\sin 120^{o}}{5}

sin 120 $^{\circ}$ = sin 60 $^{\circ}$

5 sin y = 3 sin 60 $^{\circ}$

sin y = $\frac{3 \sin 60^{o}}{5}$

$\frac{3 \times 0.866}{5}$

= $\frac{2.598}{5}$

y = sin-1 0.5196 = 30 $^{\circ}$ 18'

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Pergunta 3 Relatório

If f(x) = 2(x - 3)2 + 3(x - 3) + 4 and g(y) = 5 + y, find g [f(3)] and f[g(4)].

3 and 4

-3 and 4

-3 and -4

3 and -4

0 and 5

Detalhes da Resposta

f(x0 = 2(x - 3)2 + 3(x - 3) + 4
= (2 + 3) + (x - 3) + 4
5(x - 3) + 4
5x - 15 + 4
= 5x - 11
f(3) = 5 x 3 - 11
= 4

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Pergunta 4 Relatório

Bola choose at random a number between 1 and 300. What is the probability that the number is divisible by 4?

1

4

6

4

7

4

Detalhes da Resposta

Numbers divisible by 4 between 1 and 300 include 4, 8, 12, 16, 20 e.t.c. To get the number of figures divisible by 4, We solve by method of A.P
Let x represent numbers divisible by 4, nth term = a + (n - 1)d
a = 4, d = 4
Last term = 4 + (n - 1)4
288 = 4 + 4n - n
=

\frac{288}{4}

= 72
rn(Note: 288 is the last Number divisible by 4 between 1 and 300)
Prob. of x =

\frac{72}{288}

=

\frac{1}{4}

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Pergunta 5 Relatório

The sides of a triangle are $ (x + 4)\text{cm} $, $ x\text{cm} $ and $ (x - 4)\text{cm} $, respectively If the cosine of the largest angle is $ \frac{1}{5} $, find the value of $ x $

24cm

20cm

28cm

7cm

88

7

Detalhes da Resposta

< B is the largest since the side facing it is the largest, i.e. (x + 4)cm

Cosine B = $\frac{1}{5}$
= 0.2 given
b2 - a2 + c2 - 2a Cos B
Cos B = $\frac{a^{2} + c^{2} - b^{2}}{2 a c}$
$\frac{1}{5}$ = $\frac{x^{2} + ? (x - 4)^{2} - (x + 4)^{2}}{2 x (x - 4)}$
$\frac{1}{5}$ = $\frac{x (x - 16)}{2 x (x - 4)}$
$\frac{1}{5}$ = $\frac{x - 16}{2 x - 8}$
= 5(x - 16)
= 2x - 8
3x = 72
x = $\frac{72}{3}$
= 24

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Pergunta 6 Relatório

Factorize $6x^2 - 14x - 12$

2(x + 3)(3x - 2)

6(x - 2)(x + 1)

2(x - 3)(3x + 2)

6(x + 2)(x - 1)

(3x - 4)(2x + 3)

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Pergunta 7 Relatório

In the figure, 0 is the centre of circle PQRS and PS//RT. If PRT = 135, then PSO is

67

\frac{1}{2}

^{\circ}

22

\frac{1}{2}

^{\circ}

33

\frac{1}{2}

^{\circ}

45

^{\circ}

90

^{\circ}

Detalhes da Resposta

< R = 180

^{\circ}

- 45

^{\circ}

(sum of angles on a straight line)

< R = < P = 45 $^{\circ}$ (corresponding angles)

< PSO = < P = 45 $^{\circ}$ ( $△$ PSO is a right angle)

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Pergunta 8 Relatório

If 2x + 3y = 1 and x - y = 11, find (x + y).

5

3

8

2

11

Solution Of Linear Equations

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Pergunta 9 Relatório

Find the x co-ordinates of the points of intersection of the two equations in the graph.

1, 1

0, 4

0, 0

4, 0

4, 9

Detalhes da Resposta

If y = 2x + 1 and y = x2 - 2x + 1

then x2 - 2x + 1 = 2x + 1

x2 - 4x = 0

x(x - 4) = 0

x = 0 or 4

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Pergunta 10 Relatório

Find a factor which is common to all 3 binomial expressions $4a^2 - 9b^2,\; 8a^3 + 27b^3,\; (4a + 6b)^2$.

4a + 6b

4a - 6b

2a + 3b

2a - 3b

None

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Pergunta 11 Relatório

In the figure, TSP = PRQ, QR = 8cm, PR = 6cm and ST = 12cm. Find the length SP.

4cm

16cm

9cm

14cm

Impossible, insufficient data

Detalhes da Resposta

\frac{P Q}{6 + R T} = \frac{6}{12} = \frac{6}{P S}

(Similar triangles)

$\frac{6}{12} = \frac{8}{P S}$

PS = $\frac{12 \times 8}{6}$

= 16cm

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Pergunta 12 Relatório

In the figure, PQRSTW is a regular hexagon. QS intersects RT at V. Calculate TVS.

60

^{\circ}

90

^{\circ}

120

^{\circ}

30

^{\circ}

80

^{\circ}

Detalhes da Resposta

From the diagram, PQRSTW is a regular hexagon.

Hexagon is a six sided polygon.

Sum of interior angles of polygon = (2n - 4)90 $^{\circ}$ = [2 x 6 - 4] x 90 = 8 x 90 = 720 $^{\circ}$

each angle = $\frac{720^{o}}{6} = 120^{o}$ and TVS = $\frac{120}{2} = 60^{o}$

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Pergunta 13 Relatório

Find the angle of the sectors representing each item in pie chart of the following data 6, 10, 14, 16, 26

15o, 25o, 35o, 40o, 65o

60o, 100o, 140o, 160o, 260o

6o, 10o, 14o, 16o, 26o

30o, 50o, 70o, 80o, 130o

none of the above

Detalhes da Resposta

6 + 10 + 14 + 16 + 26 = 72

\frac{6}{72}

x 360
= 30o
Similarly others give 30o, 50o, 70o, 80o and 130o respectively

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Pergunta 14 Relatório

Simplify $ (1 + \frac{\frac{x - 1}{1}}{\frac{1}{x + 1}})(x + 2) $

(x2 - 1)(x + 2)

x2(x + 2)

x2 + 34

\frac{3 x^{2} - 1}{(x - 1)}

Change Of Subject Of A Formula/Relation

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Pergunta 15 Relatório

Two fair dice are rolled. What is the probability that both show up the same number of points.

1

36

7

36

1

2

1

3

1

6

Detalhes da Resposta

A dice has 6 faces, 2 dice has 6 x 6 = 36 combined face
Prob. of both showing the same number of points
=

\frac{6}{36}

=

\frac{1}{6}

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Pergunta 16 Relatório

If $pq + 1 = q^2$ and $t = \frac{1}{p} - \frac{1}{pq}$ express $t$ in terms of $q$

1

P-Q

1

q-1

1

q+1

1 + q

1

1-q

Detalhes da Resposta

Pq + 1 = q2......(i)
t =

\frac{1}{p}

-

\frac{1}{p q}

.........(ii)
p =

\frac{q^{2} - 1}{q}

Sub for p in equation (ii)
t =

\frac{1}{q^{2} - \frac{1}{q}}

-

\frac{1}{\frac{q^{2} - 1}{q} \times q}

t =

\frac{q}{q^{2} - 1}

-

\frac{1}{q^{2} - 1}

t =

\frac{q - 1}{q^{2} - 1}

=

\frac{q - 1}{(q + 1) (q - 1)}

=

\frac{1}{q + 1}

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Pergunta 17 Relatório

If (x - 2) and (x + 1) are factors of the expression x3 + px2 + qx + 1, what is the sum of p and q

9

-3

3

17

3

2

3

Detalhes da Resposta

x3 + px2 + qx + 1 = (x - 1) Q(x) + R
x - 2 = 0, x = 2, R = 0,
4p + 2p = -9........(i)
x3 + px2 + qx + 1 = (x - 1)Q(x) + R
-1 + p - q + 1 = 0
p - q = 0.......(ii)
Solve the equation simultaneously
p =

\frac{- 3}{2}

q =

\frac{- 3}{2}

p + q =

\frac{3}{2}

-

\frac{3}{2}

=

\frac{- 6}{2}

= -3

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Pergunta 18 Relatório

A cone is formed by bending a sector of a circle Saving an angle or 210°. Find the radius of the base of the cone. If the diameter of the circle is 12cm.

7.00cm

1.75cm

21cm

3.50cm

2.21cm

Loci

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Pergunta 19 Relatório

The cost of production of an article is made up as follows: Labour ₦70, Power ₦15, Materials ₦30, Miscellaneous ₦5. Find the angle of the sector representing Labour in a pie chart

210o

105o

106o

175o

90o

Detalhes da Resposta

Total cost of production = ₦120.00
Labour Cost =

\frac{70}{120}

x

\frac{360^{o}}{1}

= 120o

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Pergunta 20 Relatório

If the price of oranges was raised by $ \frac{1}{2} $k per orange. The number of oranges a customer can buy for ?2.40 will be less by 16. What is the present price of an orange?

2

\frac{1}{2}

\frac{2}{15}

5

\frac{1}{3}

20

Detalhes da Resposta

Let x represent the price of an orange and
y represent the number of oranges that can be bought
xy = 240k, y =

\frac{240}{x}

.....(i)
If the price of an oranges is raised by

\frac{1}{2}

k per orange, number that can be bought for ?240 is reduced by 16
Hence, y - 16 =

\frac{240}{x + \frac{1}{2}}

=

\frac{480}{2 x + 1}

=

\frac{480}{2 x + 1}

.....(ii)
subt. for y in eqn (ii)

\frac{240}{x}

- 16
=

\frac{480}{2 x + 1}

=

\frac{240 - 16 x}{x}

=

\frac{480}{2 x + 1}

= (240 - 16x)(2x + 1)
= 480x
= 480x + 240 - 32x2 - 16
480x = 224 - 32x2
x2 = 7
x =

\sqrt{7}

= 2.5
= 2

\frac{1}{2}

k

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Pergunta 21 Relatório

Thirty boys and X girls sat for a test. The mean of the boys' scores and that of the girls were respectively 6 and 8. Find 8 if the total score was 468.

38

24

36

22

41

Detalhes da Resposta

Le the number of girls = A; Total score of boys = 30

\times

6 = 180
Total score of girls = A

\times

8 = 8A

∴

180 + 8A = 468
8A = 288
A = 36

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Pergunta 22 Relatório

What is the volume of this regular three dimensional figure?

160cm2

48cm2

120cm2

40cm2

Detalhes da Resposta

Volume of the three dimensional figures = v = A x h

A = $\frac{1}{2}$ x 4 x 3

= 6cm2

V = 6 x 8

= 48cm2

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Pergunta 23 Relatório

The quadratic equation whose roots are $1 - \sqrt{13}$ and $1 + \sqrt{13}$ is

x2 + (1 -

\sqrt{13}

)x + 1 +

\sqrt{13}

= 0

x2 - 2x - 12 = 0

x2 - 2x + 12 = 0

x2 + 12 + 2x2 = 0

Detalhes da Resposta

1 -

\sqrt{13}

and 1 +

\sqrt{13}

Product of roots = (1 -

\sqrt{13}

) (1 +

\sqrt{13}

) = -12
x2 - (sum of roots) x + (product of roots) = 0
x2 - 2x + 12 = 0

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Pergunta 24 Relatório

Simplify $ \frac{\left(\frac{2}{3}-\frac{1}{5}\right)-\frac{1}{3}\text{ of }\frac{2}{5}}{3-\frac{1}{1\frac{1}{2}}} $

1

7

1

3

Detalhes da Resposta

$\frac{2}{3} - \frac{1}{5}$ = $\frac{10 - 3}{15}$
= $\frac{7}{15}$
$\frac{1}{3}$ Of $\frac{2}{5}$ = $\frac{1}{3}$ x $\frac{2}{5}$
= $\frac{2}{5}$
( $\frac{2}{3} - \frac{1}{5}$ ) - $\frac{1}{3}$ of $\frac{2}{5}$
= $\frac{7}{15} - \frac{2}{15}$ = $\frac{1}{3}$
3 - $\frac{1}{1 \frac{1}{2}}$ = 3 - $\frac{2}{3}$
= $\frac{7}{3}$
$\frac{\frac{2}{3} - \frac{1}{5} of \frac{2}{15}}{3 - \frac{1}{1 \frac{1}{2}}}$
= $\frac{\frac{1}{3}}{\frac{7}{3}}$
= $\frac{1}{3}$ x $\frac{3}{7}$
= $\frac{1}{7}$

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Pergunta 25 Relatório

The cumulative frequency function of the data below is given below by the equation $y = cf(x)$. What is $cf(5)$?

Score($n$)	Frequency($f$)
3	30
4	32
5	30
6	35
8	20

30

32

55

62

92

Detalhes da Resposta

If y = cf(x)
cf(5) = 30 + 32 + 30
= 92

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Pergunta 26 Relatório

If $ \sin \theta = \frac{x}{y} $ and $0^o < 90^o$, then find $ \frac{1}{\tan \theta} $.

x

√y²−x²

y

x

√y²+ x²

y²−x²

x

Detalhes da Resposta

\frac{1}{t a n θ}

=

\frac{c o s θ}{s i n θ}

sin

θ

=

\frac{x}{y}

cos

θ

=

\frac{\sqrt{y^{2} - x^{2}}}{y}

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Pergunta 27 Relatório

In a racing competition, Musa covered a distance 5x km in the first hour and (x + 10)km in the next hour. He was second to Nzozi who covered a total distance of 118km in the two hours. Which of the following inequalities is correct?

-1 < x < x < 0

-3 < x < 3

0

\leq

x < 18

15 < x < 18

Detalhes da Resposta

Total distance covered by Musa in 2 hrs
= x + 10 + 5x
= 6x + 10
Ngozi = 118 km
If they are equal, 6x + 10 = 188
but 6x + 10 < 118

6x < 108

= x < 18

0 < x < 18 = 0 $\leq$ x < 18

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Pergunta 28 Relatório

If $a = \frac{2x}{1-x}$ and $b = \frac{1+x}{1-x}$, then $a^2 - b^2$ in the simplest form is $\frac{3x+1}{1-x}$

\frac{3 x + 1}{1 - x}

3x²−1

(x−1)²

3x−2

x−1

3x−2

1−X

3x²−1

(x−1)

Detalhes da Resposta

a2 - b2 = (

\frac{2 x}{1 - x}

)2 - (

\frac{1 + x}{1 - x}

)2
= (

\frac{2 x}{1 - x} + \frac{1 + x}{1 - x}

)(

\frac{2 x}{1 - x} - \frac{1 + x}{1 - X}

)
= (

\frac{3 x + 1}{1 - x}

)(

\frac{x - 1}{1 - x}

)
=

\frac{3 x + 1}{x - 1}

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Pergunta 29 Relatório

A variable point p(x, y) traces a graph in a two-dimensional plane. (0, 3) is one position of P. If x increases by 1 unit, y increases by 4 units. The equation of the graph is

-3 =

\frac{y + 4}{x + 1}

4y = -3 + x

\frac{y}{x}

=

\frac{- 3}{4}

4x = y + 3

Detalhes da Resposta

P(x, y), P(0, 3) If x increases by 1 unit and y by 4 units, then ratio of x : y = 1 : 4

\frac{x}{1}

=

\frac{y}{4}

y = 4x
Hence the sign of the graph is y + 3 = 4x

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Pergunta 30 Relatório

Find the area of the shaded portion of the semicircular figure.

\frac{r^{2}}{4} (4 π - 3 \sqrt{3})

\frac{r^{2}}{4} (2 π - 3 \sqrt{3})

\frac{1}{2} r^{2} π

\frac{1}{8} r^{2} \sqrt{3}

\frac{r^{2}}{4} (4 π - 3 \sqrt{3})

Detalhes da Resposta

Asector =

\frac{60}{360} \times π r^{2}

= $\frac{1}{6} π r^{2}$

A $△$ = $\frac{1}{2} r^{2} \sin 60^{o}$

$\frac{1}{2} r^{2} \times \frac{\sqrt{3}}{2} = \frac{r^{2} \sqrt{3}}{4}$

A $shaded portion$ = Asector -
A $△$

= ( $\frac{1}{6} π r^{2} - \frac{r^{2} \sqrt{3}}{4})^{3}$

= $\frac{π r^{2}}{2} - \frac{3 r^{2} \sqrt{3}}{4}$

= $\frac{r^{2}}{4} (2 π - 3 \sqrt{3})$

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Pergunta 31 Relatório

Tunde and Shola can do a piece of work in 18 days. Tunde can do it alone in x days, whilst Shola takes 15 days longer to do it alone. Which of the following equations is satisfied by x?

x2 - 5x - 10 = 0

x2 - 20x + 360 = 0

x2 - 21x - 270 = 0

3x2 - 65x + 362 = 0

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Pergunta 32 Relatório

The table below is drawn for a graph $y = x^3 - 3x + 1$.

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y = x^3 - 3x + 1$	$1$	$-1$	$3$	$1$	$-1$	$3$	$1$

From x = -2 to x = 1, the graph crosses the x-axis in the range(s)

-1 < x < 0 and 0 < x < 1

-2 < x < -1 and 0 < x < 1

-2 < x < -1 and -1 < x < 0

0 < x < 1

Detalhes da Resposta

If the graph of y = x3 - 3x + 1 is plotted,the graph crosses the x-axis in the ranges -2 < x < -1 and 0 < x < 1

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Pergunta 33 Relatório

A man invested a total of ₦50000 in two companies. If these companies pay dividends of 6% and 8% respectively, how much did he invest at 8% if the total yield is ₦3700?

₦15000

₦29600

₦27800

₦21400

₦35000

Detalhes da Resposta

Total yield = ₦3,700
Total amount invested = ₦50,000
Let x be the amount invested at 6% interest and let y be the amount invested at 8% interest
then yield on x =

\frac{6}{100}

x and yield on y =

\frac{8}{100}

y
Hence,

\frac{6}{100}

x +

\frac{8}{100}

y
= 3,700.........(i)
x + y = 50,000........(ii)
6x + 8y = 370,000 x 1
x + y = 50,000 x 6
6x + 8y = 370,000.........(iii)
6x + 6y = 3000,000........(iv)
Eqn (iii) - Eqn (ii)
2y = 70,000
y =

\frac{70, 000}{2}

= 35,000
Money invested at 8% is ₦35,000

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Pergunta 34 Relatório

$ \dfrac{0.0001432}{1940000} = k \times 10^n $ where $ 1 \le k < 10 $ and n is a whole number. The values K and n are

7.381 qnd -11

2.34 and 10

3.871 and 2

7.831 and -11

Detalhes da Resposta

$\frac{0.0001432}{1940000}$ = k x 10n

where 1 $\leq$ k $\leq$ 10 and n is a whole number. Using four figure tables, the eqn. gives 7.38 x 10-11

k = 7.381, n = -11

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Pergunta 35 Relatório

A right circular cone has a base radius r cm and a vertical angle 2yo. The height of the cone is

r tan yo cm

r sin yo cm

r cot yo cm

r cos yo cm

r cosec yo cm

Detalhes da Resposta

\frac{r}{h}

= tan yo
h =

\frac{r}{t a n y^{o}}

= r cot yo

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Pergunta 36 Relatório

P sold his bicycle to Q at a profit of 10%. How much did the bicycle cost P?

₦200

₦196

₦180

₦205

₦150

Detalhes da Resposta

Let the selling price(SP from P to Q be represented by x
i.e. SP = x
When SP = x at 10% profit
CP =

\frac{100}{100}

+ 10 of x =

\frac{100}{110}

of x
when Q sells to R, SP = ₦209 at loss of 5%
Q's cost price = Q's selling price
CP =

\frac{100}{95}

x 209
= 220.00
x = 220
=

\frac{2200}{11}

= 200
= ₦200.00

Ler a nota de aula em Sine, Cosine And Tangent Of An Angle (WAEC)

Ler a nota de aula em Percentages (WAEC)

Ler a nota de aula em Modular Arithmetic (WAEC)

Sine, Cosine And Tangent Of An Angle

Percentages

Modular Arithmetic

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Pergunta 37 Relatório

Measurements of the diameters, in centimeters, in centimeters, of 20 copper spheres are distributed as shown below:

Class boundary in cm	frequency
3.35 − 3.45	3
3.45 − 3.55	6
3.55 − 3.65	7
3.65 − 3.75	4

What is the mean diameter of the copper spheres?

3.40cm

3.58cm

3.56cm

3.62cm

Detalhes da Resposta

\begin{array}{cc} x(mid point) & f & f x \\ 3.4 & 3 & 10.2 \\ 3.5 & 6 & 21.0 \\ 3.6 & 7 & 25.2 \\ 3.7 & 4 & 14.8 \end{array}

\sum f

= 20

\sum f x

= 71.2
mean =

\frac{\sum f x}{\sum f}

=

\frac{71.2}{20}

= 3.56

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Pergunta 38 Relatório

Simplify $ \frac{3^n - 3^{n-1}}{3^3 \times 3^n - 27 \times 3^{n-1}} $

1

6

1

27

4

3

Detalhes da Resposta

\frac{3^{n} - 3^{n - 1}}{3^{3} \times 3^{n} - 27 \times 3^{n - 1}}

=

\frac{3^{n} - 3^{n - 1}}{3^{3} (3^{n} - 3^{n - 1})}

=

\frac{3^{n} - 3^{n - 1}}{27 (3^{n} - 3^{n - 1})}

=

\frac{1}{27}

Ler a nota de aula em Algebraic Expressions (WAEC)

Ler a nota de aula em Fractions, Decimals, Approximations And Percentage (JAMB)

Algebraic Expressions

Fractions, Decimals, Approximations And Percentage

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Pergunta 39 Relatório

The larger value of y for which (y - 1)2 = 4y - 7 is

2

4

6

7

8

Detalhes da Resposta

(y - 1)2 = 4y - 7
y2 - 2xy + 1 = 4y - 7
y2 - 6y + 8 = 0
(y - 4)(y - 2)
y = 4 or 2 = 4

Ler a nota de aula em Bearings (WAEC)

Ler a nota de aula em Circles (WAEC)

Ler a nota de aula em Surds (radicals) (WAEC)

Ler a nota de aula em Quadratic Equations (WAEC)

Bearings

Circles

Surds (radicals)

Quadratic Equations

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Pergunta 40 Relatório

P varies directly as the square of Q and inversely as R. If P = 36, when Q = 3 and R = 4, find P when Q = 5 and R = 2.

72

100

90

200

125

Detalhes da Resposta

P \propto \frac{Q^{2}}{R}

∴

P = \frac{K Q^{2}}{R}

36 = \frac{K Q^{2}}{4}

K = \frac{36 \times 4}{9}

K = 16

P = \frac{16 \times 5^{2}}{2} = 200

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Pergunta 41 Relatório

Rationalize $ \frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}} $

-2 \sqrt{35}\)

4\sqrt{7}\) - 6

\sqrt{5}

-√35

4\sqrt{7}\) - 8

\sqrt{5}

√35

Detalhes da Resposta

\frac{5 \sqrt{7} - 7 \sqrt{5}}{\sqrt{7} - \sqrt{5}}

=

\frac{5 \sqrt{7} - 7 \sqrt{5}}{\sqrt{7} - \sqrt{5}}

x

\frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}}

=

\frac{(5 \times 7) + (5 \sqrt{7} \times 5) - (7 \times \sqrt{5} \times 7) (- 7 \times 5)}{(\sqrt{7})^{2}}

=

\frac{5 \sqrt{35} - 7 \sqrt{35}}{2}

=

\frac{- 2 \sqrt{35}}{2}

= -

\sqrt{35}

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Ler a nota de aula em Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

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Pergunta 42 Relatório

In the diagram, find the size of the angle marked ao

60o

80o

120o

160o

Detalhes da Resposta

2 x s = 280o(Angle at centre = 2 x < at circum)

S = $\frac{280^{o}}{2}$

= 140

< O = 360 - 280 = 80o

60 + 80 + 140 + a = 360o

(< in a quad); 280 = a = 360

a = 360 - 280

a = 80o

Ler a nota de aula em Angles (WAEC)

Angles

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Pergunta 43 Relatório

A straight lie y = mx meets the curve y = x2 - 12x + 40 in two distinct points. If one of them is (5, 5) find the other

(5, 6)

(8, 8)

(8, 5)

(7,7)

(7, 5)

Detalhes da Resposta

When y = 5, y = x2 - 12x + 40, becomes
x2 - 12x + 40 = 5
x2 - 12x + 40 - 5 = 0
x2 + 12x + 35 = 0
x2 - 7x - 5x + 35 = 0
x(x - 7) - 5(x - 7) = 0
= (x - 5)(x - 7)
x = 5 or 7

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Pergunta 44 Relatório

XYZ is a triangle and XW is perpendicular to YZ at = W. If XZ = 5cm and WZ = 4cm, Calculate XY.

5