WAEC SSCE - General Mathematics - 2015…

In the given diagram, $\bar{QT}$ and $\bar{PR}$ are straight lines, < ROS = (3n - 20), < SOT = n, < POL = m and < QOL is a right angle. Find the value of n.

35^o Azịza ziri ezi

40^o

55^o

60^o

Akọwa Nkọwa

Since < QOL is a right angle, we have: < QOT + < SOT = 90^o Since < QOT = 180^o - < POL, we have: < QOT = 180^o - m Substituting into the first equation, we get: 180^o - m + n = 90^o n - m = 90^o - 180^o = -90^o Also, from the diagram, we have: < QOT + < ROS + < SOT = 180^o Substituting < QOT = 180^o - < POL and simplifying, we get: 180^o - m + (3n - 20) + n = 180^o 4n - m = 20 We now have a system of two equations: n - m = -90^o 4n - m = 20 Solving simultaneously, we get: n = 35^o

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Gụọ akwụkwọ nkuzi na Angles And Intercepts On Parallel Lines (WAEC)

Algebraic Expressions

Angles And Intercepts On Parallel Lines

Lee Azịza

Ajụjụ 6 Ripọtì

Ada draws the graph of y = x² - x - 2 and y = 2x - 1 on the same axes. Which of these equations is she solving?

x² - x - 3 = 0

x² - 3x - 1 = 0 Azịza ziri ezi

x² - 3x - 3 = 0

x² + 3x - 1 = 0

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Gụọ akwụkwọ nkuzi na Graphs Of Linear And Quadratic Functions (WAEC)

Quadratic Equations

Graphs Of Linear And Quadratic Functions

Lee Azịza

Ajụjụ 7 Ripọtì

The probability that kebba, Ebou and Omar will hit a target are $\frac{2}{3}$, $\frac{3}{4}$ and $\frac{4}{5}$ respectively. Find the probability that only Kebba will hit the target.

$\frac{2}{5}$

$\frac{7}{60}$

$\frac{1}{30}$ Azịza ziri ezi

$\frac{1}{60}$

Akọwa Nkọwa

The probability that only Kebba will hit the target means that Kebba hits the target and the other two miss the target. We can find this probability by multiplying the probability of Kebba hitting the target with the probability of Ebou missing the target and the probability of Omar missing the target. Probability that only Kebba hits the target = Probability of Kebba hitting the target × Probability of Ebou missing the target × Probability of Omar missing the target = $\frac{2}{3}$ × $\frac{1}{4}$ × $\frac{1}{5}$ = $\frac{2}{3\times4\times5}$ = $\frac{1}{30}$ Therefore, the probability that only Kebba will hit the target is $\frac{1}{30}$.

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Probability

Lee Azịza

Ajụjụ 8 Ripọtì

In the diagram, O is the centre, $\bar{RT}$ is a diameter, < PQT = 33$^o$ and < TOS = 76$^o$. Using the diagram, find the size of angle PRS.

76^o

71^o Azịza ziri ezi

38^o

33^o

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 9 Ripọtì

Two angles of a pentagon are in the ratio 2:3. The others are 60^o each. Calculate the smaller of the two angles

72^o

100^o

120^o

144^o Azịza ziri ezi

Akọwa Nkọwa

Let's call the two angles that are in the ratio 2:3 as "2x" and "3x". The sum of the interior angles of a pentagon is given by the formula (5-2) x 180 degrees = 540 degrees. Since the other three angles are all 60 degrees each, we can find the sum of those angles by multiplying 60 by 3, which gives us 180 degrees. Now we can set up an equation to solve for the unknown angle, which is 2x + 3x. 2x + 3x + 180 = 540 Simplifying this equation, we get: 5x + 180 = 540 Subtracting 180 from both sides, we get: 5x = 360 Dividing both sides by 5, we get: x = 72 So the smaller angle, which is 2x, is equal to 2 times 72, which is 144 degrees. Therefore, the correct answer is 144 degrees.

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Angles

Lee Azịza

Ajụjụ 10 Ripọtì

Simplify: $\frac{3x - y}{xy} - \frac{2x + 3y}{2xy} + \frac{1}{2}$

$\frac{4x + 5y - xy}{2xy}$

$\frac{5y - 4x + xy}{2xy}$

$\frac{5x + 4y - xy}{2xy}$

$\frac{4x - 5y + xy}{2xy}$ Azịza ziri ezi

Akọwa Nkọwa

To simplify the expression, we first need to find a common denominator for the fractions. The lowest common multiple of the denominators xy and 2xy is 2xy. $\frac{3x - y}{xy} - \frac{2x + 3y}{2xy} + \frac{1}{2} = \frac{3x\cdot2 - y\cdot2}{xy\cdot2} - \frac{2x\cdot1 + 3y\cdot1}{2xy\cdot1} + \frac{1\cdot xy}{2\cdot xy}$ Simplifying further, we get: $\frac{6x - 2y}{2xy} - \frac{2x + 3y}{2xy} + \frac{xy}{2xy}$ Combining like terms, we get: $\frac{4x - 5y + xy}{2xy}$ Therefore, the answer is option D: $\frac{4x - 5y + xy}{2xy}$.

Lee Azịza

Ajụjụ 11 Ripọtì

Factorize completely: 6ax - 12by - 9ay + 8bx

(2a - 3b)(4x + 3y)

(3a + 4b)(2x - 3y) Azịza ziri ezi

(3a - 4b)(2x + 3y)

(2a + 3b)(4x -3y)

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 12 Ripọtì

If m = 4, n = 9 and r = 16., evaluate $\frac{m}{n}$ - 1$\frac{7}{9}$ + $\frac{n}{r}$

1$\frac{5}{16}$

1$\frac{1}{16}$

$\frac{5}{16}$

- 1$\frac{37}{48}$ Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Lee Azịza

Ajụjụ 13 Ripọtì

The sum 11011₂, 1111₂ and 10_m10_n0₂. Find the value of m and n.

m = 0, n = 0

m = 1, n = 0

m = 0, n = 1 Azịza ziri ezi

m = 1, n = 1

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 14 Ripọtì

The angle of elevation of an aircraft from a point K on the horizontal ground 30$\alpha$. If the aircraft is 800m above the ground, how far is it from K?

400.00m

692.82m

923.76m

1,600.99m Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Introductory Calculus (WAEC)

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

Introductory Calculus

Coordinate Geometry

Lee Azịza

Ajụjụ 15 Ripọtì

Describe the shaded portion in the diagram

P' $\cap$ Q $\cap$ R' Azịza ziri ezi

(P' $\cap$ R)' $\cap$ R

P' $\cap$ Q $\cap$ R

(P $\cap$ Q)' $\cap$ R

Akọwa Nkọwa

The shaded portion in the diagram represents the set of elements that belong to the complement of P and the intersection of Q and R. In other words, it represents the set of elements that are not in P, but are in both Q and R. Therefore, the correct answer is P' $\cap$ Q $\cap$ R'.

Gụọ akwụkwọ nkuzi na Logical Reasoning (WAEC)

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Logical Reasoning

Sets

Lee Azịza

Ajụjụ 16 Ripọtì

A letter is selected from the letters of the English alphabet. What is the probability that the letter selected is from the word MATHEMATICS?

$\frac{9}{13}$

$\frac{11}{26}$

$\frac{4}{13}$ Azịza ziri ezi

$\frac{1}{26}$

Akọwa Nkọwa

The word MATHEMATICS contains the letters A, C, E, H, I, M, S, and T. Since there are 26 letters in the English alphabet, the probability of selecting any letter from the alphabet is 1 out of 26. Out of the letters in the word MATHEMATICS, there are 8 letters, so the probability of selecting a letter from the word MATHEMATICS is 8 out of 26. Therefore, the probability of selecting a letter from the word MATHEMATICS is: $$\frac{8}{26}=\frac{4}{13}$$ Therefore, the correct answer is option (C).

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Probability

Surds (radicals)

Lee Azịza

Ajụjụ 17 Ripọtì

If $\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}$ is simplified as m + n$\sqrt{6}$, find the value of (m + n)

$\frac{1}{3}$

$\frac{2}{3}$

1$\frac{1}{3}$ Azịza ziri ezi

1$\frac{2}{3}$

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Surds (radicals)

Lee Azịza

Ajụjụ 18 Ripọtì

In the diagram, the shaded part is carpet laid in a room with dimensions 3.5m by 2.2m leaving a margin of 0.5m round it. Find area of the margin

4.7m² Azịza ziri ezi

4.9m²

5.7m²

5.9m²

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Lengths And Perimeters (WAEC)

Lengths And Perimeters

Lee Azịza

Ajụjụ 19 Ripọtì

If 2ⁿ = y, Find 2$^{(2 + \frac{n}{3})}$

4y$^\frac{1}{3}$ Azịza ziri ezi

4y$^-3$

2y$^\frac{1}{3}$

2y$^-3$

Akọwa Nkọwa

We know that 2$^n$ = y. Therefore, 2$^{(2 + \frac{n}{3})}$ can be written as 2$^2$ × 2$^{\frac{n}{3}}$ = 4 × 2$^{\frac{n}{3}}$. Substituting 2$^n$ = y in the above equation, we get: 2$^{(2 + \frac{n}{3})}$ = 4y$^\frac{1}{3}$ Therefore, the answer is (a) 4y$^\frac{1}{3}$.

Gụọ akwụkwọ nkuzi na Indices (WAEC)

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Indices

Logarithms

Lee Azịza

Ajụjụ 20 Ripọtì

$\begin{array}{c|c}
x & 0 & 1\frac{1}{4} & 2 & 4\\
\hline
y & 3 & 5\frac{1}{2} & &
\end{array}$
The table given shows some values for a linear graph. Find the gradient of the line

2 Azịza ziri ezi

Akọwa Nkọwa

To find the gradient of a line, we need to calculate the change in y divided by the change in x between two points on the line. Let's choose two points on the line from the given table. We can select the points (0,3) and (2,y), where y is the value of y for x = 2. The change in y is y - 3, and the change in x is 2 - 0 = 2. Therefore, the gradient of the line is: gradient = change in y / change in x = (y - 3) / 2 We don't know the value of y yet, but we can use the other point on the line to find it. Using the points (1.25, 5.5) and (2, y), we get: gradient = (y - 5.5) / (2 - 1.25) = (y - 5.5) / 0.75 Setting the two expressions for the gradient equal to each other, we get: (y - 3) / 2 = (y - 5.5) / 0.75 Solving for y gives: y = 7 Therefore, the value of y for x = 2 is 7, and the gradient of the line is: gradient = (y - 3) / 2 = (7 - 3) / 2 = 2 So the answer is 2, and the gradient of the line is 2. In simple terms, the gradient of a line tells us how steep the line is. If the gradient is positive, the line is sloping upwards from left to right. If the gradient is negative, the line is sloping downwards from left to right. The larger the absolute value of the gradient, the steeper the line.

Gụọ akwụkwọ nkuzi na Graphs Of Linear And Quadratic Functions (WAEC)

Graphs Of Linear And Quadratic Functions

Lee Azịza

Ajụjụ 21 Ripọtì

Which of the following is used to determine the mode of a grouped data?

Bar chart

Frequency polygon

Ogive

Histogram Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Representation Of Data

Integration

Lee Azịza

Ajụjụ 22 Ripọtì

$\begin{array}{c|c}
Scores & 0 - 4 & 5 - 9 & 10 - 14\\
\hline Frequency & 2 & 1 & 2\end{array}$

The table shows the distribution of the scores of some students in a test. Calculate the mean scores.

5.6

6.2

6.6

7.0 Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Integration

Statistics

Lee Azịza

Ajụjụ 23 Ripọtì

In the diagram, O is the centre. If PQ//RS and ∠ONS = 140°, find the size of ∠POM.

40^o Azịza ziri ezi

50^o

60^o

80^o

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Euclidean Geometry

Lee Azịza

Ajụjụ 24 Ripọtì

Given that logx 64 = 3, evaluate x log$_2$8

12 Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Modular Arithmetic

Logarithms

Lee Azịza

Ajụjụ 25 Ripọtì

The population of students in a school is 810. If this is represented on a pie chart, calculate the sectoral angle for a class of 7 students

32^o Azịza ziri ezi

45^o

60^o

75^o

Akọwa Nkọwa

To find the sectoral angle for a class of 7 students, we first need to calculate what percentage of the total population they represent: Percentage of class of 7 students = (Number of students in the class / Total population) x 100% Percentage of class of 7 students = (7 / 810) x 100% Percentage of class of 7 students = 0.864% (rounded to three decimal places) Next, we need to convert this percentage to an angle of the pie chart. We know that a full circle is 360 degrees, and the percentage of the class of 7 students represents a certain portion of this circle. We can use the following formula to calculate the sectoral angle: Sectoral angle = Percentage of class of 7 students x 360 degrees Plugging in the percentage we calculated earlier, we get: Sectoral angle = 0.864% x 360 degrees Sectoral angle ≈ 3.11 degrees (rounded to two decimal places) Therefore, the sectoral angle for a class of 7 students in the pie chart is approximately 3.11 degrees. So, the correct option is (a) 32^o.

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Gụọ akwụkwọ nkuzi na Mensuration (JAMB)

Representation Of Data

Sets

Mensuration

Lee Azịza

Ajụjụ 26 Ripọtì

The area of a rhombus is 110 cm$^2$. If the diagonals are 20 cm and (2x + 1) cm long, find the value of x.

5.0 Azịza ziri ezi

4.0

3.0

2.5

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Areas

Angles

Lee Azịza

Ajụjụ 27 Ripọtì

Calculate the mean deviation of 5, 3, 0, 7, 2, 1

0.0

2.0 Azịza ziri ezi

2.5

3.0

Akọwa Nkọwa

To find the mean deviation, we need to first calculate the mean of the given set of numbers: Mean = (5 + 3 + 0 + 7 + 2 + 1) / 6 = 18 / 6 = 3 Next, we need to find the absolute deviation of each number from the mean: |5 - 3| = 2 |3 - 3| = 0 |0 - 3| = 3 |7 - 3| = 4 |2 - 3| = 1 |1 - 3| = 2 To calculate the mean deviation, we add up these absolute deviations and divide by the number of numbers: Mean deviation = (2 + 0 + 3 + 4 + 1 + 2) / 6 = 12 / 6 = 2 Therefore, the answer is option (B) 2.0.

Gụọ akwụkwọ nkuzi na Measures Of Location (JAMB)

Measures Of Location

Lee Azịza

Ajụjụ 28 Ripọtì

In the diagram, VW//YZ, |WX| = 6cm, |XY| = 16cm, |YZ| = 20cm and |ZX| = 12cm. Calculate |VX|

3cm

4cm

6cm

8cm Azịza ziri ezi

Akọwa Nkọwa

Lee Azịza

Ajụjụ 29 Ripọtì

The volume of a cone of height 3cm is 38$\frac{1}{2}$cm³. Find the radius of its base. [Take $\pi = \frac{22}{7}$]

3.0cm

3.5cm Azịza ziri ezi

4.0cm

4.5cm

Akọwa Nkọwa

Lee Azịza

Ajụjụ 30 Ripọtì

A farmer uses $\frac{2}{5}$ of his land to grow cassava, $\frac{1}{3}$ of the remaining for yam and the rest for maize. Find the part of the land used for maize

$\frac{2}{15}$

$\frac{2}{5}$ Azịza ziri ezi

$\frac{2}{3}$

$\frac{4}{5}$

Akọwa Nkọwa

Let's assume the farmer has 15 parts of land. The fraction of land used for cassava is $\frac{2}{5}$ of 15, which is 6 parts. The remaining land is 15 - 6 = 9 parts. The fraction of the remaining land used for yam is $\frac{1}{3}$ of 9, which is 3 parts. Therefore, the part of the land used for maize is the remaining land after cassava and yam are planted, which is 9 - 3 = 6 parts. So the fraction of the land used for maize is $\frac{6}{15}$, which simplifies to $\frac{2}{5}$. Therefore, the answer is option (B) $\frac{2}{5}$.

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Fractions, Decimals And Approximations

Number Bases

Lee Azịza

Ajụjụ 31 Ripọtì

If $\frac{27^x \times 3^{1 - x}}{9^{2x}} = 1$, find the value of x.

$\frac{1}{2}$ Azịza ziri ezi

-$\frac{1}{2}$

-1

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Indices (WAEC)

Number Bases

Indices

Lee Azịza

Ajụjụ 32 Ripọtì

PQRT is square. If x is the midpoint of PQ, Calculate correct to the nearest degree, LPXS

53^o

55^o

63^o Azịza ziri ezi

65^o

Akọwa Nkọwa

Lee Azịza

Ajụjụ 33 Ripọtì

The dimension of a rectangular tank are 2m by 7m by 11m. If its volume is equal to that of a cylindrical tank of height 4cm, calculate the base radius of the cylindrical tank. [Take $\pi = \frac{22}{7}$]

14cm

3$\frac{1}{2}$m Azịza ziri ezi

1$\frac{3}{4}$m

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Volumes (WAEC)

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Volumes

Fractions, Decimals And Approximations

Lee Azịza

Ajụjụ 34 Ripọtì

If {X: 2 d- x d- 19; X integer} and 7 + x = 4 (mod 9), find the highest value of x

15 Azịza ziri ezi

Akọwa Nkọwa

To solve this problem, we need to use the given conditions to find the possible values of x and then determine the highest value among them. From the first condition, we know that x is an integer between 2 and 19 (inclusive) and can be represented as {X: 2 d- x d- 19; X integer}. This means that x can take on any of the following values: 2, 3, 4, ..., 18, 19. The second condition tells us that 7 + x is congruent to 4 modulo 9, which can be written as: 7 + x ≡ 4 (mod 9) To solve for x, we can subtract 7 from both sides of the congruence: x ≡ -3 (mod 9) Since we want x to be between 2 and 19, we can add or subtract multiples of 9 to -3 until we get a value within the range of possible values for x. Doing so, we obtain: x ≡ -3 (mod 9) x ≡ 15 (mod 9) Since 15 is the largest value of x that satisfies both conditions, our answer is 15.

Gụọ akwụkwọ nkuzi na Positive And Negative Integers, Rational Numbers (WAEC)

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Positive And Negative Integers, Rational Numbers

Modular Arithmetic

Lee Azịza

Ajụjụ 35 Ripọtì

In the diagram, $\bar{OX}$ bisects < YXZ and $\bar{OZ}$ bisects < YZX. If < XYZ = 68^o, calculate the value of < XOZ

68^o

72^o

112^o

124^o Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Angles And Intercepts On Parallel Lines (WAEC)

Angles And Intercepts On Parallel Lines

Lee Azịza

Ajụjụ 36 Ripọtì

The rate of consumption of petrol by a vehicle varies directly as the square of the distance covered. If 4 litres of petrol is consumed on a distance of 15km. how far would the vehicle go on 9 litres of petrol?

22$\frac{1}{2}$km Azịza ziri ezi

30km

33$\frac{1}{2}$km

45km

Akọwa Nkọwa

Since the rate of consumption of petrol varies directly as the square of the distance covered, let's assume that the rate is k$\frac{\text{litres}}{\text{km}^2}$. Then we can write the relationship as: rate of consumption = k(distance covered)$^2$ We are given that when the distance covered is 15 km, the rate of consumption is 4 litres. Substituting these values in the above equation, we get: 4 = k(15)$^2$ k = $\frac{4}{15^2}$ Now, we need to find how far the vehicle can go on 9 litres of petrol. Let's assume that the vehicle can travel a distance of x km on 9 litres of petrol. Using the same equation as before, we get: 9 = $\frac{4}{15^2}$(x)$^2$ Simplifying this equation, we get: x$^2$ = $\frac{9}{\frac{4}{15^2}}$ x$^2$ = 506.25 x = 22.5 Therefore, the vehicle can travel 22.5 km on 9 litres of petrol. So the answer is 22$\frac{1}{2}$km.

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Gụọ akwụkwọ nkuzi na Ratio, Proportions And Rates (WAEC)

Sequence And Series

Ratio, Proportions And Rates

Lee Azịza

Ajụjụ 37 Ripọtì

Make K the subject of the relation T = $\sqrt{\frac{TK - H}{K - H}}$

K = $\frac{H(T^2 - 1)}{T^2 - T}$ Azịza ziri ezi

K = $\frac{HT}{(T - 1)^2}$

K = $\frac{H(T^2 + 1)}{T}$

K = $\frac{H(T - 1)}{T}$

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Algebraic Fractions (WAEC)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Algebraic Fractions

Quadratic Equations

Lee Azịza

Ajụjụ 38 Ripọtì

The scores of twenty students in a test are as follows: 44, 47, 48, 49, 50, 51, 52, 53, 53, 54, 58, 59, 60, 61, 63, 65, 67, 70, 73, 75. Find the third quartile.

64 Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Measures Of Dispersion (JAMB)

Measures Of Dispersion

Lee Azịza

Ajụjụ 39 Ripọtì

Adding 42 to a given positive number gives the same result as squaring the number. Find the number

7 Azịza ziri ezi

Akọwa Nkọwa

Let's call the unknown number "x". According to the problem, adding 42 to x gives the same result as squaring x. We can write this as an equation: x + 42 = x^2 To solve for x, we can rearrange this equation to get it in standard quadratic form: x^2 - x - 42 = 0 Now we can factor this quadratic equation: (x - 7)(x + 6) = 0 This gives us two possible solutions: x = 7 and x = -6. However, we were given that x is a positive number, so we can discard the negative solution. Therefore, the answer is x = 7. To check our answer, we can substitute x = 7 into the original equation: 7 + 42 = 49 7^2 = 49 So adding 42 to 7 does indeed give the same result as squaring 7.

Lee Azịza

Ajụjụ 40 Ripọtì

A trader bought 100 oranges at 5 for N40.00 and 20 for N120.00. Find the profit or loss percent

20% profit

20% loss

25% profit

25% loss Azịza ziri ezi

Akọwa Nkọwa

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 41 Ripọtì

In a circle radius rcm, a chord 16$\sqrt{3}cm$ long is 10cmfrom the centre of the circle. Find, correct to the nearest cm, the value of r

22cm

17cm Azịza ziri ezi

16cm

15cm

Akọwa Nkọwa

In a circle, a radius is a line segment that connects the center of the circle to any point on the circle. A chord is a line segment that connects two points on a circle. Given that a chord of length 16$\sqrt{3}cm$ is 10cm from the center of the circle, we can use the Pythagorean theorem to find the length of the radius. First, we draw a line from the center of the circle perpendicular to the chord, creating a right triangle with the radius, the perpendicular line, and half the chord length (which is $8\sqrt{3}$ cm). Using the Pythagorean theorem: \begin{align*} r^2 &= (8\sqrt{3})^2 + 10^2 \\ r^2 &= 192 + 100 \\ r^2 &= 292 \\ r &\approx \sqrt{292} \\ r &\approx 17 \text{ cm} \end{align*} Rounding to the nearest whole number, we get the answer of 17cm. Therefore, the correct option is (b) 17cm.

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Circles

Triangles And Polygons

Lee Azịza

Ajụjụ 42 Ripọtì

In the diagram, O is the centre, $\bar{RT}$ is a diameter, < PQT = 33$^o$ and <TOS = 76$^o$. Using the diagram, calculate the value of angle PTR.

73^o

67^o

57^o Azịza ziri ezi

37^o

Akọwa Nkọwa