NURUL-ISLAM COMPREHENSIVE COLLEGE
Nurul Islam Junction, Eyita-Ojokoro Road, Eyita, Ikorodu Lagos. image
www.nurulislamschools.com MATHEMATICS
LESSON NOTE
2023/2024 ACADEMIC SESSION
SECOND TERM
CLASS: SS3
LAWAL ABDULLATEEF OLAWALE
08/01/2024
WEEK ONE
DATE: 08/01/2024
WEEK: ONE
SUBJECT: MATHEMATICS CLASS: SSS THREE
LESSON TITLE: BOND AND DEBENTURES
PERIOD: 1 and 2 DURATION: 80 minutes
LEARNING OUTCOME/OBJECTIVES: At the end of the lesson, students should be able to;
Understand Bond and debenture,
Solve problems on bonds and debentures.
LEARNING RESOURCES:
Audio Visual Resources.
Objects of different lengths or height
Model Mathematical SetWeb Resources
www.khanacademy.com www.byjus.com RESOURCES AND MATERIALS (i) New General Mathematics 2 published by (ii) MAN Mathematics 2 published by the Association of Mathematics teachers SCHEME OF WORK WEEK(S) TOPIC Review of first term work: (i) Bonds and debentures (ii) Shares (iii) Rates (iv) Income tax and (v) Value added tax. CO-ORDINATE GEOMETRY OF STRAIGHT LINE: Cartesian coordinate (ii) plotting the linear graph (iii) determine the distance between two coordinate points. (iv) Finding the mid-point of the line joining two points (v) practical application of coordinate geometry. (vi) Gradient and intercept of a straight line. COORDINATE GEOMETRY OF A STRAIGHT LINE CONTINUES: (I) Define the gradient and intercepts of a line. (ii) Find the angle between two intersecting straight lines (iii) Application of linear graphs to real-life situations. DIFFERENTIATION OF ALGEBRAIC FUNCTION: (I) Meaning of differentiation/ derived function (ii) differentiation from first principle (iii) standard derivative of some basic functions. DIFFERENTIACONTINUESLGEBRAIC FUNCTION CONTINEUS: Rules of differentiation such as: (a) Sum and difference (b) Product rule (c) Quotient rule. (d) Application of real situations such as Maximal, Minima velocity, Acceleration, and rate of change. INTEGRATION AND EVALUATION SIMPLE ALGEBRAIC FUNCTION: (i) definition (ii) Method of integration: (a) substitution method (b) partial fraction method (c) part. (iii) Application of integration in calculating the area under the curve (iv) Use of Simpson’s rule to find the area under the curve. 7-12. Revision and Mock Examination. CONTENT:
Lesson objectives: At the end of the lesson students should be able to; Understand Bonds and Debentures
BONDS Bonds are debt financial instruments that both public and private sector companies use to raise funds for their operations. Government agencies, financial institutions as well as private enterprises issue these instruments to investors. Bonds are secured by their physical assets. The holder of these bonds is the lender, while the issuer of these bonds is the borrower. The borrower can issue these bonds to the lender, only by promising to pay back the loan at a specific maturity date with a fixed interest rate. This interest rate is generally lower than debentures because the physical assets of a company secure bonds whereas the debentures are unsecured instruments. The process of determining these bonds is called bond valuation. It is used to determine the theoretical price, fair price, or intrinsic price of the bonds. Bond Terminology Example Find the EMI for a principal amount of Rs. 1,00,000 for 12 months at a 10% rate of interest. Here, P = Rs. 1,00,000 r = 10% and n = 12 months. So, the EMI calculated will be, Rs. 8792. Practice Questions Q. Rahul plans on buying a 3-year bond having a Rs. 1000 par value at an interest rate of 10%. Find the price at which one can purchase the bond after maturation at par and if Rahul requires a return of 14%. Rs. 915 Rs. 907 Rs. 902 Rs. 917 Answer: Rs. 907 Q. Find the EMI for a property that is mortgaged with a principal amount of Rs. 5,00,000 at a rate of interest, 3.5% for a period of 10 years. true657 Rs. 1647 Rs. 1523 Rs. 1547 Answer: Rs. 1547 What is ‘Debenture’ Debenture Debentures are other forms of debt that don’t have to be paid back. Due to the lack of collateral, debentures rely on the creditworthiness and reputation of the person or company that issued them. Companies and governments usually issue debentures to get cash or money.
UnderstandingDebentures Like most bonds, debentures can pay coupon payments, and interest payments made at regular intervals. Like other types of bonds, debentures are written down in an indenture. An indenture is a legal contract between the bond issuer and the bondholders. The contract spells out the details of a debt offering, such as when the debt will be paid off, when interest or coupon payments will be made, how interest will be calculated, and other details. Both governments and businesses can make debentures.
Most of the time, governments sell bonds with maturities of more than ten years. Since the government is backing these bonds, they are considered low-risk investments.
Debentures are also used as long-term loans by companies. Still, corporate debentures are not backed by anything. Instead, only the health of the company’s finances and its creditworthiness are used as collateral. These debt instruments come with an interest rate and can be redeemed or paid back on a set date. Usually, a company pays the interest on its loans before giving dividends to its stockholders. Compared to other types of loans and financial instruments, debentures have lower interest rates and longer payback terms, which is good for businesses. What is a debenture formula? Coupon Rate = (Total Annual Coupon Payment/Par Value of the Bond) *100 Example: Compute the maturity value of a #100,000 debenture which pays 1/2 % interest rate over 8 years. Solution. Interest = 100,000 + 44,000 = #144,00 Bonds Debentures Definition Bonds are debt financial instruments issued by large corporations, financial institutions, and government agencies that are backed up by collaterals or physical assets. Debentures are debt financial instruments issued by private companies, but any collaterals or physical assets do not back them up. Owner The owner of a bond is called a bondholder. The owner of a debenture is called a debenture holder. Collateral Bonds are secured by the collateral or physical assets of the issuing company. Debentures do not get secured by the collateral or physical assets of the issuing company. Lenders purchase these instruments solely based on the reputation of the issuing company. Tenure Bonds are long-term investments and their tenure is generally higher than debentures. Debentures are generally short to medium-term investments and their tenure is usually lower than bonds. Issuer Large corporations, financial institutions, and government agencies issue these bonds for their long-term capital requirements. Private companies generally issue debentures for their immediate capital requirements. Rate of Interest The bonds carry a fixed or floating interest rate that is generally lower than debentures because they are more stable in terms of repayment, and they get backed by collateral of the issuing company. The debentures carry a fixed or floating interest rate that is generally higher than bonds because they are less stable in terms of repayment, and they are also not backed by collateral. Priority During Liquidation If the company is on the verge of liquidation, the bondholders are given priority over debenture holders for repayment of capital and interest amount. If the company is on the verge of liquidation, the debenture holders are given second priority over bondholders for repayment of capital and interest amount. Payment Structure The payment of interest for bonds is on an accrual basis. The issuing company pays this amount on a monthly, half-yearly, or yearly basis and this payment is not dependent on the performance of the company. The payment of interest for bonds is done on a periodical basis and depends on the company’s performance. Risk Bonds are less riskier than debentures because they have the security of the physical assets of the issuing company. Debentures are riskier than bonds because they do not have the security of the physical assets of the issuing company. Conclusion There are several differences between bonds and debentures. However, both are important when it comes to raising capital to finance the short and long-term needs of a corporation. Lenders who prefer low-risk investments when compared to shares put their money in financial instruments like bonds and debentures.
DATE: 08/01/2024 WEEK: ONE SUBJECT: MATHEMATICS CLASS: SSS Three LESSON TITLE: SHARES AND INCOME PERIOD: 3 and 4 DURATION: 80 minutes LEARNING OUTCOME/OBJECTIVES: At the end of the lesson, students should be able to; Solve problems on Shares and income LEARNING RESOURCES: Audio Visual Resources. Objects of different lengths or heights Model MathematicResources Web Resources www.khanacademy.com www.byjus.com RESOURCES AND MATERIALS (i) New General Mathematics 2 published by (ii) MAN Mathematics 2 published by CONTENT: Lesson objectives: At the end of the lesson students should be able to; Solve problems with Shares and income
DEFINITION OF SHARES Shares are sold by companies to members of the public. Those who purchase shares intend to earn money from their investment. The amount which a company decides to pay from its profit to its shareholders is called dividends. Example 1. If a shareholder bought a share worth N75,000 on a face value of N1 ordinary share at N1.50k each. (a) How many shares did the shareholder buy? (b) If a dividend of 10% is declared, how much dividend was received? Solution: (a) Number of shares bought = N75,000/N1.50k = N50,000 shares (b) 10% dividends means 10% of N1 = 10k per share Therefore, Dividend = 50,000 x 10k = N5,000.00 Example 2. A man earns N20,000 per annum and he is given N8,000 as tax-free. He paid tax at the rate of 10% on the first N5,000; 15% on the next N5,000 and 20% on the remaining. Find the man’s (a) taxable income (b) total income tax per annum (c) net pay Solution: Annum earning (income) = N20,000 Tax free income = N3,000 Taxable income = N20,000 – N8,000 = N12,000 Income tax on 1st N5,000 = 10% of N5,000 = N500 Income tax on next N5,000 = 15% of N5,000 = N750 Income tax on the remaining N2,000 = 20% of N2,000 = N400 Total income tax per annum = N(500 + 750 + 400) = N1,650 Net pay = Annual income – Total income tax = N20,000 – N1,650 = N18,350 EVALUATION: 1. Mr. Aliu paid N2,400 in N1 ordinary shares of a company sold at N2.50k How many shares did he buy? How much dividend did he get if the dividend declared is N20k per share?
CONCLUSION: The teacher goes around to assess the student’s work and gives corrections on the board for students to copy. ASSIGNMENT: Find the monthly payment on a loan of N1,200,500 to be amortized for 4 year period at the rate of 8%. 2. Mr. Aliu paid N3,400 in N1 ordinary shares of a company sold at N2.50k How many shares did he buy? How much dividend did he get if the dividend declared is N15k per share?
DATE: 08/01/2024 WEEK: ONE SUBJECT: MATHEMATICS CLASS: SSS THREE LESSON TITLE: Income and value Added tax PERIOD: 5 DURATION: 40 minutes LEARNING OUTCOME/OBJECTIVES: At the end of the lesson, students should be able to; (i) Solve problems on Income tax LEARNING RESOURCES: Audio Visual Resources. Objects of different lengths or height Model Mathematical Set
Web Resourceswww.khanacademy.com www.byjus.com RESOURCES AND MATERIALS (i) New General Mathematics 2 published by (ii) MAN Mathematics 2 published by CONTENT: Lesson objectives: At the end of the lesson students should be able to; Solve problems with Income Tax Computation of income and return If the number of shares held by a shareholder = n, the rate of dividend = r% per annum, and face value = F then Annual income = nrF/100 Annual return = annual income/investment in shares × 100% Calculation of the number of shares To find the number of shares held by a person, we use the following. Number of shares held = investment/investment for the share Or, annual income/income from one share Or, total face value/face value of one share Solved examples 1. Michael buys $ 100 shares at a $ 20 premium in a company paying 15% divided. Find the MV of 200 shares, his annual income, and his percentage income.
Solution:
Market value of one share = $ 100 + $ 20 = $ 120 Therefore, the market value of 200 shares = 200 × $ 120 = $ 24,000 Annual income = Number of shares × Rate of dividend × nominal value or face value of 1 share = 200×15 / 100×100 = $ 3000 $ 3000 is the income obtained on investing $ 24000 Therefore, the percentage income = 3000 / 24000 × 100 % = 12.5 %
ASSIGNMENT 1. Alexander bought some shares of $ 10 at a par value which yields 10% divided per annum. At the end of one year, he received $ 2000 as a dividend. How many shares did he buy? Let the number of shares be n. Then Annual income or dividend = nrF / 100 ⟹ $ 2000 = n×10×$10 / 100 ⟹ n = 2000×100 / 10×10 = 2000 Therefore, the required number of shares = 2000 2. A man got N5,000,000 housing loan at an annual interest rate of 10%. If the loan is amortized for 10 years term. What is the monthly payment the man is expected to pay? 3. A man earns a taxable income of N100,000 per annum. If he is taxed at the rate of 10% per annum on the first N15,000. 15% on the next N15,000 and 20% on the remaining, find his total income tax per annum. NURUL-ISLAM COMPREHENSIVE COLLEGE
Nurul Islam Junction, Eyita-Ojokoro Road, Eyita, Ikorodu Lagos.
www.nurulislamschools.org MATHEMATICS
LESSON NOTE
2023/2024 ACADEMIC SESSION
SECOND TERM
CLASS: SS3
LAWAL ABDULLATEEF OLAWALE
15/01/2024
WEEK TWO
DATE: 15/01/2024 WEEK: TWO SUBJECT: MATHEMATICS CLASS: SSS THREE LESSON TITLE: COORDINATE GEOMATRY OF A STRAIGHT LINE PERIOD: 1 and 2 DURATION: 80 minutes LEARNING OUTCOME/OBJECTIVES: At the end of the lesson, students should be able to; (i) Calculate the distance between two points (ii) Calculate the mid-point between. LEARNING RESOURCES: Audio Visual Resources. Objects of different lengths or height Model Mathematical Set Web Resources www.khanacademy.com www.byjus.com RESOURCES AND MATERIALS (i) New General Mathematics 2 published by (ii) MAN Mathematics 2 published by the Association of Mathematics teachers
CONTENT:
Lesson objectives: At the end of the lesson students should be able to; Calculate the distance between two points.
RECTANGULAR COORDINATE SYSTEM The rectangular coordinate system consists of two real number lines that intersect at a right angle. The horizontal number line is called the x-axis, and the vertical number line is called the y-axis. These two number lines define a flat surface called a plane, and each point on this plane is associated with an ordered pair of real numbers (x, y). The first number is called the x-coordinate, and the second number is called the y-coordinate. The Intersection of the two axes is known as the origin, which corresponds to the point (0,0).
An ordered pair (x, y) represents the position of a point relative to the origin. The x-coordinate represents a position to the right of the origin if it is positive and to the left of the Origin if it is negative. The y-coordinate represents a position above the origin if it is positive and below the origin if it is negative. Using this system, every position (point) in the plane is uniquely identified. For example, the pair (2,3)(2,3) denotes the position relative to the origin as shown:
This system is often called the Cartesian coordinate system, named after the French mathematician René Descartes (1596– 1650). The x- and y-axes break the plane into four regions called quadrants, named using roman numerals I, II, III, and IV, as pictured. In quadrant I, both coordinates are positive. In quadrant II, the x-coordinate is negative and the y-coordinate is positive. In quadrant III, both coordinates are negative. In quadrant IV, the x-coordinate is positive and the y-coordinate is negative.
Example 1 Plot the ordered pair (−3,5) and de termine the quadrant in which it lies. Solution: The coordinates x = −3 and y = 5 indicate a point 3 units to the left of and 5 units above the origin.
Answer: The point is plotted in quadrant II (QII) because the x-coordinate is negative and the y-coordinate is positive. Ordered pairs with 00 as one of the coordinates do not lie in a quadrant; these points are on one axis or the other (or the point is the origin if both coordinates are 00). Also, the scale indicated on the x-axis may be different from the scale indicated on the y-axis. Choose a scale that is convenient for the given situation. Example 3 Plot this set of ordered pairs: {(4,0),(−6,0),(0,3),(−2,6),(−4,−6)}. Solution: Each tick mark on the x-axis represents 22 units and each tick mark on the y-axis represents 33 units.
Example 4 Plot this set of ordered pairs: {(−6,−5),(−3,−3),(0,−1),(3,1),(6,3)}. Solution:
In this example, the points appear to be collinear, or to lie on the same line. The entire chapter focuses on finding and expressing points with this property.
CLASS WORK Plot the set of points {(5,3),(−3,2),(−2,−4),(4,−3)} and indicate in which quadrant they lie. Answer Graphs are used in everyday life to display data visually. A line graph consists of a set of related data values graphed on a coordinate plane and connected by line segments. Typically, the independent quantity, such as time, is displayed on the x-axis and the dependent quantity, such as distance traveled, on the y-axis.
ASSIGNMENT Give the coordinates of points A,B,C,D, and E. 1.
DATE: 15/01/2024 WEEK: TWO SUBJECT: MATHEMATICS CLASS: SSS Three LESSON TITLE: COORDINATE GEOMETRY OF A STRAIGHT LINE PERIOD: 3 and 4 DURATION: 80 minutes LEARNING OUTCOME/OBJECTIVES: At the end of the lesson, students should be able to; Determine the length of a line LEARNING RESOURCES: Audio Visual Resources. Objects of different lengths or heights Model Mathematic Resources Web Resources www.khanacademy.com www.byjus.com RESOURCES AND MATERIALS (i) New General Mathematics 2 published by (ii) MAN Mathematics 2 published by CONTENT: Lesson objectives: At the end of the lesson students should be able to; Determine the length of a line
Example 6 The following line graph shows the number of mathematics and statistics bachelor’s degrees awarded in the United States each year since 1970.
How many mathematics and statistics bachelor’s degrees were awarded in 1975?
In which years were the number of mathematics and statistics degrees awarded at the low of 11,000?Solution: a. The scale on the x-axis represents time since 1970, so to determine the number of degrees awarded in 1975, read the y-value of the graph at x = 5.
The y-value corresponding to x = 5 is 18. The graph indicates that this is in thousands; there were 18,000 mathematics and statistics degrees awarded in 1975. b. To find the year a particular number of degrees was awarded, first look at the y-axis. In this case, 11,000 degrees is represented by 11 on the y-axis; look to the right to see in which years this occurred.
The y-value of 11 occurs at two data points, one where x = 10 and the other where x = 30. These values correspond to the years 1980 and 2000, respectively. Answers: In the year 1975, 18,000 mathematics and statistics degrees were awarded. In the years 1980 and 2000, the lows of 11,000 mathematics and statistics degrees were awarded.
DISTANCE FORMULA Frequently you need to calculate the distance between two points in a plane. To do this, form a right triangle using the two points as vertices of the triangle and then apply the Pythagorean theorem. Recall that the Pythagorean theorem states that if given any right triangle with legs measuring a and b units, then the square of the measure of the hypotenuse c is equal to the sum of the squares of the legs: a2+b2=c2. In other words, the hypotenuse of any right triangle is equal to the square root of the sum of the squares of its legs.
Example 7 Find the distance between (−1,2) and (3,5). Solution: Form a right triangle by drawing horizontal and vertical lines through the two points. This creates a right triangle as shown below:
The length of leg b is calculated by finding the distance between the x-values of the given points, and the length of leg a is calculated by finding the distance between the given y-values. a = 5 – 2 = 3units b = 3 − (−1) = 3+1=4units Next, use the Pythagorean theorem to find the length of the hypotenuse. C = 32+42 √=9+16 √=25 √=5units Answer: The distance between the two points is 5 units. Generalize this process to produce a formula that can be used to algebraically calculate the distance between any two given points.
Given two points, (x1,y1) and (x2,y2), then the distance, d, between them is given by the distance formula:
Example 8 Calculate the distance between (−3,−1) and (−2,4). Solution: Use the distance formula.
It is a good practice to include the formula in its general form as a part of the written solution before substituting values for the variables. This improves readability and reduces the chance for errors. Answer:
EXERCISE Calculate the distance between (−7,5) and (−1,13).
Example 10 Do the three points (1,−1),(3,−3)(1,−1),(3,−3), and (3,1)(3,1) form a right triangle? Solution: The Pythagorean theorem states that having side lengths that satisfy the property a2+b2=c2 is a necessary and sufficient condition of right triangles. In other words, if you can show that the sum of the squares of the leg lengths of the triangle is equal to the square of the length of the hypotenuse, then the figure must be a right triangle. First, calculate the length of each side using the distance formula .
EXERCISE Graph the given set of ordered pairs. {(−4,5),(−1,1),(−3,−2),(5,−1)} {(−15,−10),(−5,10),(15,10),(5,−10)} {(−2,5),(10,0),(2,−5),(6,−10)} {(−8,3),(−4,6),(0,−6),(6,9)} {(−1.2,−1.2),(−0.3,−0.3),(0,0),(0.6,0.6),(1.2,1.2)} State the quadrant in which the given point lies. (−3,2) (−12,−15) (17.3,1.9) x<0 and y<0 x<0 and y>0
DATE: 15/01/2024 WEEK: TWO SUBJECT: MATHEMATICS CLASS: SSS THREE LESSON TITLE: COORDINATE GEOMATRY OF A STRAIGHT LINE PERIOD: 5 DURATION: 40 minutes LEARNING OUTCOME/OBJECTIVES: At the end of the lesson, students should be able to; (i) Determine the mip-point of a straight line. LEARNING RESOURCES: Audio Visual Resources. Objects of different lengths or height Model Mathematical Set
Web Resourceswww.khanacademy.com www.byjus.com RESOURCES AND MATERIALS (i) New General Mathematics 2 published by (ii) MAN Mathematics 2 published by CONTENT: Lesson objectives: At the end of the lesson students should be able to; Determine the mid-point of a straight line Midpoint Formula The point that bisects the line segment formed by two points, (x1,y1) and (x2,y2), is called the midpoint and is given by the following formula:
The midpoint is an ordered pair formed by finding the average of the x-values and the average of the y-values of the given points. Example 11 Calculate the midpoint between (−1,−2) and (7,4). Solution: First, calculate the average of the x- and y-values of the given points.
Next, form the midpoint as an ordered pair using the averaged coordinates.
To verify that this is indeed the midpoint, calculate the distance between the two given points and verify that the result is equal to the sum of the two equal distances from the endpoints to this midpoint.
ASSIGNMENT Find the midpoint between (−6,5) and (6,−11) Key Takeaways Use the rectangular coordinate system to uniquely identify points in a plane using ordered pairs (x,y). Ordered pairs indicate position relative to the origin. The x-coordinate indicates position to the left and right of the origin. The y-coordinate indicates position above or below the origin. The scales on the x-axis and y-axis may be different. Choose a scale for each axis that is appropriate for the given problem. Graphs are used to visualize real-world data. Typically, independent data is associated with the x-axis and dependent data is associated with the y-axis. The Pythagorean theorem gives us a necessary and sufficient condition of right triangles. Given a right triangle, then the measures of the sides satisfy a2+b2=c2. Conversely, if the sides satisfy a2+b2=c2, then the triangle must be a right triangle. The distance formula, D = , is derived from the Pythagorean theorem and gives us the distance between any two points, (x1,y1) and (x2,y2), in a rectangular coordinate plane. The midpoint formula, is derived by taking the average of each coordinate and forming an ordered pair.
ASSIGNMENT Calculate the area of the shape formed by connecting the following set of vertices. {(0,0),(0,3),(5,0),(5,3)} {(−1,−1),(−1,1),(1,−1),(1,1)} {(−2,−1),(−2,3),(5,3),(5,−1)} {(−5,−4),(−5,5),(3,5),(3,−4)} {(0,0),(4,0),(2,2)} Calculate the distance between the given two points. (−5,3) and (−1,6) (6,−2) and (−2,4) (0,0) and (5,12) (−6,−8) and (0,0) (−7,8) and (5,−1) (−1,−2) and (9,22) (−1,2) and (−72,−4) (−12,13) and (52,−113) (−13,23) and (1,−13) (12,−34) and (32,14) Calculate the area and the perimeter of the triangles formed by the following set of vertices. {(−4,−5),(−4,3),(2,3)} {(−1,1),(3,1),(3,−2)} {(−3,1),(−3,5),(1,5)} {(−3,−1),(−3,7),(1,−1)} Find the midpoint between the given two points. (−1,6) and (−7,−2) (8,0) and (4,−3) (−10,0) and (10,0) (−3,−6) and (−3,6) (−10,5) and (14,−5) (0,1) and (2,2) (5,−3) and (4,−5) (0,0) and (1,1) (−1,−1) and (4,4) (3,−5)(3,−5) and (3,5)(3,5)
Given the right triangle formed by the vertices (0,0),(6,0), and (6,8), show that the midpoints of the sides form a right triangle.
Given the isosceles triangle formed by the vertices (−10,−12),(0,12), and (10,−12) show that the midpoints of the sides also form an isosceles triangle.
Calculate the area of the triangle formed by the vertices (−4,−3),(−1,1), and (2,−3). (Hint: The vertices form an isosceles triangle.)
Calculate the area of the triangle formed by the vertices (−2,1),(4,1), and (1,−5).
Calculate the area and the perimeter of the triangles formed by the following set of vertices. {(−4,−5),(−4,3),(2,3)}{(−4,−5),(−4,3),(2,3)} {(−1,1),(3,1),(3,−2)}{(−1,1),(3,1),(3,−2)} {(−3,1),(−3,5),(1,5)}{(−3,1),(−3,5),(1,5)} {(−3,−1),(−3,7),(1,−1)} NURUL-ISLAM COMPREHENSIVE COLLEGE
Nurul Islam Junction, Eyita-Ojokoro Road, Eyita, Ikorodu Lagos.
www.nurulislamschools.org MATHEMATICS
LESSON NOTE
2023/2024 ACADEMIC SESSION
SECOND TERM
CLASS: SS3
LAWAL ABDULLATEEF OLAWALE
22/01/2024
WEEK THREE
DATE: 22/01/2024 WEEK: THREE SUBJECT: MATHEMATICS CLASS: SSS THREE LESSON TITLE: COORDINATE GEOMATRY OF A STRAIGHT LINE PERIOD: 1 and 2 DURATION: 80 minutes LEARNING OUTCOME/OBJECTIVES: At the end of the lesson, students should be able to; (i) Calculate the distance between two points (ii) Calculate the mid-point between. LEARNING RESOURCES: Audio Visual Resources. Objects of different lengths or height Model Mathematical Set Web Resources www.khanacademy.com www.byjus.com RESOURCES AND MATERIALS (i) New General Mathematics 2 published by (ii) MAN Mathematics 2 published by the Association of Mathematics teachers
CONTENT:
Lesson objectives: At the end of the lesson students should be able to; Calculate the distance between two points.
general differentiation formula? Answer: There are six general differentiation formulas which are as follows:
Question: What is the difference between differentiation and derivative? Answer: Derivative of a function is the rate of change of the output value with respect to its input value. On the other hand, the differential is the actual change of a function. Most importantly, differentiation is the process of finding a derivative. Question: What is the derivative of 0? Answer: The derivative of 0 is 0 because in general, we have the following rule for finding the derivative of a constant function, f(x) = a.
DATE: 15/01/2024 WEEK: THREE SUBJECT: MATHEMATICS CLASS: SSS Three LESSON TITLE: COORDINATE GEOMETRY OF A STRAIGHT LINE PERIOD: 3 and 4 DURATION: 80 minutes LEARNING OUTCOME/OBJECTIVES: At the end of the lesson, students should be able to; Determine the length of a line LEARNING RESOURCES: Audio Visual Resources. Objects of different lengths or heights Model Mathematic Resources Web Resources www.khanacademy.com www.byjus.com RESOURCES AND MATERIALS (i) New General Mathematics 2 published by (ii) MAN Mathematics 2 published by CONTENT: Lesson objectives: At the end of the lesson students should be able to; D
DATE: 15/01/2024 WEEK: THREE SUBJECT: MATHEMATICS CLASS: SSS THREE LESSON TITLE: COORDINATE GEOMATRY OF A STRAIGHT LINE PERIOD: 5 DURATION: 40 minutes LEARNING OUTCOME/OBJECTIVES: At the end of the lesson, students should be able to; (i) Determine the mip-point of a straight line. LEARNING RESOURCES: Audio Visual Resources. Objects of different lengths or height Model Mathematical Set
Web Resourceswww.khanacademy.com www.byjus.com RESOURCES AND MATERIALS (i) New General Mathematics 2 published by (ii) MAN Mathematics 2 published by CONTENT: Lesson objectives: At the end of the lesson students should be able to; D NURUL-ISLAM COMPREHENSIVE COLLEGE
Nurul Islam Junction, Eyita Ojokoro road, Eyita, Ikorodu Lagos.
www.nurulislamschools.org MATHEMATICS
LESSON NOTE
2023/2024 ACADEMIC SESSION
SECOND TERM
CLASS: SS3
LAWAL ABDULLATEEF OLAWALE
29/01/2024
WEEK FOUR
DATE: 29/01/2024 WEEK: FOUR SUBJECT: MATEMATICS CLASS: SSS THREE LESSON TITLE: . PERIOD: 1 and 2 DURATION: 80 minutes LEARNING OUTCOME/OBJECTIVES: At the end of the lesson, students should be able to; Derive the roots of the method. LEARNING RESOURCES: Audio Visual Resources. Object of different length or height Model Mathematical Set Web Resources www.khanacademy.com www.byjus.com RESOURCES AND MATERIALS (i) New General Mathematics 2 published by (ii) MAN Mathematics 2 published by Association of Mathematics teachers
CONTENT
DATE: 29/01/2024 WEEK: FOUR SUBJECT: MATEMATICS CLASS: SSS THREE LESSON TITLE: . PERIOD: 3 and 4 DURATION: 80 minutes LEARNING OUTCOME/OBJECTIVES: At the end of the lesson, students should be able to; Use the graphical method to solve quadratic Equation. LEARNING RESOURCES: Audio Visual Resources. Object of different length or height Model Mathematical Set Web Resources www.khanacademy.com www.byjus.com RESOURCES AND MATERIALS (i) New General Mathematics 2 published by (ii) MAN Mathematics 2 published by CONTENT: Lessons objectives: At the end of the lesson students should be able to;
DATE: 29/01/2024 WEEK: FOUR SUBJECT: MATEMATICS CLASS: SSS THREE LESSON TITLE: PERIOD: 5 DURATION: 40 minutes LEARNING OUTCOME/OBJECTIVES: At the end of the lesson, students should be able to; Solve application problems involving quadratic equation. LEARNING RESOURCES: Audio Visual Resources. Object of different length or height Model Mathematical Set
Web Resourceswww.khanacademy.com www.byjus.com RESOURCES AND MATERIALS (i) New General Mathematics 2 published by (ii) MAN Mathematics 2 published by CONTENT: Example: What is ddx(5x−2)3 ? The Chain Rule says: the derivative of f(g(x)) = f’(g(x))g’(x) (5x−2)3 is made up of g3 and 5x−2: f(g) = g3 g(x) = 5x−2 The individual derivatives are: f’(g) = 3g2 (by the Power Rule) g’(x) = 5 d/dx(5x−2)3 = (3g(x)2)(5) = 15(5x−2)2