Grade 10: Mathematics

In this unit, 'Real numbers', you learn to apply Euclid's division lemma and fundamental theorem of arithmetic. You work with notes, a worked line and a self-mark. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Real numbers' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: apply Euclid's division lemma and fundamental theorem of arithmetic. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, State Euclid's division lemma: for positive integers a and b there are unique q and r with a = bq + r and 0 <= r < b. To find the HCF of 225 and 135, divide the larger by the smaller: 225 = 135 x 1 + 90.
2. Step 2, Repeat with the divisor and remainder. 135 = 90 x 1 + 45. Then 90 = 45 x 2 + 0. The remainder is now 0, so the last divisor, 45, is the HCF of 135 and 225.
3. Step 3, Use the lemma to prove a fact. Any integer is 3q, 3q + 1 or 3q + 2. Squaring gives 9q^2 = 3(3q^2), then 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1, then 9q^2 + 12q + 4 = 3(3q^2 + 4q + 1) + 1. So a square is 3m or 3m + 1, never 3m + 2.
4. Step 4, Apply the fundamental theorem of arithmetic: every composite number is a product of primes in one unique order. Factor 360: 360 = 2 x 180 = 2 x 2 x 90 = 2 x 2 x 2 x 45, so 360 = 2^3 x 3^2 x 5.
5. Step 5, Find HCF and LCM from prime factors. 12 = 2^2 x 3 and 18 = 2 x 3^2. HCF takes the smaller powers: 2 x 3 = 6. LCM takes the larger powers: 2^2 x 3^2 = 36. Check HCF x LCM = 6 x 36 = 216 = 12 x 18.
6. Step 6, Your turn: use Euclid's algorithm on 84 and 36, then find their LCM. Answer: 84 = 36 x 2 + 12 and 36 = 12 x 3 + 0, so HCF = 12. LCM = 84 x 36 / 12 = 252.

Activity. About 30 to 40 minutes of class work, then one short practice task for home. Keep evidence: a worked task and a corrected copy.

Standard to meet. Apply Euclid's division lemma and fundamental theorem of arithmetic: at least 8 of 10 marks on a short task, with one line of reasoning for each answer.

In this unit, 'Polynomials', you learn to find zeroes and relate them to coefficients. You work with notes, a worked line and a self-mark. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Polynomials' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: find zeroes and relate them to coefficients. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Find zeroes of p(x) = x^2 - 5x + 6 by factorising: (x - 2)(x - 3) = 0, so the zeroes are 2 and 3. On a graph, the parabola cuts the x-axis at (2, 0) and (3, 0).
2. Step 2, Relate to coefficients: for ax^2 + bx + c the sum of zeroes is -b/a and the product is c/a. Here sum = 5 = -(-5)/1 and product = 6 = 6/1. Both agree with 2 + 3 and 2 x 3.
3. Step 3, Build a polynomial: zeroes with sum -3 and product -10. Use x^2 - (sum)x + product = x^2 + 3x - 10. Check by factorising: (x + 5)(x - 2), zeroes -5 and 2, sum -3, product -10.
4. Step 4, Find an unknown: x^2 - 6x + k has a zero at 2. Substitute: 4 - 12 + k = 0, so k = 8. Then the other zero is 6 - 2 = 4 by the sum rule, and 2 x 4 = 8 matches the product.
5. Step 5, Quadratic with a leading coefficient: for 2x^2 - 8x + 6 the sum of zeroes is 8/2 = 4 and the product is 6/2 = 3. The zeroes are 1 and 3, since 2 - 8 + 6 = 0 and 18 - 24 + 6 = 0.
6. Step 6, Your turn: find the sum and product of the zeroes of 3x^2 - 12x + 9, then the zeroes. Answer: sum 4, product 3; factorise 3(x - 1)(x - 3), so zeroes 1 and 3.

Activity. About 30 to 40 minutes of class work, then one short practice task for home. Keep evidence: a worked task and a corrected copy.

Standard to meet. Find zeroes and relate them to coefficients: at least 8 of 10 marks on a short task, with one line of reasoning for each answer.

In this unit, 'Pair of linear equations in two variables', you learn to solve by graph, substitution and elimination. You work with notes, a worked line and a self-mark. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Pair of linear equations in two variables' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: solve by graph, substitution and elimination. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Elimination: x + y = 10 and x - y = 4. Add the equations: 2x = 14, so x = 7. Then y = 10 - 7 = 3. Check in the second: 7 - 3 = 4. The solution is (7, 3).
2. Step 2, Substitution: 2x + 3y = 12 and y = x - 1. Replace y: 2x + 3(x - 1) = 12, so 5x - 3 = 12, 5x = 15, x = 3. Then y = 2. Check: 6 + 6 = 12 and 2 = 3 - 1.
3. Step 3, Word problem: 3 pens and 2 notebooks cost 36; 2 pens and 5 notebooks cost 57. So 3p + 2n = 36 and 2p + 5n = 57. Multiply the first by 5 and the second by 2: 15p + 10n = 180 and 4p + 10n = 114.
4. Step 4, Subtract: 11p = 66, so p = 6. Then 3 x 6 + 2n = 36 gives n = 9. Check in the second equation: 2 x 6 + 5 x 9 = 12 + 45 = 57. A pen costs 6 and a notebook 9.
5. Step 5, Graph meaning: x + y = 10 and x - y = 4 are two lines that meet at exactly one point, (7, 3). But x + y = 5 and 2x + 2y = 12 are parallel, so there is no solution.
6. Step 6, Your turn: solve 3x + y = 11 and x + y = 5. Answer: subtract to get 2x = 6, so x = 3 and y = 2. Check: 9 + 2 = 11.

Activity. About 30 to 40 minutes of class work, then one short practice task for home. Keep evidence: a worked task and a corrected copy.

Standard to meet. Solve by graph, substitution and elimination: at least 8 of 10 marks on a short task, with one line of reasoning for each answer.

In this unit, 'Quadratic equations', you learn to solve by factors, completing the square and the formula. You work with notes, a worked line and a self-mark. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Quadratic equations' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: solve by factors, completing the square and the formula. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Factors: x^2 - 5x + 6 = 0. Find two numbers with product 6 and sum -5, namely -2 and -3. So (x - 2)(x - 3) = 0 and x = 2 or x = 3. Check x = 2: 4 - 10 + 6 = 0.
2. Step 2, Completing the square: x^2 + 6x - 7 = 0. Write x^2 + 6x + 9 = 16, so (x + 3)^2 = 16. Then x + 3 = 4 or -4, giving x = 1 or x = -7. Check x = 1: 1 + 6 - 7 = 0.
3. Step 3, Formula: 2x^2 + 3x - 5 = 0. The discriminant D = 3^2 - 4 x 2 x (-5) = 9 + 40 = 49. Then x = (-3 plus or minus 7) / 4, so x = 1 or x = -5/2. Check: 2 + 3 - 5 = 0.
4. Step 4, Nature of roots: D > 0 gives two real roots, D = 0 gives equal roots, D < 0 gives none. For x^2 - 4x + 4 = 0, D = 16 - 16 = 0, so a repeated root x = 2.
5. Step 5, Word problem: a rectangle's length is 3 more than its width and area is 40. So w(w + 3) = 40, w^2 + 3w - 40 = 0, (w + 8)(w - 5) = 0. Width 5, length 8. Reject w = -8.
6. Step 6, Your turn: solve x^2 - 7x + 12 = 0 and 3x^2 - x - 2 = 0. Answer: x = 3 or 4; and D = 1 + 24 = 25, x = (1 plus or minus 5) / 6, so x = 1 or -2/3.

Activity. About 30 to 40 minutes of class work, then one short practice task for home. Keep evidence: a worked task and a corrected copy.

Standard to meet. Solve by factors, completing the square and the formula: at least 8 of 10 marks on a short task, with one line of reasoning for each answer.

In this unit, 'Arithmetic progressions', you learn to find nth term and sum. You work with notes, a worked line and a self-mark. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Arithmetic progressions' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: find nth term and sum. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Identify: in the AP 3, 7, 11, 15, the first term a = 3 and the common difference d = 4. Check d with 7 - 3 = 4 and 15 - 11 = 4.
2. Step 2, nth term: a_n = a + (n - 1)d. For n = 10: 3 + 9 x 4 = 39. So the 10th term is 39. List check: 3, 7, 11, 15, 19, 23, 27, 31, 35, 39 is ten terms.
3. Step 3, Find n: which term equals 99? 3 + (n - 1) x 4 = 99, so (n - 1) x 4 = 96, n - 1 = 24, n = 25. The 25th term is 99.
4. Step 4, Sum: S_n = n/2 x (first + last). S_10 = 10/2 x (3 + 39) = 5 x 42 = 210. Alternative formula n/2 x (2a + (n - 1)d) = 5 x (6 + 36) = 210, which agrees.
5. Step 5, Word problem: a stack has 20 logs in the bottom row and one fewer in each row up, down to 1. That is the sum 1 + 2 + ... + 20 = 20/2 x (1 + 20) = 210 logs.
6. Step 6, Your turn: for the AP 5, 9, 13, ..., find the 12th term and the sum of 12 terms. Answer: a_12 = 5 + 11 x 4 = 49; S_12 = 6 x (5 + 49) = 324.

Activity. About 30 to 40 minutes of class work, then one short practice task for home. Keep evidence: a worked task and a corrected copy.

Standard to meet. Find nth term and sum: at least 8 of 10 marks on a short task, with one line of reasoning for each answer.

In this unit, 'Triangles', you learn to apply similarity and the Pythagoras theorem. You work with notes, a worked line and a self-mark. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Triangles' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: apply similarity and the Pythagoras theorem. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Basic proportionality: in triangle ABC, DE is parallel to BC with D on AB and E on AC. If AD = 4, DB = 6 and AE = 6, then AD/DB = AE/EC gives 4/6 = 6/EC, so EC = 9.
2. Step 2, Check: the ratio AD : DB = 2 : 3 and AE : EC = 6 : 9 = 2 : 3. Equal ratios confirm DE is parallel to BC. Also AD/AB = 4/10 = AE/AC = 6/15 = 2/5.
3. Step 3, Area ratio: similar triangles with sides in ratio 2 : 3 have areas in ratio 4 : 9. If the smaller triangle has area 16 cm2, the larger has 16 x 9/4 = 36 cm2.
4. Step 4, Shadow problem: a 6 m pole casts a 4 m shadow while a tower casts a 20 m shadow at the same time. The triangles are similar, so height / 20 = 6 / 4, and the height is 30 m.
5. Step 5, Pythagoras: a right triangle with legs 5 and 12 has hypotenuse the root of 25 + 144 = 169, which is 13. The converse: 8, 15, 17 gives 64 + 225 = 289 = 17^2, so it is a right triangle.
6. Step 6, Your turn: DE is parallel to BC, AD = 3, DB = 5 and AE = 6. Find EC. And two similar triangles have sides in ratio 3 : 4, the small area being 27. Answer: EC = 10; the larger area is 27 x 16/9 = 48.

Activity. About 30 to 40 minutes of class work, then one short practice task for home. Keep evidence: a worked task and a corrected copy.

Standard to meet. Apply similarity and the Pythagoras theorem: at least 8 of 10 marks on a short task, with one line of reasoning for each answer.

In this unit, 'Coordinate geometry', you learn to use distance, section and area formulae. You work with notes, a worked line and a self-mark. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Coordinate geometry' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: use distance, section and area formulae. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Distance formula: between (2, 3) and (8, 11), distance = root of (8 - 2)^2 + (11 - 3)^2 = root of 36 + 64 = root of 100 = 10 units.
2. Step 2, Midpoint: for (-2, 4) and (6, -2) the midpoint is ((-2 + 6)/2, (4 - 2)/2) = (2, 1). Check: distance from (-2, 4) to (2, 1) is 5, and from (2, 1) to (6, -2) is 5.
3. Step 3, Section formula: find the point dividing (1, 2) and (7, 8) in the ratio 1 : 2. x = (1 x 7 + 2 x 1)/3 = 9/3 = 3 and y = (1 x 8 + 2 x 2)/3 = 12/3 = 4. The point is (3, 4).
4. Step 4, Check the section: distance from (1, 2) to (3, 4) is root 8, and from (3, 4) to (7, 8) is root 32 = 2 root 8. The ratio is 1 : 2, which agrees.
5. Step 5, Area of a triangle with vertices (1, 1), (5, 1), (3, 5): half of the absolute value of 1(1 - 5) + 5(5 - 1) + 3(1 - 1) = half of |-4 + 20 + 0| = 8 square units. Check: base 4, height 4, area 8.
6. Step 6, Your turn: find the distance between (-1, 2) and (4, 14) and the midpoint of (4, 6) and (10, 2). Answer: root of 25 + 144 = 13; midpoint (7, 4).

Activity. About 30 to 40 minutes of class work, then one short practice task for home. Keep evidence: a worked task and a corrected copy.

Standard to meet. Use distance, section and area formulae: at least 8 of 10 marks on a short task, with one line of reasoning for each answer.

In this unit, 'Introduction to trigonometry and its applications', you learn to use ratios, identities and heights and distances. You work with notes, a worked line and a self-mark. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Introduction to trigonometry and its applications' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: use ratios, identities and heights and distances. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Ratios: in a right triangle, sin A = 3/5 means opposite = 3 and hypotenuse = 5. The adjacent side is the root of 25 - 9 = 4, so cos A = 4/5 and tan A = 3/4.
2. Step 2, Identity check: sin^2 A + cos^2 A = 9/25 + 16/25 = 1. Also tan A = sin A / cos A = (3/5) / (4/5) = 3/4. Both match, so the ratios are consistent.
3. Step 3, Standard values: sin 30 = 1/2, cos 60 = 1/2, tan 45 = 1. Evaluate sin 30 + cos 60 = 1/2 + 1/2 = 1. And tan 45 x sin 30 = 1/2.
4. Step 4, Height problem: from 20 m away the angle of elevation of a tower top is 30 degrees. Height = 20 x tan 30 = 20 / root 3 = about 11.55 m. Draw the right triangle first.
5. Step 5, Equal angle: at 45 degrees elevation and 30 m from the base, tan 45 = 1 gives height = 30 m. A ladder of 10 m at 60 degrees to the ground reaches 10 x sin 60 = 5 root 3 = about 8.66 m.
6. Step 6, Your turn: if tan theta = 5/12, find sin theta and cos theta. Answer: the hypotenuse is root of 25 + 144 = 13, so sin theta = 5/13 and cos theta = 12/13.

Activity. About 30 to 40 minutes of class work, then one short practice task for home. Keep evidence: a worked task and a corrected copy.

Standard to meet. Use ratios, identities and heights and distances: at least 8 of 10 marks on a short task, with one line of reasoning for each answer.

In this unit, 'Circles and constructions', you learn to use tangent properties and construct tangents. You work with notes, a worked line and a self-mark. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Circles and constructions' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: use tangent properties and construct tangents. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Tangent and radius: a tangent is perpendicular to the radius at the point of contact. A circle with radius 5 cm has centre O, and P is 13 cm from O. Then PT^2 = 13^2 - 5^2 = 144.
2. Step 2, Length of tangent: PT = 12 cm. Check with the 5, 12, 13 triple. The two tangents from P have the same length 12, so the figure is symmetric about the line OP.
3. Step 3, Angle between tangents: if the angle at the centre between the two radii to the points of contact is 120 degrees, the angle between tangents is 180 - 120 = 60 degrees, since the quadrilateral has two right angles.
4. Step 4, Construct tangents from P, 7 cm from the centre of a circle of radius 3 cm: join OP, find its midpoint M, draw a circle centre M with radius MO. It cuts the given circle at T and T'. Join PT and PT'.
5. Step 5, Verify the construction: angle OTP is in a semicircle, so it is 90 degrees, and PT is a tangent. Length PT = root of 49 - 9 = root of 40, about 6.32 cm. Measure to check.
6. Step 6, Your turn: find the tangent length from a point 17 cm from the centre of a circle of radius 8 cm. Answer: PT = root of 289 - 64 = root of 225 = 15 cm.

Activity. About 30 to 40 minutes of class work, then one short practice task for home. Keep evidence: a worked task and a corrected copy.

Standard to meet. Use tangent properties and construct tangents: at least 8 of 10 marks on a short task, with one line of reasoning for each answer.

In this unit, 'Areas related to circles and surface areas and volumes', you learn to solve problems on sectors and combined solids. You work with notes, a worked line and a self-mark. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Areas related to circles and surface areas and volumes' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: solve problems on sectors and combined solids. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Sector: radius 14 cm and angle 90 degrees. Area = 90/360 x pi r^2 = 1/4 x 22/7 x 196 = 154 cm2. Arc length = 1/4 x 2 x 22/7 x 14 = 22 cm.
2. Step 2, Another sector: radius 21 cm, angle 60. Area = 1/6 x 22/7 x 441 = 231 cm2 and arc = 1/6 x 2 x 22/7 x 21 = 22 cm. Perimeter of the sector adds two radii, 22 + 42 = 64 cm.
3. Step 3, Combined solid: a capsule is a cylinder of radius 7 and height 10 with a hemisphere on one end. Cylinder volume = 22/7 x 49 x 10 = 1,540 cm3.
4. Step 4, Hemisphere volume = 2/3 x pi r^3 = 2/3 x 22/7 x 343 = 2,156/3 = about 718.67 cm3. Total volume = 1,540 + 718.67 = about 2,258.67 cm3.
5. Step 5, Surface of the capsule: curved cylinder 2 x 22/7 x 7 x 10 = 440, hemisphere 2 pi r^2 = 308, and the flat circular end 154. Total 440 + 308 + 154 = 902 cm2.
6. Step 6, Your turn: find the area of a sector with radius 7 cm and angle 120 degrees. Answer: 120/360 x 22/7 x 49 = 1/3 x 154 = about 51.33 cm2.

Activity. About 30 to 40 minutes of class work, then one short practice task for home. Keep evidence: a worked task and a corrected copy.

Standard to meet. Solve problems on sectors and combined solids: at least 8 of 10 marks on a short task, with one line of reasoning for each answer.

In this unit, 'Statistics and probability', you learn to find mean, median and mode of grouped data and calculate probability. You work with notes, a worked line and a self-mark. You see it, say it, then do it alone.

• Show it with objects or a picture: build or draw the idea in 'Statistics and probability' before using symbols.
• Say the rule in your own words, then write one worked line, naming each step.
• Use it: find mean, median and mode of grouped data and calculate probability. Start with one easy question, then one with a real-life story.
• Check it: use a second method, estimate, or work backwards to test your answer.

Step by step.

1. Step 1, Grouped mean: classes 0-10, 10-20, 20-30, 30-40, 40-50 have f = 5, 8, 12, 10, 5 and midpoints 5, 15, 25, 35, 45. Then fx = 25, 120, 300, 350, 225 and N = 40.
2. Step 2, Mean = sum fx / N = 1,020 / 40 = 25.5. Check: it lies in the middle class region 20-30, which is reasonable because the biggest frequency is there.
3. Step 3, Mode: the modal class is 20-30 with f1 = 12, f0 = 8, f2 = 10. Mode = 20 + (12 - 8) / (24 - 8 - 10) x 10 = 20 + 4/6 x 10 = 26.67.
4. Step 4, Median: N/2 = 20. Cumulative frequencies 5, 13, 25 show the median class is 20-30. Median = 20 + (20 - 13)/12 x 10 = 20 + 5.83 = 25.83. The three averages are close.
5. Step 5, Probability: a card is drawn from a pack of 52. P(king) = 4/52 = 1/13. P(a red face card) = 6/52 = 3/26. For a die, P(a prime) = 3/6 = 1/2, since 2, 3, 5 are prime.
6. Step 6, Your turn: find the probability of a multiple of 3 when a die is rolled, and of drawing a spade from 52 cards. Answer: 2/6 = 1/3; and 13/52 = 1/4.

Activity. About 30 to 40 minutes of class work, then one short practice task for home. Keep evidence: a worked task and a corrected copy.

Standard to meet. Find mean, median and mode of grouped data and calculate probability: at least 8 of 10 marks on a short task, with one line of reasoning for each answer.

On the slide: L1 Basic · Count and name shapes

Notes: Count objects up to 20 and name a circle, square and triangle. Notes: give ten buttons and two trays; learners count aloud, move three across and say how many are left (7), then hunt the room for one shape of each kind.

On the slide: L2 Awareness · Add, subtract and measure

Notes: Add and subtract within 100 and measure a length with a ruler. Notes: work 46 + 27 = 73 on a number line, jumping 20 then 7; learners measure a pencil in centimetres and explain how they checked their answer a second way.

On the slide: L3 Foundation · Fractions and decimals

Notes: Compare fractions and decimals and use the four operations with whole numbers. Notes: shade 3/4 and 6/8 on equal strips to show they match, then add 0.5 + 0.25 = 0.75 and write it as 75 per cent of one whole.

On the slide: L4 Practitioner · Equations and percentages

Notes: Solve one-step and two-step equations and percentage problems, showing each step. Notes: practice case is 3x + 5 = 20, so x = 5, checked by substituting back; then a 15 per cent discount on 400 leaves 340, because 400 x 0.85 = 340.

This course: Grade 10 Mathematics is at this level. Its units include Real numbers; Polynomials; Pair of linear equations in two variables; Quadratic equations and more.

On the slide: L5 Specialist · Geometry, data and graphs

Notes: Use the right-angled triangle rule, the mean and a straight-line graph to solve a problem. Notes: legs of 6 and 8 give a longest side of 10; the data 4, 6, 6, 8, 11 have mean 7; plot y = 2x + 1 and read y = 7 at x = 3.

On the slide: L6 Advanced · Proof, calculus and algebra

Notes: Prove results and use calculus and matrices to model change. Notes: differentiate f(x) = x^3 - 3x to get 3x^2 - 3, so turning points sit at x = 1 and x = -1; the second derivative 6x shows a minimum then a maximum. Learners prove the square root of 2 is irrational.

On the slide: L7 Super expert · Research-level modelling

Notes: Formulate and defend a proof or statistical model, stating assumptions and limits. Notes: for 25 values with mean 50 and standard deviation 10, the standard error is 2 and a 95 per cent interval is about 45.9 to 54.1 (t = 2.064). Learners check residuals and write a referee-style critique.