National 5 Course Notes. Scientific Notation (or Standard Form) This is a different way of writing very large and very small numbers in the form:-

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1 National 5 Course Notes Scientific Notation (or Standard Form) This is a different way of writing very large and very small numbers in the form:- a x 10 n where a is between 1 and 10 and n is an integer (positive or negative whole number). Example:- Write these numbers in scientific notation. (a) (b) = 2.34 x 10 6 = x 10-4 Rule: Put a decimal point after the first non-zero to get a, then count the number of places the decimal point has moved to get n. Example:- Write these numbers in full. (a) 6.3 x 10 4 (b) 8.87 x 10-2 = = Volume of Solids Remember to use cubic units for volume, i.e. mm 3, cm 3, m 3, etc. Special Prisms

2 V = (½bh) x h = (½ x 6 x 11) x 14 = 33 x 14 = 462 cm 3 You need to know all of the above formulae but the next ones are given in the formula sheet. The Sphere Example:- The Cone V = 1 3 πr2 h

3 Example:- Find the volume of this cone. V = 3.14 x 15 2 x 40 3 V = 9420 cm 3 Remember to use the radius and not the diameter! 30 2 = 15 Composite Volumes Split the shape up into separate simple shapes then add/subtract to find volume of composite shape. Example V cone = 3.14 x 6 2 x 18 3 V cylinder = 3.14 x 6 2 x 30 V hemisphere = (4 x 3.14 x 6 3 3) 2 V cone = cm 3 V cylinder = cm 3 V hemisphere = cm 3 Total Volume = = cm 3 Liquid Volumes Remember:- 1 cm 3 = 1 millilitre (ml) 1000 ml = 1 litre

4 Changing the Subject of a Formula

5 Simultaneous Equations

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7 The Straight Line The general form of the straight line is y = mx + c, where m is the gradient and c is the y-intercept (where the line cuts the y-axis). The formula for the gradient is m = y 2 y 1 x 2 x 1. Pick 2 points on the line (x 1, y 1 ) and (x 2, y 2 ) and substitute into the equation, then solve. Example Find the gradient of a straight line passing through the points (-3, 1) and (2,11). (x 2, y 2 ) Solution:- Use m = y 2 y 1 x 2 x 1 and substitute values in. So, m = ( 3) = 10 5 = 2. (x 1, y 1 ) Example A straight line has equation y = 5x 3. Write down the gradient and y- intercept of the line. Solution:- Gradient m = 5 and y-intercept c = -3.

8 If you know the gradient and one point on a straight line, you can work out the equation of the line using the formula:- y b = m(x a). Example:- Find the equation of the straight line with gradient of 3, where the point (2, 4) lies on the line. (a, b) Solution:- y b = m(x a) => y 4 = 3(x 2) m => y 4 = 3x 6 => y = 3x => y = 3x 2 Example:- Find the equation of this line. Solution - Pick 2 points on the line (1,3) and (3,7) Calculate the gradient m = = 4 2 = 2. Pick a 3 rd point (2,5) Sub into y b = m(x a) y 5 = 2(x 2) y 5 = 2x 4 y = 2x y = 2x + 1 Example:- Write down the gradient and y-intercept of 2y + 6x 2 = 0. Solution Re-arrange equation into the form y = mx + c, so 2y + 6x 2 = 0 2y + 6x = 2 take the -2 to the other side 2y = -6x + 2 take the 6x to the other side y = -3x + 1 divide everything by 2 So m = -3 and c = 1. N.B. Lines with a positive gradient slope upwards from left to right. Lines with a negative gradient slope downwards from left to right. Parallel lines have equal gradients. Horizontal lines have a gradient of 0 and equation y = the y-intercept. Vertical lines have an undefined gradient and equation x = the x-intercept.

9 Circles You need to know the following formulae:- d = 2r C = πd A = πr 2 d = C π r = A π where d = diameter, r = radius, π = 3.14, C = circumference and A = area. Example Calculate the length of the arc. Length of an Arc arc length = angle 360 x πd Solution:- arc length = 45 x 3.14 x 10 = 3.925cm 360 Remember to use the diameter d = 2r = 2 x 5 = 10cm Example Find the angle at the centre of the circle. Solution:- Find the circumference of the circle: C = πd = 3.14 x 14 = 43.96cm Re-arrange the arc length formula to give: angle = arc length circumference x 360o. = x 360 = = 147.4o.

10 Example:- Find the area of the sector. Area of a Sector sector area = = angle 360 x πr2 sector area = 135 x 3.14 x = = 35.6cm 2 Example:- Find the size of the angle at the centre of the circle. Solution - Find the area of the circle: A = 3.14 x 40 2 = 5024m 2 Re-arrange the sector formula: angle = sector area circle area x 360o Exam Type Questions = x 360o = = 301 o Area = x 402 = = 488.9cm 2 You are being asked to calculate the length of the major arc. Arc length = 250 x 80 = = 55.6cm 360

11 Solution QC = radius = 15cm AC = AB BC = = 12cm Use Pythagoras to find AQ AQ 2 = QC 2 AC 2 = = = 81 AQ = 81 = 9cm so PQ = 2 x 9 = 18cm Algebra

12 Factorising You must factorise in the following order:- 1. Common Factor e.g. 10ab 25b 2 = 5b(2a 5b) N.B. always remove the Highest Common Factor 2. Difference of Two Squares e.g. 9x 2 25 = (3x 5)(3x + 5) Look for 2 terms that you can take the square root of and a subtraction sign. 3. Trinomial/Quadratic e.g. x 2 10x + 24 = (x 4)(x 6) Look for 2 numbers which add to the middle number (-10) and multiply to the end number (24). Examples:- 1. ab 2 ac 2 = a(b 2 c 2 ) = a(b c)(b + c) common factor then diff of 2 squares 2. 7x x + 28 = 7(x 2 + 4x + 4) = 7(x + 2)(x + 2) = 7(x + 2) 2 common factor then trinomial 3. 3x 2 5x 2 = (3x + 1)(x 2). Method 3 x -2 = -6 -> factors which add to -5 are -6 and 1 Re-write 3x 2 5x 2 as 3x 2 6x - x 2 Split it into 3x 2 6x and x 2 Factorise each part by removing a common factor 3x 2 6x = 3x(x-2) and x - 2 = 1(x - 2). So brackets become (3x + 1)(x 2). Put 2 common factors of 3x and 1 together. Both brackets should match after you remove common factor! Equations and Inequalities Remember the rule:- CHANGE SIDE, CHANGE SIGN! All the letter terms go on the left, numbers on the right. Examples 1. 3x + 5 = 2x x + 4 = 2x x - 20 = 10x x = 9-5x =>3x 2x = 3 5 => 5x 2x = 10 4 => 11x 10x = => x + 5x = 9 6 => x = -2 => 3x = 6 => x = 5 => 6x = 3 => x = 2 => x = ½

13 Solve inequations in the same way, but remember if you divide by a negative, you need to change the direction of the inequality sign. Examples 1. 7x + 3 > 3x x 1 < 9x 57 => 7x 3x > 29 3 => x 9x < => 4x > 26 => -8x < -56 => x > 6.5 => x > = 7. Change direction of inequality sign as dividing by a negative Pythagoras Example

14 Similarity Method:- 1. Find the scale factor ( new old ) 2. Find the area factor by squaring the scale factor or volume factor by cubing the scale factor. 3. Find the missing value. Examples 1. Find the area of the larger rectangle. Scale factor = 20 4 = 5 Area factor = 5 2 = 25 Area = 25 x 12 = 300cm 2

15 2. Find the volume of the bigger cylinder. Scale Factor = 16 8 = 2 Volume Factor = 2 3 = 8 Volume = 8 x 250 = 2000ml Percentages - Compound Interest To get the multiplier Increase interest rate, e.g. 8% increase = = 1.08 Decrease interest rate, e.g. 8% decrease = = 0.92 Example Sarah puts 300 in the bank and leaves it in for 5 years. The interest rate she receives is 2.3%. How much does she have in the bank after 5 years? Method - amount x (1 + %) n, where n is the number of years. Solution x ( ) 5 = 300 x = = Appreciation/Depreciation This is where the value of something increases (appreciation) or decreases (depreciation). Example Depreciation - Tim bought a computer three years ago for Each year since then it has depreciated by 15 % of its value at the start of the year. How much is the computer worth now? Solution 1400 x (1 0.15) 3 = 1400 x = = Example Appreciation Zoe bought a flat in 2000 for It increased in value by 7% each year until she sold it in How much did she sell it for (to 3 significant figures)? Solution x = = 98400

16 Reverse Percentages Method Example 1. What % are we dealing with now? 2. Find what 1% is worth 3. Now multiply by 100 to find out the original 100% Since Jim bought his house, it has appreciated in value by 12% and is now worth How much did Jim pay for his house originally? Solution % -> % -> = % -> 500 x 100 = Fractions Changing Top Heavy Fractions to Mixed Numbers divide numerator by denominator. Write the remainder as the new numerator. Example Change 25 4 to a mixed number. Solution = 6 r 1. So 25 4 = 61 4 Changing Mixed Numbers to Top Heavy Fractions multiply whole number by denominator then add answer to numerator. Example - Change 7 4 to a top heavy fraction. 9 Solution - 7 x = = 67. So = 67 9 Adding/Subtracting Fractions The denominators have to be the same before you can add/subtract fractions. You will need to change one or both fractions to have a common denominator. Examples = = Multiply top and bottom by same number (usually denominator of other fraction) = = Borrow from whole number so subtraction is possible! = = = = = Remember to simplify answers as far as possible, i.e. simplify fractions and change top heavy fractions to mixed numbers.

17 Multiplying Fractions Rule Top x Top, Bottom x Bottom. Change mixed numbers to top heavy fractions being doing multiplication then change back to mixed number at the end. Examples x 3 8 = x = = 6 40 = 3 20 = = = = Dividing Fractions Rule Similar to Multiplying Fractions but flip the second fraction, then multiply as above. Examples = = = = = = = = = 4 11 Statistics Mean = add up all the numbers number of numbers Median = the middle number in an ordered dataset Mode = the most common number Range = highest lowest value Example Find the mean, median, mode and range of the following dataset. 23, 56, 24, 23, 12, 28. Solution Mean = = = 27.7 (to 1 dp) Median = 23.5 (put data in order - 12, 23, 23, 24, 28, 56) Mode = 23 Range = = 44.

18 Five-Figure Summary The five figure summary consists of the following L = the lowest figure in the data set Q 1 = quartile 1 (median of lower half of dataset) Q 2 = quartile 2 which is the same as the median Q 3 = quartile 3 (median of upper half of dataset) H = the highest figure Example What is the 5-figure summary for this dataset? L Q 1 Q 2 Q 3 H L = 2, Q 1 = 5, Q 2 = 6.5, Q 3 = 8, H = 10 Semi-Interquartile Range (SIQR) SIQR = Q 3 Q 1 2 = = 1.5 Standard Deviation This is a measure of spread. The lower the standard deviation, the less spread out the scores/data points are.

19 Scattergraphs Example Draw a line of best fit onto the scattergraph and determine its equation. Solution - Try to get your line to go through as many points as possible, with as many points above as below the line. Now find the equation of the line:- c = 10 (read off the graph) Pick 2 points on the line (must be on the graph too), e.g. (0,10) and (60, 60). Calculate gradient m = = So y = 5 x Estimate the Maths mark of a student who got 70% in Physics. Solution Put x = 70 into equation => y = 5 x = = 68%.