maths revision (foundation)


types of number and bodmas

integers- another name for a whole number- either a positive or negative number, or zero

examples: -365, 0, 1, 17, 989, 1 234 567 890

rational or irrational

rational numbers can be written as fractions.rational numbers come in 3 different forms: integers e.g. 4(=4/1), -5(=-5/1), -12(=-12/1). fractions p/q where p and q are (non-zero) integers, e.g. 1/4, -1/2, 7/4. terminating or recurring decimals e.g. 0.125 (=1/8), 0.333333... (=1/3)

irrational- they cant be written as fractions, they are never ending, non-repeating decimals. square roots of +ve integers are either integers or irrational (e.g. suquare root 2 and 3 are irrational, but square root 4 = 2 isnt). surds are numbers or expressions containing irrational roots. pi is also irrational.

bodmas- brackets, other, division, multiplication, addition, subtraction. Bodmas tells you the order in which to work something out.

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multiples, factors and prime

multiples- the multiples of a number are just its time table.

factor- the factors of a number are all the numbers that divide into it.

method: start off with 1 x itself, then try 2 x, then 3 x and so on

prime number- is a number that doesn't divide by anything except for 1 and itself. (the only exception is 1, which is not a prime number)

finding prime factors- the factor tree: any number can be broken down into a string of prime factors all multiplied together- this is called 'prime factor decomposition' or 'prime factorisation'

                                                  420                                                         so 400=2x2x3x5x7

                            42                                         10                                               =2(2)x3x5x7

                        7          6                              2             5

                               2        3                         

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LCM (least common multiple)- the smallest number that will divide by all the numbers in find the LCM you need to create a prime factor tree for both of the numbers.for example for 18 the prime factors you will find will be: 2x3x3 and for 30 it is: 2x3x5, then place it in a venn diagram, like the image. after you have done that times all the numbers together, so the LCM is 90.

HCF (highest common factor)- the biggest number that will divide into all the numbers in the question.example: to then find the HCF of 18 and 30 only times the numbers they have in common, so the answer would be 6.

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the number on the top of the fraction is the numerator, and the number on the bottom is the denominator.

cancelling down- to cancel down or simplify a fraction, divide top and bottom by the same number, till they dont go down any further.

mixed numbers- mixed numbers are things like 31/3, with an integer part and a fraction part. improper fractions are ones where the top number is larger then the bottom. you need to convert between the two.

multiplying- multiply top and bottom separately. it usually helps to cancel down first if you can.

dividing- turn the second fraction upside down and then multiply.

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fractions 2

common denominators- good for ordering fractions by size, and for adding or subtracting fractions. you need to find a number that all the denominators divide into- this will be your common denominator, the simplest way is to find the LCM of the denominators.

adding, subtracting (sort the denominators first)- make sure the denominators are the same, times diagonally then subtract or add the two top numbers. eg. 2/5 + 4/7= 20+14/35 = 34/35

fractions of something: example- what is 9/20 of 360? '9/20 of' means 9/20 x ? so multiply the 'something' by the top of the fraction, and divide it by the bottom. 9/20 of 360= (360/ 20) x9 =£18 x 9 =£162

expressing a fraction: write 180 as a fraction of 80- just write the first number over the second and cancel down. 180/80 = 9/4

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fractions, decimals and percentages

                  divide                           x by 100

fraction                           decimal                        percentage

  35%   e.g. 0.35 x 100            e.g. 7/20 is 7 divide 20     

percentage                     decimal                        fraction

   / 100

terminating decimals- the digits after the decimal point go on top and a power of 10 goes on the bottom depending on gow many digits there are after the decimal point. e.g. 0.6= 9/10 0.12= 12/100 0.345= 345/1000

recurring decimals- have a pattern of numbers that repeat forever. e.g. 1/3= 0.33333... , it doesnt have to be a single number that repeats. e.g. 0.143143143... repeating is marked with dots on top, then everything from the first dot to the seciond dot is repeating bit. e.g. 0.2(.)5= 0.2555... 0.2(.)5(.)= 0.252525...

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rounding numbers

decimal places- identify the position of the last digit from the number of decimal places, then look at the next digit to the right called the decider. if the decider is 5 or more, then round up the last digit, if the decider is 4 or less, then leave the last digit how it is. there must be no more digits after the last digit.

significant figures (s.f)- the method for s.f is identical to that for decimal places except that locating the last digit is more difficult. the 1st significant figure of any number is simply the first digit which isnt a 0. the 2nd, 3rd, 4th etc.significant figures follow on immediately after the first, regardless of being zeros or not zeros.

0.002309                                           2.03070

2=1st  3=2nd  0=3rd  9=4th               2=1st  0=2nd  3=3rd  0=4th

after rounding the last digit, end zeros must be filled in up to, but not beyond, the decimal point.

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estimating calculations- estimate the value of 127.8 + 41.9 / 56.5 x 3.2 = 130 + 40 / 60 x 3 = 170 / 180 = 0.9444... = 1. keep rounding up so there is no decimal point. a cylindrical glass has a height of 18cm and a radius of 3cm. find an estimate in cm3 for the volume of the glass. the formula for the volume of a cylinder is V= pi.r(2)height. round the numbers to 1 s.f pi=3.14159... =3 (1 s.f), height=20cm (1 s.f) and radius= 3cm (1 s.f) put it into the formula V=pi r(2) h = 3x3 (2) x 20 = 3x9x20= 540 cm (3). use your answer to estimate the number of glasses that called be filled from a 2.5 litre bottle of lemonade. 2.5 litres=2500cm(3) 2500/540=2500/500 = 5 glasses.

estimating square roots- find two square numbers, one either side of the number you're given. decide which number is closest to, and make a sensible estimate of the digit after the decimal point. estimate the value of 87 square root to 1 d.p. 87 is between 81 (=9(2)) and 100 (=10(2)) its closer to 81, so its square root will be closer to 9 than 10: square root 87= 9.3.

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upper and lower bounds- when a measurement is rounded to a given unit, the actual measurement can be anything up to half a unit bigger or smaller. the mass of a cake is given as 2.4 kg to the nearest 0.1. find the interval within which m, the actual mass of the cake, lies.

lower bound = 2.4 - 0.05 = 2.35 kg

upper bound = 2.4 + 0.05 = 2.45kg                         so the interval is 2.35kg < m < 2.45kg

the actual value is greater than or equal to the lower bound but strictly less than the upper bound. the actual mass of the cake could be exactly 2.35kg.

5 has been rounded to the nearest whole number, it could have been 4.5 < x < 5.5

a number can also be truncated (cut off) 2.1 has been truncated to 1 dp 2.1 < x < 2.2

32.7 g has been truncated to 1dp 32.7 < x < 32.8

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upper and lower bounds- when a measurement is rounded to a given unit, the actual measurement can be anything up to half a unit bigger or smaller. the mass of a cake is given as 2.4 kg to the nearest 0.1. find the interval within which m, the actual mass of the cake, lies.

lower bound = 2.4 - 0.05 = 2.35 kg

upper bound = 2.4 + 0.05 = 2.45kg                         so the interval is 2.35kg < m < 2.45kg

the actual value is greater than or equal to the lower bound but strictly less than the upper bound. the actual mass of the cake could be exactly 2.35kg.

5 has been rounded to the nearest whole number, it could have been 4.5 < x < 5.5

a number can also be truncated (cut off) 2.1 has been truncated to 1 dp 2.1 < x < 2.2

32.7 g has been truncated to 1dp 32.7 < x < 32.8

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upper and lower bounds- when a measurement is rounded to a given unit, the actual measurement can be anything up to half a unit bigger or smaller. the mass of a cake is given as 2.4 kg to the nearest 0.1. find the interval within which m, the actual mass of the cake, lies.

lower bound = 2.4 - 0.05 = 2.35 kg

upper bound = 2.4 + 0.05 = 2.45kg                         so the interval is 2.35kg < m < 2.45kg

the actual value is greater than or equal to the lower bound but strictly less than the upper bound. the actual mass of the cake could be exactly 2.35kg.

5 has been rounded to the nearest whole number, it could have been 4.5 < x < 5.5

a number can also be truncated (cut off) 2.1 has been truncated to 1 dp 2.1 < x < 2.2

32.7 g has been truncated to 1dp 32.7 < x < 32.8

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standard form

standard form is useful for writing very big or very small numbers in a more covenient way, e.g. 56000 000 000 would be 5.6 x 10(10) in standard form. 0.000 000 003 45 would be 3.45 x 10(-9)

a number written in standard form must always be in exactly this form: A x 10(n) A= must always be between 1 and 10.  10(n)= this number is just the number of places the decimal point moves.

three rules: 1)the front number must always be between 1 and 10. 2) the power of 10, n, is how far the decimal point moves. 3) n is positive for big numbers, n is negative for small numbers.

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standard form 2

calculations with standard form

multiplying and dividing- rearrange to put the front numbers and the powers of 10 together. multiply or divide the front numbers, and use to power rules to multiply or divide the powers of 10. make sure the answer is still in standard form.

find (2 x 10(3)) x (6.75 x 10(5)) =  (2 x 6.75) x (10(3) x 10(5)) = 13.5 x 10(3+5) = 13.5 x 10(8)= 1.35x10(9)

adding and subtracting- make sure the powers of 10 are the same- you'll probably need to rewrite one of them. add or subtract the front numbers. convert the answer to standard form.

calculate (9.8 x 10(4)) + (6.6 x 10(3))= (9.8 x 10(4)) + (0.66 x 10(4)) = (9.8 + 0.66) x 10(4) =    10.46 x 10(4) = 1.046 x 10(5)

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algebra basics

negative numbers- + + makes +.  + - makes -.  - + makes -.  - - makes +

1) multiplying or dividing. e.g. -2 x 3 = -6  -8 / -2 = +4. 2) two signs are together e.g. 5 - -4 = 5 + 4 =9

letters multiplied together- 1) abc means a x b x c. 2) gn(2) means g x n x n. 3) (gn)(2) means g x g x n x n. 4) p(q-r)(3) means p x (q-r) x (q-r) x (q-r).

terms- a term is a collection of numbers, letters and brackets, all multiplied/divided together.

terms are seperated by + and - signs. everyterm has a + or - infront of it.

simplifying or 'collecting like terms': to simplify an algebraic expression, you combine like terms- terms that have the same combination of letters.

simplify 2x - 4 + 5x + 6= 7x + 2

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powers and roots

the seven easy rules:

1) when multiplying, you add the powers e.g. 3(6) x 3(4) = 3(6+4)= 3(10).

2) when dividing, you subtract the powers e.g. 5(4) / 5(2) = 5(4-2)= 5(2).

3) when raising one power to another, you multiply them e.g. (3(2))(4)= 3(2x4)= 3(8).

4)  x(1)= x, anything to the power 1 is just itself e.g. 3(1)=3.

5) x(0)=1, anything to the power 0 is just 1 e.g. 5(0)=1. 6)

6) 1(x)=1, 1 to any power is still just 1 e.g. 1(23)=1. 

7) fractions-apply the power to both top and bottom e.g. (1 3/5)(3)= (8/5)(3)= 8(3)/5(3)= 512/125. 

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multiplying out brackets

single brackets: the main thing to remember when multiplying out brackets is that the thing outside the bracket multiplies each seperate term inside the bracket. e.g. 4a(3b - 2c) = (4a x 3b) + (4a x -2c) = 12ab - 8ac

double brackets: multiply everything in the first bracket by everything in the seond bracket. e.g. expand and simplify (2p - 4) (3p + 1) = 6p(2)-12p+2p-4= 6p(2)-10p-4

(3x + 5)(2) = (3x + 5) (3x + 5) = 9x(2)+15x+15x+25= 9x(2)= 30x+25

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factorising- putting brackets in

1)take out the biggest number that goes into all the terms

2) for each letter in turn, take out the highest power (eg. x, x(2) etc.) that will go into every term.

3)open the bracket and fill in all the bits needed to reproduce each term.

4)check your answer by multiplying out the bracket and making sure it matches the original expression.

example: factorise 3x(2) + 6x = 3x(x + 2)

biggest number that'll divide into 3 and 6 (3). highest power of x that will go into both terms (x)

factorise 8x(2)y + 2xy(2) = 2xy(4x + y)

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solving equations

rearrange until you have x=number- And mixture of x's and numbers.1) first, rearrange the equation so that all the x's are on one side and the numbers are one side and the numbers are on the other. combine terms where you can. 2) then divide both sides by the number multiplying x to find the value of x. example: solve 5x+4=8x- 5. (+5) 5x+4+5=8x-5+5 = 5x+9=8x. (-5x) 5x+9-5x=8x-5x = 9=3x. (/3) 9/3=3x/3 = 3=x

multiply out brackets first- If your equation has brackets in it. 1) multiply them out before rearranging. 2) solve it in the same way as above.example: solve 3(3x-2)=5x+10 = 9x-6=5x+10. (-5x) 9x-6-5x=5x+10-5x = 4x-6=10. (+6) 4x-6+6=10+6 = 4x=16. (/4)4x/4=16/4 = x=4

get rid of fractions- 1) you need to get rid of fractions before anything else. 2) to get rid of fractions, multiply every term of the equation by whatever is on the bottom of the fraction. if there are two fractions, you'll need to multiply by both denominators. example: 1) solve x+2 / 4 =4x-7. (x4) 4(x+2) / 4=4(4x)-4(7) = x+2=16x-28 = 30=15x = 2=x. 2) solve 3x+5 / 2=4x+10 / 3. (x2) (x3) 2x3x(3x+5) / 2=2x3(4x+10) / 3 = 3(3x+5)=2(4x+10) = 9x+15=8x+20 = x=5.

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solving equations 2

solving equations using the 6-step method: 

1)get rid of any fractions. 2) multiply out any brackets. 3) collect all the x-terms o one side and all number terms on the other. 4) reduce it to the form 'ax=b' (by combining terms). 5) finally divide both sides by A to give 'x=  ', and thats your answer. 6) if you had 'x(2)=  ' instead, square root both sides to end up with 'x=+  '.

dealing with squares 

at step 5, and then step 6 is to take square roots. there's one important thing to remember: whenever you take the square root of a number, the answer can be positive or negative. example: there are 75 tiles on a roof. each row contains three times the number of tiles as each column. how many tiles are there in one column? (let the number of tiles in a column be x. write an equation for the total number of tiles in terms of x) 3x * x =75 = 3x(2)=75. (/3) x(2)=25. (square root) x=5

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rearranging formulas

rearrange formulas with the soving equations method

is very similar to solving equations. 1) get rid of any square root signs by squaring both sides. 2) get rid of any fractions. 3) multiply out any brackets. 4) collect all the subject terms on one side and all non-subject terms on the other. 5) reduce it to the form 'ax=b' (by combining like terms). you might have to do some factorising here too. 6) divde both sides by A to give 'x=  '. 7) if your left with 'x(2)=   ', square root both sides to get 'x= +  ' (dont forget the +).

if the subject appears in a fraction

you wont always need to use the 7 steps so just ignore the ones that dont apply. example: make b the subject of the formula a=5b+3 / 4. (x4) 4a=4(5b+3) / 4 = 4a=5b+3. (-3) 5b=4a-3. (/5) b=4a-3 / 5.

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rearranging formulas 2

if theres a square or square root involved

if the subject appears as a square or in a square root. example: 1) make u the subject of the formula v(2)=u(2)+2as. (-2as) u(2)=v(2)-2as. (square root) u=+square root(v(2)-2as). 2) make n the  subject of the formula 2(m+3)=square root (n+5). (square it) (2(m+3))(2)=(squareroot(n+5))(2) = 4(m(2)+6m+9)=n+5 = 4m(2)+24m+36=n+5. (-5) n=4m(2)+24m+31

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factorising quadratics

factorising a quadratic

1)factorising a quadratic means putting it into 2 brackets. 2) the standard format for quadratic equations is: ax(2)+bx+c=0. 3) if a=1, the quadratic is much easier to deal with. 4) as well as factorising a quadratic, you might be asked to solve the equation. this just means finding the values of x that make each bracket 0.

factorising method when a=1

1) always rearrange into the standard format: x(2)+bx+c=0. 2) write down the two brackets with the x's in: (x    )(x   )=0. 3) then find 2 numbers that multiply to give 'c' (the end number) but also add/subtract to give 'b' (coefficient of x). 4) fill in the +/- signs and make sure they work out properly. 5) as an essential check, expand the brackets to make sure they give the origenal equation. 6) finally, solve the equation by setting each bracket equal to 0. example: solve x(2)-x=12. 1)rearrange into standard format x(2)-x-12=0. 2) write down the initial brackets. (x   )(x   )=0 3) find the right pairs of numbers that multiply to give c (=12) and add or subtract to give b (=1) 3x4 add/subtract to give 7 or 1. (x   3)(x   4)=0 = (x+3)(x-4)=0. solve equation by setting each bracket equal to 0. (x+3)=0 = x=-3. (x-4)=0 = x=4.

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finding the nth term of a linear sequence

this method works for linear sequences- ones with common difference (where the terms increase or decrease by the same amount each time). linear sequences are also known as arithmetic sequences. example: find an expression for the nth term of the sequence that starts 5, 8, 11, 14...

n:             1          2          3           4               the commo difference is 3, so '3n' is in the formula

term:        5          8          11         14

                     +3         +3         +3

5-3=2. you have to +2 to get to the term

so the answer is 3n+2

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sequences 2

deciding if a term is in a sequence

you might be given the nth term and asked if a certain value is in the sequence. the trick here is to set the expression equal to that value and solve to find n. if n is a whole number, the value is in the sequence. example: the nth term of a sequence is given by n(2)-2. a) find the 6th term in the sequence. 6(2)=36-2 = 34. b) is 45 a term in this sequence? set it equal to 45... n(2)-2=45 = n(2)=47 n=square root 47= 6.8556... n is not a whole number, so 45 is not in the squence.

other types of sequences 

you could be asked to continue a sequence that isnt linear or quadratic. these sequences usually involve doing something to the previous terms in order to find the next one. example: find the next two terms in each of the following sequences. a) 0.2, 0.6, 1.8, 5.4, 16.2... the rule for this sequence is 'multiply the previous term by 3' so the next two terms are: 16.2x3=48.6 and 48.6x3=145.8. (this is an example of a geometric sequence) b) 1, 1, 2, 3, 5 the rule for this one is 'add the two previous terms', so the next two terms are. 3+5=8 and 5+8=13. (this is known as the fibonacci sequence).

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the inequality symbols:  > means 'greater than'       < means 'less than'      ≥ means 'greater than or equal to'     ≤ means 'less than or equal to'

algebra with inequalities: the key with inequalities, is you solve them like regular equations but with a big exception. whenever you multiply or divide by a negative number, you must flip the inequality sign. example: 1) x is an ingeger such that -4 < x ≤ 3. find all the possible values of x. work out what each bit of the inequality is telling you: -4 < x means 'x is greater than -4', x ≤ 3 means 'x is than than or equal to 3'. now just write down all the values that x can take. (remember, integers are just +ve or -ve whole numbers) -3, -2, -1, 0, 1, 2, 3.

you can show inequalities on number lines- all you have to remember is that you use an open circle for > or < and a coloured in circle for ≤ and ≥. show the inequality -1 < x ≤ 4.(

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simultaneous equations

six steps for easy simultaneous equations.

example: solve the simultaneous equations 2x=6-4y and -3-3y=4x. 1) rearrange both equations into the form ax+by=c, and label the two equations (1) and (2). 2x + 4y = 6  (1)  4x +3y = -3  (2).   2) match up the numbers in front (the 'coefficients') of either the x's or y's in both equations. you may need to multiply one or both equations by a suitable number. relabel them (3) and (4)(1)x2 4x + 8y =12  (3)  4x + 3y =-3  (4) . 3) add or subtract the two equations to eliminate the terms with the same coefficient. (3) - (4) = 5y=15. 4) solve the resulting equation. 5y=15 = y=3. 5) substitute the value you've found back into the equation (1) and solve it. 2x+(4x3)=6 = 2x+12=6 = 2x=-6 = x=-3. substitute both these values into equation (2) to make sure it works. if it doesn't the you have done something wrong and you'll have to do it all again.

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proof with inequalities

example: Ellie says, "if x > y then x(2)> y(2)" is she correct? explain your answer.

try some values for x and y- x=2, y=1 and x(2)=4 > 1=y(2). x=5, y=2 and x(2)=25 > 4=y(2). at first glance ellie is correct. but x=-1, y=-2 but x(2)=1 < 4=y(2). so ellie is wrong as the statement does not hold for all values of x and y.

geometric proof

prove that the sum of the exterior angles of a triangle is 360 degrees.

first sketch the triangle with angles a, b and c. then the exterior angles are: 180 degrees-a, 180 degrees-b and 180 degrees-c. so their sum is: (180-a) + (180-b) + (180-c) = 540 degree-(a+b+c)=540-180 (as the angles in a triangle add up to 180 degrees) = 360

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evaluating functions- just put the numbers in:

just shove the numbers into the function machine.

example: find the value of y when x=36 is put into the function machine.

x ------ /3 ------ +6 ------ y     36 / 3 = 12

             machine A             12 + 6 + 18

the function represented by the machine is y = (x / 3) + 6

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straight lines and gradients

learn to spot these straight line equations

if an equation has a y and/or x but no higher powers (like x(2) or x(3)), then its a straight line equation.

vertical and horizontal lines: 'x = a' and 'y = a'. 'x = a' is a vertical line through 'a' on the x-axis.       'y = a' is a horizontal line through 'a' on the y-axis.

the main diagonal through the origin: 'y=x'. 'y=x' is the main diagonal that goes uphill from left to right.

other sloping lines through the origin: 'y=ax'. the value of 'a' is the gradient.

the gradient is the steepness of the line.

the larger the gradient, the steeper the slope. a negative gradient tells you it slopes downhill. you can find it by dividing the change in y by the change in x.

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using 'y = mx + c' is the most straightforward way of dealing with straight-line equations. the first thing you have to do is rearrange the equation into standard format, like this: where 'm'=gradient of the line. 'c'=y-intercept.  y=2+3x --- y=3x+2 --- (m=3, c=2).  x-y=0 --- y=4/5x-3/5 --- (m=4/5, c=-3/5).

finding the equation of a straight-line graph. example: find the equation of the line on the graph in the form y=mx+c. 1) find 'm' (gradient) and 'c' (y-intercept). 'm'= change in y/change in x = 15/30 = 1/2. 2) use these to write the equation in the form y=mx+c. y= 1/2x + 15.

finding the equation of a line through two points

if your given two points on a line you can find the gradient, then you can use the gradient and on eof the points to find the equation of the line. example: find the equation of the straight line that passes through (-2, 9) and (3, -1). give your answer in the form y=mx+c. 1) use the two points to find 'm'. m=change in y/change in x=-1-9/3-(-2)=-10/5=-2. so y=-2x+c. 2) substitute one of the points into the equation you've just found. substitute (-2, 9) into eqn: 9=-2(-2)+c = 9=4+c. 3) rearrange the equation to find 'c'. c=9-4 = c=5. 4) substitute back into y=mx+c: y=-2x+5.

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drawing straight line graphs

the 'table of 3 values' method-example: draw the graph of y=-2x+4 for values of x from -1 to 4. 1)draw up a table with three suitable values of x. 2) find the y-values by putting each x-value into the equation: when x=4, y=-2x+4 = (-2x4)+4=-4. 3) plot the points and draw the line. the tables gives the points (0, 4), (2, 0) and (4, -4).

using y = mx + c-example: draw the graph of 4y-2x=-4. 1) get the equation into the form y=mx+c. 4y-2x=-4 --- y=1/2x-1. 2) put a dot on the y-axis at the value of c. 'c'=1. 3) using m, go across and up or down a certain number of units. make another dot, then repeat this step a few times in both directions. go 2 along and 1 up, because 'm'=+1/2. 4) when you have 4 or 5 dots, draw a straight line through them. 5) finally check that the gradient looks right. a gradient of +1/2 should be quite gentle and uphill left to right- which it is.

the 'x=0, y=0' method- example: sketch the straight line y=3x-5 on the diagram. 1)set x=0 in the equation, and find y- this is where it crosses the y-axis. y=3x-5. when x=0, y=-5. 2) set y=0 in the equation and find x- this is where it crosses the x-axis. when y=0, 0=3x-5. so x=5/3. 3) mark on the two points and draw a line passing through them.

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find the mid-point using the average of the end points.

to find the mid-point of a line segment, just add the x-coordinates and divide by two, then do the same for the y-coordinates.

example: points A and B are given by the coordinates (7, 4) and (-1, -2) respectively. M is the mid-point of the line segment AB. find the coordinates of M. add the x-coordinate of A to the x-coordinate of B and divide by two to find the x-coordinate of the midpoint. do the same with the y-coordinates. (7 + -1 / 2 , 4 + -2 / 2) = (6/2 , 2/2) = (3, 1). so the mid point of AB has coordinates (3,1).

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parallel lines

parallel lines have the same gradient

parallel lines all have the same gradient, which means their y=mx+c equations all have the same value of m. so the lines: y= 2x + 3, y= 2x and y= 2x - 4 are all parallel.

example: line j has a gradient of -0.25. find the equation of line k, which is parallel to line j and passes through point (2, 3). 1) first find the 'm' value for line k. lines j and k are parallel so their gradients are the same m= -0.25. 2) substitute the value for 'm' into y = mx + c to give you the 'equation so far'.  y= -0.25x + c. 3) substitute the x and y values for the given point on line k and solve for 'c'.  when x=2, y=3: 3=(-0.25x2) + c --- 3= -0.5 +c. c=3.5. 4) write out the full equation  y= -0.25x + 3.5.

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quadratic graphs

plotting quadratics

example: complete the table of values for the equation y = x(2) + 2x - 3 and then plot the graph.

x  -4  -3  -2  -1   0  1  2

5    0  -3  -4  -3  0  5

1) substitute each x-value into the equation to get each y-value. e.g. y=(-4)(2) + (2x-4) -3 = 5.       2) plot the points and join them with a completely smooth curve.

dont let one point drag your graph off in a different direction. you will never get spikes or lumps.

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harder graphs

x(3) graphs: y = ax(3) + bx(2) + cx + d

all x(3) graphs have a wiggle in the middle, sometimes its a flat wiggle, sometimes its more pronounced. -x(3) graphs always go down from the top left, + x(3) ones go up from bottom left.


example: draw the graph of y= x(3) + 4x(2) for values of x between -4 and +2.

start by making a table of values:

x  -4  -3  -2  -1   0   1    2           plot the points and join them with a lovely smooth curve. dont use a

y   0   9   8    3   0   5   24          ruler.

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real-life graphs

graphs can show billing structures

many bills are made up of two charges - a fixed charge and a cost per unit charge. e.g. you might pay £11 each month for your phone line, and then be charged 3p for each minute of calls you make.

example: this graph shows how a broadband bill is calculated.

a) how many gigabytes (GB) of internet usage are included in the basic monthly cost? 18 GB. the first section of the graph is horizontal. you're charged £24 even if you dont usre the internet during the month. its only after you have used 18 GB that the bill starts rising.

b) what is the cost for each additional gigabyte (to the nearest 1p)? gradient of sloped section=cost per GB vertical change/horizontal change = 11/19 = £0.5789... per GB. to the nearest 1p is £0.58.

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distance-time graphs

1) at any point, gradient=speed. 2)  the steeper the graph, the faster its going. 3) flat sections are where it stopped. 4)if the gradient's negative, its coming back.

example: henry went out for a ride on his bike. after a while he got a puncture and stopped to fix it. this graph shows the first part of henrys journey.

a)at what time did henry leave home? he left home at the point where the line starts at 8:15. b)how far did henry cycle before getting a puncture? the horizontal part of the graph is where henry stopped 12km. c) what was Henry's speed before getting a puncture? using the speed formula is the same as finding the gradient. speed= distance/time = 12km/0.5 hours = 24km/h. d) at 9:30 henry turns round and cycles home at 24km/h complete the graph to show this. you have to work out how long it will take henry to cycle the 18km home. time = distance/speed = 18km/24km/h = 0.75 hours. 0.75 x 60 mins = 45 mins.

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gradients of real-life graphs

the gradient of a graph represents rate

(y-axis units) per (x-axis units)

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writing ratios as fractions- to write a ratio as a fraction just put one number over the other. e.g. if apples and oranges are in the ratio 2:9 then we say there are 2/9 as many apples as oranges or 9/2 times as many oranges as apples.

reducing ratios to their simplest form- divide all the numbers in the ratio by the same thing. example: write the ratio 15:18 in its simplest form. for the ratio 15:18, both numbers have a factor of 3, so divide them by 3. we cant reduce this anymore. so the simplest form of 15:18 is 5:6.

more awkward cases- if the ratio conatins decimals or fractions- multiply. example: simplify the ratio 2.4:3.6 as far as possible. 1) multiply both sides by 10 to get rid of the decimal parts. 2)now divide to reduce the ratio to its simplest form. 2.4:3.6 x10 = 24:36 /12 = 2:3.

if the ratio has mixed units- convert to the smaller unit. example: reduce the ratio 24mm:7.2cm to its simplest form. 1)convert 7.2cm to mm. 2) simplify the resulting ratio. once units on both sides are the same, get rid of them. 24mm:7.2cm = 24mm:72mm /24 = 1:3.

to get the ratio to the form 1:n or n:1- just divide. example: reduce 3:56 to the form 1:n. divide both sides by 3: 3:56 /3 = 1:56/3 = 1:18 2/3 (or 1:18.6)

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ratios 2

scaling up ratios- if you know the ratio between parts and the actual size of one part, you can scale up the ratio to find the other parts. example: mortar is made from mixing sand and cement in the ratio 7:2. how many buckets of mortar will be made if 21 buckets of sand are used in the mixture? you need to multiply by 3 to go from 7 to 21 on the left hand side (LHS)- so do that on both sides. 7:2 x3 = 21:6. so 21 buckets of sand and 6 buckets of cement are used. amount of mortar made = 21+6=27 buckets.

part:whole ratios- example: mrs miggins owns tabby cats and ginger cats. the ratio of tabby cats to the total number of cats is 3:5. a) what fraction of mrs miggins cats are tabby cats? the ratio tells you that for every 5 cats, 3 are tabby cats. part/whole = 3/5. b) what is the ratio of tabby cats to ginger cats? 3 in every 5 cats are tabby, so 2 in every 5 are ginger. for every 3 tabby cats there are 2 ginger cats. 5-3=2. tabby:ginger = 3:2. c) mrs miggins has 12 tabby cats. how many ginger cats does she have? scale up the ratio from part b) to find the number of ginger cats. tabby:ginger = 3:2 x4 = 12:8. there are 8 ginger cats

proportional division- example: jess, mo and greg share £9100 in the ratio 2:4:7. how much does mo get? 1)add up all the parts. 2+4+7= 13 parts. 2)divide to find one part. £9100/13=£700 (=1 part). 3) multiply to find the amounts, want to find out mo's. 4x£700=£2800.

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ratios 3

changing ratios

example: in an animal sanctuary there are 20 peacocks, and the ratio of peacocks to pheasants is 4:9. if 5 of the pheasants fly away, what is the new ratio of peacocks to pheasants? give your answer in its simplest form. 1) find the origenal number of pheasants. peacocks:pheasants = 4:9 x5 = 20:45. 2)work out the number of pheasants remaining. 45-5= 40 peasants left. 3)write the new ratio of peacocks to pheasants and simplify. peacocks:pheasants = 20:40 /20 = 1:2

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direct and inverse proportion

direct proportion- 1) two quantities, A and B, are in direct proportion if increasing one increases the other one proportionally. so if quantity A is doubled, so is quantity B. 2) remember the golden rule for direct proportion. divide for one, then times for all. example: hannah ays £3.60 per 400g of cheese. she uses 220g of cheese to make 4 cheese pasties. how much would the cheese cost if she wanted to make 50 cheese pasties? in 1 pasty there is: 220g/4 = 55g of cheese, so in 50 pasties there is: 55g x 50 = 2750g of cheese. 1g of cheese would cost: £3.60/400 = 0.9. so 2750g of cheese would cost: 0.9 x 2750 = 2475 = £24.75.

inverse proportion- 1) two quantities, C and D, are in verse proportion if increasing one quantity causes the other quantity to decrease proportionally. so if quantity C is doubled, quatity D is halved. 2) the rule for finding inverse proportion is: times for one, then divide for all. example: 4 bakers can decorate a 100 cakes in 5 hours. a) how long will oit take 10 bakers to decorate the same number of cakes? 100 cakes will take 1 baker: 5x4=20 hours. so 100 cakes will take 10 bakers: 20/10=2 hours for 10 bakers. b) how long will it take 11 bakers to decorate 220 cakes? 1000 cakes will take 1 baker: 20 hours. 1 cake will take 1 baker: 20/100=0.2 hours. 220 cakes will take 1 baker: 0.2x220=44 hours. 220 cakes will take 11 bakers: 44/11=4 hours

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type 1- "find x% of y": turn the percentage into a decimal/fraction, then multiply. e.g. 15% of £46= 15/100 x £46= £6.90

a shopkeeper had a box of 140 chocolate bars. he sold 60% of the chocolate bars for 62p each and he sold the other 40% at 2 for £1. how much did the box of chocolate bars sell for in total? 1) find 60% and 40% of 140. 60% of 140 bars= 0.6x140=84 bars. 40% of 140 bars= 0.4x140=56 bars. 2) so he sold 84 bars for 62p each and 56 bars at 2 for £1. total sales= (84x0.62)+(52/2)= £80.08.

type 2- "find the new amount after a % increase/decrease": find the multiplyer- the decimal thaat represents the percentage change. A %increase has a multiplyer greater than 1. A %decrease has a multiplier less than 1.

type 3- "express x as a percenatge of y": divide x by y, then multiply by 100. e.g. 209 as a percentage of 400=209/400x100=52.25%.

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percentages 2

type 1- finding the percentage change: percentage change= change / original x 100. you end up with a percentage rather than an amount, as you did for type 3 previously. a trader buys 6 watches at £25 each. he scratches one of them, so he sells that one for £11. he sells the other 5 for £38 each. find his profit as a percentage. 1) here the 'change' is profit, so the formula looks like this: percentage profit= profit/original x 100. 2) work out the actual profit (amount made- amount spent): profit= (£38x5)+£11-(6x£25)=£51. 3) calculate the percentage profit: percentage profit= 51/6 x 25 x 100= 34%.

type 2-finding the orignal value: 1)write out the amount in the question as a percentage of the origenal value. 2) divide to find 1% of the origenal value. 3) multiply by 100 to give the origenal value (=100%). a house increases in value by 10.5% to £132 600. find what it was worth before the rise. 1)an increase of 10.5% means £132 600 represents 110.5% of the original value. 2) divide by 110.5 to find 1% of the original value. 132,600= 110.5% / 110.5. 1200 = 1%. 3) then multiply by 100. 1200=1% x100. 120 000 = 100%. so the original value was £120,000.

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percentages 3

simple interest: certain percentage of the original amount only is paid at regular intervals (usually once a year). so the amount of interest is the same very time it's paid. regina invests £380 in an account which pays 3% simple interest each year. how much interest will she earn in 4 years? 1)work out the amount of interest earned in one year: 3% = 3/100=0.03. 3% of £380=0.03 x £380= £11.40. 2) multiply by 4 to get the total interest for 4 years. 4x£11.40= £45.60.

working with percentages: 1) sometimes there isn't a set method you can follow to answer percentage questions. 2) you'll have to use what you've learnt on the last couple of pages and do a bit of thinking for youself. 80% of the members of a gym are male. 35% of the male members are aged 40 and over. what percentage of gym members are males over 40 years old? 1) the percentage of male members under 40 is: 100%-35%=65%. 2) the percentage of gym members that are male and under 40 is: 65% of 80%=0.65x80%=52%.

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compound growth and decay

the formula: N = No x (multiplier)n number of days/hours/years

N (amount after n days/hours/years.) N0 (initial amount.) multiplier (percentage change multiplier)

examples: compound interest- interest added on each time, and the next nterest is calculated using the new total rather than the original amount. daniel invests £1000 in a savings account which pays 8% compound interest per annum. how much will there be after 6 years? use the formula: amount=1000(1.08)6= £1586.87. 'per annum' means 'each year'.

depreciation are about things which decrease in value over time. susan has just bought a car for £6500. a) if the car depreciates by 9% each year, how much will it be worth in 3 years time? use the formula: value=6500(0.91)3=£4898.21. b) how many complete years will it be before the car is worth less than £3000? use the formula again but this time you now dont know n. value=6500(0.91)n. use trial and error to find how many years it will be before the value drops below £3000. if n=8, 6500(0.91)8=3056.6414... n=9, 6500(0.91)9= 2781.5437... it will be 9 years before the car is worth less than £3000. 

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unit conversions

metric and imperial units- common metric conversions: 1cm=10mm. 1 tonne=1000kg. 1m=100cm. 1 litre=1000ml. 1km=1000m. 1 litre=1000cm3. 1kg=1000g. 1cm3=1ml. common imperial conversions: 1 yard=3 feet. 1 foot=12 inches. 1 gallon=8 pints. 1 stone=14 pounds. 1 pound=16 ounces. common metric-imperial conversions: 1kg-2.2 pounds. 1 foot=30cm. 1 gallon=4.5 litres. 1 mile=1.6km.

converting units- to convert between units, multiply or divide by the conversion factor. convert 10 pounds into kg. 2.2 pounds=1kg, so 10 pounds=10/2.2=4.5kg.

converting area and volume measurements:1m2 = 100cm x 100cm = 10,000cm2. 1cm2 = 10mm x10mm = 100m2. 1m2 = 100cm x 100cm x 100cm = 1,000,000cm3. 1cm3 = 10mm x 10mm x 10mm = 1000mm3. convert 9m2 to cm2. to change area measurements from m2 to cm2 multiply by 100 twice. 9 x 100 x 100 = 90,000cm2.

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speed, density and pressure

speed = distance / time. speed is the distance travelled per unit time. a car travels 9 miles at 36 miles per hour. how many minutes does it take? write down the formula, put in the values and calculate: time=distance/speed = 9 miles/36 mph = 0.25 hours = 15 minutes.

density = mass / volume. density is the mass per unit volume of a substance. a giant 'wunda-choc' bar has a density of 1.3 g/cm3. if the bars volume is 1800 cm3, what is the mass of the bar in kg? write down the formula, put in the values and calculate: mass=density x volume = 1.3g/cm3 x 11800cm3 = 2340g = 2.34kg.

pressure = force / area. pressure is the amount of force acting per unit area. it's usually measured in N/m2, or pascals (Pa). a cylindrical barrel with a weight of 200N rests on horizontal ground. the radius of the circular face resting on the ground is 0.4m. calculate the pressure exerted by the barrel on the ground to 1 d.p. work out the area of the circular face: pi x 0.4(2) = 0.5026...m2. write down the pressure formula, put in the values and calculate: pressure=force/area=200n/0.5026...m2=397.8873... N/m2= 397.9 N/m2 (1.d.p)

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5 simple rules- 1)angles that add up to 180 degree, a+b+c=180. 2) angles on a straight line add up to 180 degree, a+b+c=180. 3) angles in a quadrilateral add up to 360 degree. 4) angles round a point add up to 360 degree, a+b+c+d=360. 5) isosceles triangles have 2 sides the same and 2 angle the same.


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parallel lines

angles around parallel lines- when a line crosses two parallel lines, it forms special sets of angles. 1) the two bunches of angles formed at the points of intersection are the same. 2) there are only two different angles invol ed, and they add up to 180 degree. 3) vertically opposite angles (ones opposite each other) are equal.

alternate, allied and corresponding angles. you need to spot characteristic Z, C, U and F shapes:

alternate angles- are the same. they are found in a Z-shape.

allied angles-add up to 180 degree. they are found in a C- or U-shape.

corresponding angles- are the same. found in an F-shape.

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geometry problems

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regular polygons- equilateral triangles, 3 sides. square 4 sides. pentagon, 5 sides. hexagon, 6 sides. heptagon, 7 sides. octagon, 8 sides, nonagon, 9 sides. decagon, 10 sides.

interior and exterior angles

for an polygon (regular or irregular) formulas: sum of exterior angles=360 degree. sum of interior angles= (n-2)x180. (n is the number of sides).

for regular polygons only: exterior angles=360 / n. interior angles=180-exterior angle.

example: the interior angle of a regular polygon is 165 degree. how many sides does the polygon have? first, find the exterior angle of the shape: exterior angle = 180-165=15. use this value to find the number of sides: exterior angle = 360/n so n= 360/exterior angle= 360/15=24 sides.

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triangles and quadrilaterals

triangles- 1)equilateral triangles, 3 equal sides and 3 equal angles of 60 degree. 3 lines of symmetry, rotational symmetry order 3. 2) right-angled triangles, 1 right angle, no lines of symmetry, no rotational symmetry. 3) isosceles triangles, 2 sides the same, 2 angles the same, 1 line of symmetry, no rotational symmetry. 4) scalene triangles, all three sides different, all three angles different, no symmetry.

quadrilaterals- 1) square, 4 equal angles of 90 degree, 4 lines of symmetry, rotational order 4, diagonals are same length and cross at right angles. 2) rectangle, 4 equal sides of 90 degree, 2 lines of symmetry, rotational symmetry order 2, diagonals are same length. 3) rhombus, 4 equal sides, 2 pairs of equal angles, 2 lines of symmetry, rotational symmetry order 2, diagonals cross at right angles. 4) parallelogram, 2 pairs of equal sides, 2 pairs of equal angles, no lines of symmetry, rotational symmetry order 2. 5) trapezium, 1 pair of parallel lines, no lines of symmetry, no rotational symmetry. 6) kite, 2 pairs of equal sides, 1 pair of equal angles, 1 line of symmetry, no rotational symmetry. diagonals cross at right angles.

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the four transformations

translations- the amount the shape moves by is givin as a vector written (x,y)- where x is the horizontal movement and y is the vertical movement (up) if the shape moves left or down, x and y will be negative. shapes are congruent under translation.

rotation-  you must give three details: the angle of rotation, direction of rotation )usually 90 or 180), direction of rotation (clockwise or anticlockwise), the centre of rotation. shapes are congruent under rotation.

reflections- you must give the equation of the mirror line. shapes are congruent

enlargements- you must specify: the scale factor and the centre of enlargement. scale factor= new length/old length.

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area- triangles and quadrilaterals

areas of triangles and quadrilaterals- area of triangle= 1/2 x base x vertical height. area of parallelogram= base x verticle height. area of trapezium= average of parallel sides x distance between them (verticle height)

use the formulas to solve problems-  there is often a bit of algebra into area and perimeter questions

area and circumferance of circles- area of circle= pi x (radius)2. circumferance= pi x diameter or 2 x pi x radius.

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3d shapes- surface area and volume

vertices, faces and edges-  vertices (are the corners), faces (is the square), and edge (is the edges around each face)

surface area- the total area of all the faces added together. surface area of solid=area of net. spheres, cones and cylinders have surface area formulas: sphere= 4 x pi x radius2. cone= pi x radius x slant height + pi x radius2. surface area of a cylinder= 2 x pi x radius x height + 2 x pi x radius2.

volumes of prisms- it has a constant area of cross-section. volume of prism= cross-sectional area x length. cylinder= pi x radius2 x height.

volumes of spheres- volume= 4/3 x pi x radius3. a hemisphere is half a sphere. volume of hemisphere= 2/3 x pi x radius3.

volumes of pyramids and cones- volume of pyramid= 1/3 x base area x vertical height. volume of cone= 1/3 x pi x radius2 x height.

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projections show a 3d shape from different viewpoints- three different viewpoints include: front elevation, side elevation and plan.

1. front elevation- the view you would see from directly infront.

2.plan- the view you would see from directly above.

3. side elevation- the view you would see from directly to one side.

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triangle construction

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loci and construction

a locus- a line or region that shows all the points which fit a given rule.

the four types of loci:

1)the locus of points which are 'a fixed distance from a given point'. this locus is a circle.

2) the locus of points which are 'a fixed distance from a given line'. this locus is a sausage shape. it has straight sides and ends which are perfect semicircles.

3) the locus of points which are 'equidistant from two given lines'. keep the compass setting the same while you make all four marks, make sure you lave your compass marks showing, you get two equal angles- this locus is an angle bisector.

4)the locus of points which are 'equidistant from two given points'. this locus is all points which are the same distance from A as they are from B. this time the locus is actually the perpendicular bisector of the line joining the two points.

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loci and construction 2

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loci and construction- worked examples

finding a locus that satisfies lots of rules:

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1) 'from'- find the word 'from' in the question, and put your pencil on the diagram at the point you are going 'from'. 2) north line- at the point you are going from, draw in a north line. 3) clockwise- now draw in the angle clockwise from the north line to the line joining the two points.

bearings questions and scale drawings:

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pythagoras theorem

pythagoras' theorem- a2 + b2 = c2

1) only works for right-angled triangles. 2) pythagoras uses two sides to find the third side. 3) make sure you get the numbers in the right place. c is the longest side (hypotenuse) and it's always opposite the right angle. 4) always check your answer is sensible.

use pythagoras to find the distance between points- 1) draw a sketch to show the right-angled triangle. 2) find the lengths of the shorter sides of the triangle by subtracting the coordinates. 3) use pythagoras to find the length of the hypotenuse.

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probability basics

all probabilities are between 0 and 1- 1) probabilities are always between 0 and 1. the higher the probability of something, the more likely it is. 2) a probability of zero means it will never happen and a probability of 1 means it definitely will. probabilities can be given as fractions, decimals or percenages.

you can find some probabilities using a formula- this formula only works if all the possible outcomes are equally likely. probability= number of ways for something to happen / total number of possible outcomes.

probabilities add up to 1- 1) if only one possible result can happen at a time, then the probabilities of all the results add up to 1. 2) so since something must either happen or not happen. p(event happens) + p(event doesn't happen) =1.

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counting outcomes

listing all outcomes- a sample space diagram shows all the possible outcomes. it can be a simple list, but a two-way table works well if there are two actvities going on.

example: the spinners are spun (spinner 1 has numbers, 3, 4, 5. spinner 2 has numbers, 1, 2, 3). a) make a sample space diagram showing all the possible outcomes. 1) all the scores from one spinner go along the top. all the scores from the other spinner go down the side. 2) add the two scores together to get the different possible totals. b) find the probability of spinning a total of 6. there are 9 possible outcomes altogether, 3 ways to score 6. p(total=6) = number of ways to score 6 / total number of possible outcomes = 3/9 = 1/3.

+     3     4     5     

1     4     5     6     

    5     6     7

3     6     7     8

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probability experiments

fair or biased- the probability of rolling a three on a normal dice is 1/6- you know that each of the 6 numbers on the dice is equally likely to be rolled, and there is only one 3. but this only works if it is a fair dice. if the dice is biased then each number wont have an equal chance of being rolled.

do the experiment again and again- you need to keep repeating the experiment and count how many times each outcome happens. then you calculate the relative frequency: relative frequency = frequency / number of times you tried the experiment. the more times you do the experiment, the more accurate your estimate of the probability should be.

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probability experiments 2

record results in a frequency trees- when an experiment has two or more steps, then you can record your results using frequency trees.

use probability to find an "expected frequency"- 1)if you estimate how many times you'd expect something to happen if you do an experiment n times. 2)this expected frequency is based on the probability of the result happening. expected frequency of a result = probability x number of trials.

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sets and venn diagrams

showing sets on venn diagrams- 1)sets are just collections of things (numbers) we call these things elements. 2)sets can be written in different ways but they'll always be in a pair of curly brackets { }. 3)n(A) just means 'the number of elements in set A'. 4)on a venn diagram, each set is represented by a circle containing the elements of the set or the number of elements in the set. the universal set, is the group of things that the elements of the sets are selected from. it's everything inside the rectangle. the union of sets A and B, (A U B), contains all the elements in either set A or set B. the intersection of sets A and B, (A n B), contains all the elements in both set A and set B, where the circles overlap. the complement of set A contains all members of the universal set that aren't in set A.


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sampling and bias

use a sample to find out about a population- 1) the whole group you want to find out about is called the population. it can be a group of anything. 2) often you cant survey the whole population, e.g. because it is too big. so you select a smaller group from the population, called a sample. 3) it's really important that your sample fairly represents the whole popuation. for a sample to be representative, it needs to be: a random sample, and big enough. the bigger the sample the more reliable it should be.

simple random sampling- 1)assign a number to every member of the population. 2) create a list of random members. 3) match the random numbers to members of the population.

you need to spot problems with sampling methods- a biased sample is one that doesn't properly represent the whole population. to spot bias think: when, where, and how it is taken. and how many members are in it.

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collecting data

different types of data- qualitative data: is descriptive. it uses words, not numbers. quantitative data measures quantities using numbers. there are two types of quantitive data. discrete data: it's discrete if the numbers can only take certain exact values. continuous data: if the numbers can take any value in a range.

you can organise data into classes- to record data in a table, you often need to group it into classes to make it more manageable. discrete classes should have 'gaps' between them. continuous data classes should have no 'gaps', so are often written using inequalities.

qusetionnaires should be designed carefully- your questions should be: clear and easy to understand, easy to answer and fair- not leading or biased.

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mean, median, mode and range

the four definitions- mode= most common. median= middle value (when values are in order of size). mean= total of items / number of items. range= difference between highest and lowest.

find the median, mode, mean and range of theses numbers: 2, 5, 3, 2, 6, -4, 0, 9, -3, 1, 6, 3, -2, 3. the median is the middle value, so rearrange the numbers in order if size. when there are two middle numbers, the median is halfway between the two. -4, -3, -2, 0, 1, 2, 2, 3, 3, 3, 5, 6, 6, 9. median= 2.5. mode is the most common value. mode= 3. mean= total of items / number of items. -4-3-2+0+1+2+2+3+3+3+5+6+6+9 / 14 = 31/14 = 2.214... = 2.21 (3 s.f.). range= distance from lowest to highest value. 9-(-4)=13

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frequency tables- finding averages

find averages from frequency tables-

1)the mode is just the category with the most enteries.

2)thr range is found is found from the extremes of the first column.

3) the medium is the category containing the middle value.

4) to find the mean, you have to work out a third column yourself. the mean is then: 3rd column total / 2nd column total.

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grouped frequency tables

it groups together the classes. no gaps between classes: use inequality symbols to cover all possible values. here, 10 would go in the 1st class, but 10.1 would go in the 2nd class. to find the mid-interval values: add together the end values of the class and divide by 2. e.g. 5 + 10 / 2 =7.5.

height (h millimetres)     frequency

5 < h < 10                         12

10 < h < 15                       15

find averages from grouped frequency tables- unlike ordinary frequency tables, you dont know the actual data values, only the classes they are in. so you have to estimate the mean.

1)add a 3rd column and enter the mid-interval value for each class. 2) add a 4th column to show 'frequency x mid-interval value' for each class.

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scatter graphs

scatter graphs show correlation-1) if you can draw a line of best fit pretty close to most of your data points, the two things are correlated. if the points are randomly scattered, and you cant draw a line of best fit, then theres no correlation. 2)strong correlation is when your points make a fairly straight line. weak correlation is when your points don't line up quite so nicely. 3) if the points form a sloping uphill from the left to the right, then theer is a positve correlation- both things increase or decrease together. if the line slopes downhill from left to right, then there is a negative correlation.

use a line of best fit to make predictions- 1) you can use the line of best fit to make estimates. predicting values within the range should be fairly reliable. if you extend your line outside the range of data your prediction might be unreliable. 2) you also need to watch out for outliners- data points that dont fit the general pattern. outliners can drag your line of best fit away from the other values, so its best to ignore them.

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