Year 11 Prompt Booklet Higher (A* – C grades) Name:
Grade C PROMPT sheet
C4 Round to one significant figure
C1 Understand & use proportionality To increase a quantity by 5% Multiply the quantity by 1.05 (100+5 = 105) To decrease a quantity by 5% Multiply the quantity by 0.95 (100–5) = 95
These all have ONE significant figure 300 80 2 0.7 0.05 0.003 C4 Estimate answers to calculations
C2 Calculate using proportional change To increase £240 by 15% (100+15 = 115) = 1.15 x £240 = £276 To decrease £240 by 15% (100-15 = 85) = 0.85 x £240 = £204
C2 Multiply & divide numbers 0-1 Multiply e.g. 0.2 x 0.4 Ignore decimal points & multiply numbers 2 x 4 = 8 Count the number of decimal places (2) The answer will have this many (2) 0.2 x 0.4 = 0.08 (2 decimal places) Divide e.g. 8 ÷ 0.2 Multiply both by 10 80 ÷ 2 = 40
makes whole
C2 4 rules of fractions Add & subtract Denominators must be the same Multiply Multiply numerators; multiply denominators Divide Invert fraction after ÷ Multiply numerators; multiply denominators
Round each number to 1sf first e.g. 423 x 28 = 400 x 30 = 12000 = 20 568 600 600 e.g. 3.26 x 11.8 = 3 x 10 = 30 = 300 = 50 0. 58 0.6 0.6 6 e.g. 8.3 x 35.6 = 8 x 40 = 320 = 640 0.49 0.5 0.5 (÷0.5 = doubling the number being divided)
C5 Use a calculator efficiently Know your keys x2 x3 x
3
(-)
C6 Expand brackets and simplify Multiply everything inside the bracket by what is outside Then collect like terms together
3(x + 2) + 2(x – 5) =3x + 6 + 2x – 10 =5x - 4 Watch for the negative sign in front of the bracket It changes the sign inside the bracket
3(x + 2) - 2(x – 5) =3x + 6 - 2x + 10 =x + 16
C7 Draw a straight line graph
To draw a graph of x + y = 7
Think of x and y coordinates that add to make 7
e.g. (4,3) (3,4) (2,5) (1,6) (0,7) (-1,8) .... These are usually put into a table:
x y
-1 8
0 7
1 6
2 5
3 4
4 3
T
hen the
points are plotted and joined
y
To draw a graph of y = 2x -1
Some coordinates are usually given in a table You have to fill in the rest by following the rule of the equation ‘ whatever x is, multiply by 3 then -2’
x y
-3 -7
-2 -5
-1 -3
0 -1
1 1
2 3 2x2-1
2x0-1 2x-2-1
Then the points are plotted and joined
3 5
To find the gradient of a line
The gradient of a line is its ‘slope’ It is measure by vertical ÷ horizontal
Gradient = 6 ÷ 4 = 1.5
C8 Solve inequalities in one variable a < b means a is less than b a ≤ b means a is less than or equal to b a > b means a is greater than b a ≥ b means a is greater than or equal to b
C9 Rearrange a formula
e.g.
Inequalities can be treated like equations The solution can be shown on a number line e.g.1
l -1 e.g. 2
2x – 4 < 2 (+4 to each side) 2x < 6 (÷2 each side) x <3 l 0
l 1
l 3
l 4
l 0
l 1
l 2
e.g.
l 3
-7 ≤ 2x – 1 < 3 (+1 to each part) -6 ≤ 2x < 4 (÷2 each side) -3 ≤ x < 2
l l l -3 -2 -1
l 0
l 1
l 2
Once numbers have replaced letters:
Remember the order of operations BIDMAS Remember the rules for signs
- x -=+ - x +=-
3 7 11 15 19 23 .... +4 +4 +4 +4 +4
The term to term rule is +4 nth term = 4n -1 The nth term can be used to find the term in any position
e.g. 10th term means n=10 10th term = 4x10 – 1 = 39
C11 Plot quadratic functions
Graphs of quadratic equations have x2 in and look like this:
l 3 4
C9 Substitute numbers into expressions
C10 Find the nth term of a linear sequence If the 1st difference is constant, it is linear
2x – 7 ≤ 5x + 2 (-2x each side) -7 ≤ 3x + 2 (-2 each side) -9 ≤ 3x (÷3 each side) -3 ≤ x (swap around) x ≥ -3 (swap inequality symbol)
l l l -3 -2 -1 e.g. 3
l 2
Use the same balancing steps as when you solve equations Make ‘t’ the new subject in: v = u + at (-u from each side) v–u= at (÷a each side) v–u= at a a t=v–u a
-- = + +- = -
or
To draw the graph of y = x2 + 4
Fill the table by following the rule Then join the points with a smooth curve
x y
-3 13
-2 8
(-2)2 + 4
-1 5
0 4
1 5
2 8 22 + 4
3 13
C12 Pythagoras Theorem
C14 Locus of point
For this right angled triangle:
LOCUS is the path or region a point covers as it moves according to a rule
hypotenuse a c
Fixed distance from a point – circle
Equal distance from two points perpendicular bisector
Equal distance from two intersecting lines – angle bisector
Perpendicular from a point to a line
b a = b2 + c2 2
If finding the hypotenuse ADD the squares of the other 2 sides Then square root If finding a shorter side SUBT the squares of the other 2 sides Then square root
C13 Find lengths, areas & volumes Formulae to learn: Area of rectangle = l x w
l w
Area of triangle = b x h 2
h b
P
Area of parallelogram = b x h h b Area of trapezium = ½(a + b)xh
a h
b Area of circle = π x r
2
r
Circumference = π x d d
Volume = Area of cross-section x length cross section length
C15 Bounds of measurement
If 23cm is rounded to nearest whole cm 23 is between the whole numbers 22 and 24 l
l
l
22
23
24
22.5
23.5
lower bound
upper bound
C16 Compound Measures
C18 Graphical representation
These triangles are useful Cover the quantity you are trying to find What is uncovered is the formula to use
D
S
Scatter diagrams – used to investigate correlation
e.g. Positive Correlation
M
T
D~Distance S~Speed T~Time
D
V
M~Mass D~Density V~Volume
Examples
Strong positive
Weak positive
If it shows correlation , draw a line of best fit on it Points which do not fit the trend are called OUTLIERS and should be ignored The line can be used to predict data Line of best fit
Speed = Distance Time
Time = Distance Speed
Distance = Speed x Time
C17 Plan a Statistical Enquiry
Questions should be simple The answers need to be ‘yes or ‘no’ or a ‘number’ or from a choice of answers Tick boxes are useful Avoid responses open to interpretation Avoid leading questions Avoid open-ended questions Avoid biased questions Ensure the sample is large enough Ensure the sample will give valid results
Negative
No correlation
Frequency polygon – plot mid-points of bars & join
Histogram
Frequency polygon
C19 Estimate mean Time (t sec)
C22 Examine results of an enquiry Justify choice of presentation
x
f
fx
60 < t ≤ 70
65
12
780
70 < t ≤ 80
75
22
1650
80 < t ≤ 90
85
23
1955
90 < t ≤ 100
95
24
2280
100 < t ≤ 110
105
19
1995
∑f = 100 ∑fx = 8660
Mean = ∑fx = 8660 = 86.6sec ∑f
100
Modal class = 90 < t ≤ 100 (because this class interval has the largest frequency i.e. 24)
Median = ½ (100 + 1) th = 50.5th = 80 < t ≤ 90
C20 Compare distributions
Compare an average using mean, median or mode. Compare spread using the range (the higher the range, the bigger the spread of data)
Frequency polygons and stem & leaf diagrams are often used to compare 2 distributions on the same diagram
C21 Understand relative frequency This is the name given to an estimate of probability from an experiment or a survey Relative probability = No. times an outcome occurs Total number of trials
A scatter diagram would be used to find out if there is any correlation or relationship between two sets of data A frequency polygon would be used to compare two sets of data
Grade B PROMPT sheet B/1 Change recurring decimal to fraction If x = 0.4444444 10x = 4.4444444 9x = 4 x=4 9
If x = 0.54545 100x =54.545454 99x = 54 x = 54 99
B/2 Repeated percentage change To increase £12 by 5% per year for 4 yr = 1.054 x £12 To decrease £50 by 12% per year for 4 yr = 0.884 x £50 B/2 To find the original quantity ~If an amount has been increased by 5% Original amount = new amount ÷ 1.05 ~If an amount has been decreased by 12% Original amount = new amount ÷ 0.88 B/3 Standard Form ~ a x 10n a is between 1 & 10; n is an integer ~ When mult/div in standard form, work out number part separate from the power of 10 part e.g. 3x105 x 4x103 = 12 x108 = 1.2 x 109 ~ With a calculator use EXP or x10x B/4 Factorise a quadratic expression x2 – 3x – 4 = (x – 4)(x + 1) x2 – 25 = (x – 5)(x + 5) Difference of 2 squares
B/5 Expand 2 brackets Use F O I L F L (x + 3)(x – 2) I O F O I L 2 x – 2x + 3x – 6 =x2 + x – 6
B/6 Change the subject of a formula
Isolate the new subject Use balancing
Make c new subject Make x new subject f = 3c – 4 ax + bx = ay 3c – 4 = f (+4) x(a + b) = ay 3c = f + 4 (÷3) x = ay c =f+4 a+b 3
B/7 Evaluate algebraic formulae Rewrite the formula with numbers replacing letters
WITH A CALCULATOR
Use fraction key
or
Use (-) key for negative numbers
WITHOUT A CALCULATOR
Remember the rules for negative numbers
-+ = -- = +
-x+=-x-=+
B8 Solve simultaneous equations by an algebraic method
e.g.
Make the number of ys the same Add or subtract to eliminate the ys Same signs ~ subtract Different signs ~ add Find the value of x Substitute the value of x to find y 2x – 3y = 11 5x + 2y = 18
First plot the straight line. Decide which side of the line to shade. Leave the region required unshaded. e.g. x ≤3 y > –2 y < x
(x2) (x3)
4x – 6y = 22 15x + 6y = 54 Add the two equations to eliminate y 19x = 76 x = 4 Substitute x = 4 into one of the equations 5x + 2y = 18 5x4 + 2y = 18 20 + 2y = 18 2y = -2 y = -1
B8 Solve simultaneous equations graphically
B/9 Represent inequalities graphically
Draw the graphs of the equations Find where they cross
R B/10 Identify graphs
Learn the basic shapes of graphs
Linear graphs – straight line - equation in x Quadratic graph – parabola – equation in x2 Cubic graph – S—shape – equation in x3 Reciprocal graph – equation e.g y = 3 x
B/11 Effect of adding/multiplying by a constant on a graph Original graph y = x2 New equation y = x2 + 2 y = x2 - 2 y = 2x2 y= ½ x2
Solution is x = 2, y = 3
Change in graph Move up 2 Move down 2 Stretch from x-axis in ydirection – scale factor 2 Stretch from x-axis in ydirection – scale factor ½
B/12 Coordinates in 3D
B/14 Trigonometry
In 3D there are 3 axes, x, y and z The coordinates of a point are (x, y, z)
SOH
CAH
TOA
Example O S H
y 2
R
1 Q 1
O
1
A 2
3
x
2 z
4
C3
C
O H
T
A
EXAMPLES
S
P
A
B
sin x = 4 cos28o = x tan 28 = 5 5 5 x sin x = 0.8 x = 5xcos28 o x= 5 x = sin-1(0.8) x = 4.4 tan 28 x = 53.1o x = 9.4
On the grid the vertex P is (2, 1, 3) B/15 Difference between formulae for length, area and volume
B13 Similarity If one shape is an enlargement of the other, we say they are similar.
Corresponding angles are equal Corresponding sides have proportional lengths
Example – these 2 triangles are similar
4.8m
6m x 8m
Scale factor = 6 ÷ 4.8 = 1.25 X = 8 ÷ 1.25 = 6.4cm N.B. Always draw the 2 triangles separately and the same way up – it is easier to spot the sides that correspond to each other
Numbers and π have no dimensions Length x length = area Length x length x length = volume
Example: 5abc + 3a2b (Ignore the numbers) axbxc + axaxb volume + volume volume
B/16 Median, quartiles & interquartile range from cumulative frequency graph
B/18 Add or multiply two probabilities P(A or B) = p(A) + p(B)
200
P(A and B) = p(A) x p(B)
Cumulative frequency 160
Upper Quartile
120
If you get an answer to a probability question that is more than one, you have most certainly added instead of multiplied
Median
80
Lower quartile
B/19 Tree Diagrams
40
When going along the branches. MULTIPLY the probabilities
0 10
20
30
40
50
60
Mark
Median = 38 marks Upper quartile = 43 marks Lower quartile = 30 marks Interquartile range = 43 – 30 = 13 marks
If more than one route is wanted, ADD the probabilities Example: The probability that Jane is late = 0.2 Day 1
B/16 Box plot LQ
M
Day 2 late – 0.2 x 0.2 = 0.04
UQ
0.2 late 0.8 0.2
B/17 Compare distributions.0000
Mean, median & mode compare size Range & interquartile range compare spread Distributions can be compared visually using a box plot
not – 0.2 x 0.8 = 0.16 late
0.8
late– 0.8 x 0.2 = 0.16 0.2 not late
0.8 not– 0.8x 0.8 = 0.64 late
To find the probability of late on only one day: day1 late
= =
&
day2 not late
0.16 0.32
OR
+
day1 & not late
0.16
day 2 late
Grade A/A* PROMPT sheet
Rationalising the denominator This is the removing of a surd from the denominator of a fraction by multiplying both the numerator & the denominator by that surd
A/1 Use fractional & negative indices
In general:
a
Rules when working with indices:
ax x ay = a(x + y) a3 x a2 = a(3 + 2) = a5 23 x 22 = 2(5) = 32
ax ÷ ay = a(x - y) a7 ÷ a3 = a(7 - 3) = a4 37 ÷ 33 = 3(4) = 81
(ax)y = a (x y) (a3)2 = a6 (23)2 = 26 = 64
a0 = 1 y0 = 1 80 = 1
a-x = 1
ax/y = ( a )x
=
a2/5 = ( 5 a )2 322/5 = ( 5
1
A/2 Manipulate and simplify surds 25 is NOT a surd because it is exactly 5 3 is a surd because the answer is not exact A surd is an irrational number To simplify surds look for square number factors
25 3 = 5 3 Rules when working with surds:
b =
ab
3 x 15 =
45 =
9x5 = 9 x
5 =3 5
m a + n a = (m+n) a 2 5 + 3 5 = (2+3) 5 =5 5
a = b 72 = 20
x
12
(Multiply both top & bottom by √12)
4 3 2 3 = = 2 2
3
A/3 Upper & lower bounds
( a )x
a x
6
32 )2 =22
y
(Multiply both top & bottom by √b)
8
a-x/y =
75 =
b
12 12 6 12 12 = = = 12 2
a3
b a b = b
b
12
a
23
x
6
x
2-3 = 1 = 1
=
Example
y
a-3 = 1
=
b a
72 20
Square number =
36 2 4 5
=
6 2 2 5
=
Square number
e.g. if 1.8 is rounded to 1dp Upper bound = 1.8 + ½(0.1) = 1.85 Lower bound = 1.8 – ½(0.1) = 1.75
Calculating using bounds Adding bounds Maximum = Upper + upper Minimum = Lower + lower Subtracting bounds Maximum = Upper - lower Minimum = Lower – upper Multiplying Maximum = Upper x upper Minimum = Lower x lower
a b
If ‘a’ is rounded to nearest ‘x’ Upper bound = a + ½x Lower bound = a – ½x
3 2 5
Dividing Maximum = Upper ÷ lower Minimum = Lower ÷ upper
A/4 Direct and inverse proportion
A/6 Solve quadratic equations by formula
The symbol means: ‘varies as’ or ‘is proportional to’
x = -b ± √b2 – 4ac 2a Example
Direct proportion If: y x or y x2 or y x3 Formulae: y = kx or y = kx2 or y = kx3 Example
y is directly proportional to x When y = 21, then x = 3
To solve:
3x2 + 4x – 2 = 0 a = 3 b = 4 c = -2
x = -b ± √b2 – 4ac 2a
(find value of k first by substituting these values)
y x
y = kx
21= k x 3 k=7 y = 7x (Now this equation can be used to find y, given x)
Inverse proportion If: y 1 or y 1 or y 1 x x2 x3 Formulae: y = k or y = k or y = k 2 x x x3 Example
a is inversely proportional to b When a = 12 and b = 4 a 1 a = k b b 12= k 4 k = 48 a = 48 b
A/5 Solve quadratic equation by factorising Put equation in form ax2 + bx + c = 0 2x2 – 3x – 5=0 Factorise the left hand side (2x – 5)(x + 1) = 0 Equate each factor to zero 2x – 5 = 0 or x + 1 = 0 x = 2.5 or x = -1
x = -4 ±
√(-4)2 – 4(3)(-2) 2(3)
= -4± √16+24 6 = -4± √40 6 x = -4 +
√40 6
OR -4 - √40 6
x = 0.39(2dp) OR
-1.72 (2dp)
A/7 Solve quadratic equation by completing the square
Make the coefficient of x2 a square 2x2 + 10x + 5 = 0 (mult by 2) 4x2 + 20x + 10 = 0 Add a number to both sides to make a perfect square 4x2 + 20x + 10 = 0 (Add 15) 4x2 + 20x +25 = 15 (2x + 5)2 = 15 Square root both sides 2x + 5 = ± √15 (-5 from both sides) 2x =-5 ± √15 x =-5 + √15 OR -5 - √15 2 2 x = -0.56 OR -4.44 (2dp)
A/8 Simplify algebraic fractions Adding & subtracting algebraic fractions Example 1
A/10 Solve simultaneous equations ~ one is a quadratic Rewrite the linear with one letter in terms of the other Substitute the linear into the quadratic Example
x+3+x–5 4 3
(common denominator is 12)
= 3(x + 3) + 4(x – 5) 12 12 = 3x + 9 + 4x – 20 12 = 7x – 11 12 Example 2 5 - 3 (common denominator is (x+1)(x+2) (x + 1) (x + 2) = 5(x + 2) – 3(x + 1) (x+1)(x+2) = 5x + 10 – 3x – 3 (x+1)(x+2) = 2x + 7 (x+1)(x+2) Simplifying algebraic fractions Example 2x2 + 3x + 1 (factorise) x2 -3x – 4 = (2x + 1)(x + 1) (cancel) (x – 4)(x + 1) = (2x + 1) (x – 4)
x + y = 4 (find one letter in terms of the other y=4-x 2 x + y2 = 40 (substitute y=4 –x) x2 + (4-x)2 = 40 (Expand (4-x)2) x2 + 16 – 8x + x2 = 40 2x2 – 8x + 16 = 40 (-40 from each side) 2x2 -8x – 24 = 0 (÷2 both sides) x2 – 4x – 12 = 0 (factorise) (x – 6)(x + 2) = 0 x = 6 or x = -2
A/10 Solve GRAPHICALLY simultaneous equations ~ one is a quadratic Draw the two graphs and find where they intersect Example y=2x2-4x-3 y=2x-1
y 14 12 10 8 6
A/9 Solve equations with fractions
4 2
x + 4 = 1 Common denominator (2x-3)(x+1) 2x – 3 x+1 x(x+1)+ 4(2x-3) =1 (2x-3)(x+1) x2 + x + 8x - 12 =1 (2x-3)(x+1) x2 + 9x – 12 =1(2x-3)(x+1) x2 + 9x -12 = 2x2 –x -3 (-x2 from both sides) 2 9x -12 = x –x -3 (-9x from each side) 2 -12 = x –10x -3 (+12 to each side) 0 = x2 –10x + 9 (factorise) (x + 9)(x + 1) = 0 x = -9 or x = -1
–2
–1
0
1
2
3
4
x
–2 –4 –6
Solutions are x = -0.3 and x = 3.3 (points of intersection)
Sometimes the equation has to be adapted~ rearrange the equation to solve so that the equation of the graph drawn is on the left. On the right is the other equation to be drawn
A/11 Graph of Exponential function
A/13 Graphs of trigonometric functions
The graph of the exponential function is: y = ax Example y = 2x
LEARN THE SHAPES OF THE GRAPHS Graph of y=sin x
9 8 7 6 5 4 3 2 1 0 -4
-2
-1 ≤ sin x ≤ 1 0
2
4
Graph y = cos x
It has no maximum or minimum point It crosses the y-axis at (0,1) It never crosses the x-axis
A/12Graph of the circle The graph of a circle is of the form:
x2 + y2 = r2 where r is the radius and the centre is (0,0)
-1 ≤ cos x ≤ 1 Graph y = tan x
y 6 5 4 3 2 1
–6
–5
–4
–3
–2
–1
O
1
2
3
4
5
6
x
–1 –2 –3 –4 –5 –6
This a circle of radius 5 and a centre (0,0) The graph of this circle is
x2 + y2 = 52 x2 + y2 = 25
Tan x is undefined at 900, 2700 .... Solutions to trigonometrical equations can be found on the calculator and by using the symmetry of these graphs Example: If sin x = 0.5 x = 300 , 1500 , (See the solutions on sin graph above or from calculator)
A/14 Transformation of functions f(x) means ‘a function of x’ e.g. f(x) = x2 – 4x + 1 f(3) means work out the value of f(x) when x = 3 e.g. f(3) = 32 – 4x3 + 1 = -2 In general for any graph y = f(x) these are the transformations
y = f(x) + a y = f(x+ a) y = -f(x) y = f(-x) y = af(x)
y = f(ax)
Translation 0 a Translation -a 0 Reflection in the x-axis Reflection in the y-axis Stretch from the x-axis Parallel to the y-axis Scale factor=a Stretch from the y-axis Parallel to the x-axis Scale factor= 1 a
A/15 Change the subject of a formula The subject may only appear once Use balancing to isolate the new subject Example : To make ‘x’ the new subject A = k(x + 5) (multiply both sides by 3) 3 3A = k(x + 5) (Expand the bracket)
The subject may appear twice Collect together all the terms containing the new subject & factorise to isolate it Example: to make ‘b’ the new subject a = 2 – 7b (multiply both sides by (b – 5) b-5 a(b – 5) = 2 – 7b (Expand the bracket) ab – 5a = 2 – 7b 7b + ab – 5a = 2
(+7b to both sides) (+5a to both sides) To leave terms in b together
7b + ab = 2 + 5a
(factorise the left side) To isolate b
b(7 + a) = 2 + 5a (÷(7 + a) both sides) (7 + a) (7 + a) b = 2 + 5a (7 + a)
A/16 Enlarge by a negative scale factor With a negative scale factor:
The image is on the opposite side of the centre The image is also inverted
Example : Enlargement scale factor -2 about 0 y
3A = kx + 5k (-5k from both sides) T
3A – 5k = kx (÷ k both sides) 3A – 5k = kx k k x = 3A – 5k k
O
x
A/17 Congruence
Congruent shapes have the same size and shape, one will fit exactly over the other. 2 triangles are congruent if any of these 4 conditions are satisfied on each triangle
A/18 Similarity & enlargement
For similar shapes when:
Length scale factor = k Area scale factor = k2 Volume scale factor = k3 Example
4cm
~The corresponding sides are equal ~ SSS
A
6cm
B
If height of A = 4cm & height of B = 6cm Length scale factor = 6 ÷ 4 = 1.5 If surface area of A = 132cm2 Surface area of B = 132 x 1.52 = 297cm3 If volume of A = 120 cm3 Volume of B = 120 x 1.53 = 405cm3
~2 sides & the included angle are equal ~ SAS
A/19 Finding lengths & angles in 3D
Identify the triangle in the 3D shape containing the unknown side/angle Use Pythagoras and trigonometry as appropriate
P ~2 angles & the corresponding side are equal ~ ASA
12 cm
4 cm 3 cm
Q
D
~Both triangles are right-angled, hypotenuses are equal and another pair of sides are equal ~ RHS
5 cm
B 10 cm
A 8 cm C
A/20 Sine Rule (non-right angled triangles) To find an angle use: sin A = sin B a b To find a side use: a b sin A = sin B
=
A/20 Cosine Rule (non-right angled triangles)
a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C
sin C c OR
c sin C
=
cos A = b2 + c2 – a2 2bc cos B = a2 + c2 – b2 2ac
Use SINE RULE when given: two sides and a non-included angle any two angles and one side Example: To find side b
cos C = a2 + b2 – c2 2ab
C
b
a=30cm
400
520
A
Example: To find angle C A
B b sin B b sin 520 b b
=
a sin C
=
30 sin 400
=
30 x sin 520 sin 400
Use COSINE RULE when given: 3 sides 2 sides and the included angle
c= 5cm
B
= 36.8 cm (1dp)
A 350 c= 7.1cm a= 9cm
sin C = sin A c a sin C = sin 350 7.1 9 sin C = sin 350 x 7.1 9 sin C = 0.4524..... C = sin-1(0.4524.....) C = 28.90(1dp)
a=8cm
C
cos C = a2 + b2 – c2 2ab 2 cos C = 8 + 62 – 52 2x8x6 cos C = 0.78125...... C = cos-1(0.78125......) C = 38.60(1dp)
Example: To find angle C
B
b= 6cm
C
Example: To find side a A 1100 c= 4.2cm
B
a=?
b=3.7cm
C
a2 = b2 + c2 – 2bc cos A a2 = 3.72 + 4.22 – 2x3.7x4.2 cos1100 a2 = 41.96 a = 6.48(2dp)
A/20 Area of triangle –height not known Area = ½ ab sinC Area = ½ bc sinA Area = ½ ac sinB
VOLUME -FRUSTUM OF A CONE
H
H r
Example
R
A
R
CONE
=
0
38 a= 8cm
FRUSTUM OF CONE
Volume of frustum = Volume of whole cone – volume of cone removed
b=6cm
B
r
1 2 1 2 πR H πr h 3 3
C
A/21 Pyramid & Sphere – Surface Area
Area = ½ ab sinC = ½ x 8 x 6 x sin 380 = 14.8 cm2(1dp)
CURVED SURFACE AREA ~Curved surface area of a cylinder = 2πrh r
A/21 Pyramid & Sphere –Volume
h
VOLUME - PYRAMID Volume of Pyramid =
1 x area of cross-section x height 3
e.g. cone
~Curved surface of a cone = πrl
h h w r Volume =
1 x πr2h 3
l
Volume =
1 xlxwxh 3
h
l r
[NB To find ‘l’ use Pythagoras’ Theorem
VOLUME - SPHERE
l 2 = h 2 + r 2]
4 3 Volume of Sphere = πr 3
~Curved surface of a sphere = 4πr2 r r
A/22 Length of arc & area of sector
A/23 Circle properties
sector arc segment
r
The angle at the centre = 2 x the angle at the circumference
θ
The angle in a semi-circle is a right angle
Length of arc = θ x 2πr 3600
r θ
Angles in the same segment are equal
Opposite angles of a cyclic quadrilateral add up to 1800
Area of sector = θ x πr2 3600
A/22 Area of segment radius of circle=7cm
640
The perpendicular from the centre to a chord bisects the chord
Tangents from a point to a circle are equal
7cm Area of segment = area of sector – area of triangle = θ x πr2 - ½ ab sinθ 0 360 = 640 x π x 72 - ½ 7x7x sin640 3600 = 5.35cm2 (1dp)
The angle between a tangent and a chord is equal to the angle in the alternate segment
A/24 Vectors
Vector addition Adding graphically, the vectors go nose to tail
Vector notation
a=
This vector can be written as 3 2 B
or a or AB
3 2
b=
-2 2
C
b
a
a+b
B
A A vector has magnitude(length) & direction(shown by an arrow) Magnitude can be found by Pythagoras Theorem
a
AB = √32 + 22 = √13 =3.6
A The combination of these two vectors:
AB + BC = AC = a + b
3 + -2 2 2
A parallel vector with same magnitude but opposite direction B
=
a
1 4
P
Vector subtraction Adding graphically, the vectors go nose to tail
A Q
a=
Vector PQ is equal in length to AB but opposite in direction so we say:
3 2
b=
-2 2
PQ = -a
B
-b
A parallel vector with same direction but different magnitude B
D
a
a A
C
a-b
The combination of these two vectors:
A
AB - BC = AC = a - b
C
3 - -2 2 2
Vector CD is twice (scalar 2) the magnitude but same direction so we say:
CD = 2a
A negative scalar would reverse the direction
=
5 0
AC is called the RESULTANT vector
The sum of vectors M B
P
Example
N
A
AB = AP + PM + MB The vector AB is equal to the sum of these vectors or it could be a different route: go via N
AB = AN + NB
start point
end point
An inspector wants to look at the work of a stratified sample of 70 of these students. Language
Number of students
Greek
145
Spanish
121
German
198
French
186
Total
650
No. from Greek = 145 x 70 ≈ 16 650 No. from Spanish = 121 x 70 ≈ 13 650 No. from German = 198 x 70 ≈ 21 650
A/25 Sampling The sample is: a small group of the population. an adequate size representative of the population Simple random sampling Everyone has an equal chance e.g. pick out names from a hat Systematic sampling Arranged in some sort of order e.g. pick out every 10th one on the list Stratified sampling Sample is divided into groups according to criteria These groups are called strata A simple random sample is taken from each group in proportion to its size using this formula: No from each group = Stratum size x Sample size Population
No. from French = 186 x 70 ≈ 20 650 This only tells us ‘how many’ to take - now take a random sample of this many from each language
A/26 Histograms
A/27 Probability – the ‘and’ ‘or’ rule
Class intervals are not equal Vertical axis is the frequency density The area of each bar not the height is the frequency Frequency = class width x frequency density Frequency density = frequency ÷ class width
P(A or B) = p(A) + p(B) Use this addition rule to find the probability of either of two mutually exclusive events occurring e.g. p(a 3 on a dice or a 4 on a dice) =
To draw a histogram Calculate the frequency density Example
Age (x years)
Class width
f
Frequency density
0 < x 20
20
28
28÷20 = 1.4
20 < x 35
15
36
36÷15 = 2.4
35 < x 45
10
20
20÷10 = 2
45 < x 65
20
30
30÷20 = 1.5
1 1 2 + = 6 6 6
P(A and B) = p(A) x p(B) Use this multiplication rule to find the probability of either of both of two independent events occurring e.g. p(Head on a coin and a 6 on a dice) =
1 1 1 x = 2 6 12
Scale the frequency density axis up to 2.4 Draw in the bars to relevant heights & widths
A/28 Probability – Tree diagram for successive dependent events
To interpret a histogram
When events are dependent, the probability of the second event is called a conditional event because it is conditional on the outcome of the first event
20 Frequency density 16 (number of books 12 per £)
Example 2 milk and 8 dark chocolates in a box Kate chooses one and eats it. (ONLY 9 left now) She chooses a second one This can be shown on a tree diagram
8 4 0 0
5
10
15 20 25 30 Price (P) in pounds (£)
35
40
1st
NOTE: On the vertical axis each small square = 0.8
f = width x height
0<P5
5 x 8 = 40
5 < P 10
5 x 12 = 60
10 < P 20
10 x 5.6 = 56
20 < P 40
20 x 1.6 = 32
M
1 2
Price (P) in pounds (£)
2nd
2milk 8dark
M
10
MM= 2 x 1= 2 10 9 90
9 8 9
8 10
2
D
D
M
9 7 9
MD= 2 x 8 = 16 10 9 90
DM= 8 x 2 =16 10 9
D
90
DD= 8 x 7 = 56 10 9
90