Algebraic fractions involve polynomials, usually expressed in terms of a polynomialdivided by other polynomials in– they may need to be added, subtracted, multiplied or divided. The rules are the same as for ordinary fractions:
To add or subtract, we make a common denominator then subtract the numerators.
To multiply, we just “multiply across”, so that the end result is
To divide, turn the dividing, second fraction, upside down, and multiply.
Adding Algebraic Fractions
The new denominator is
In the diagram above the arrows indicate multiplication
To illustrate, I will expand the brackets in the numerator and add
The numeratorcan be factorised intoThere is a common factor in numerator and denominator which can be cancelled.
Subtracting Algebraic Fractions
The new denominator is
In the diagram above the arrows indicate multiplication
To illustrate, I will expand the brackets in the numerator and subtract
There is no common factor.
Multiplying Algebraic Fractions
Just multiply across
There are no common factors so we cannot cancel.
Dividing Algebraic Fractions
Turn the second, dividing fraction upside down and multiply as above:
]]>Shape
Picture
Number of Sides
Interior Angle
Sum of Interior Angles
Exterior Angle
Triangle
3
60
180
120
Square
4
90
360
90
Pentagon
5
108
540
72
Hexagon
6
120
720
60
Octagon
8
135
1080
45
n agon
To see why the angles in a triangle add tonotice that a triangle has three sides so n=1 and the angles add to 180, a square can be cut into two triangles, each with internal angles 180, so all the internal angles of both triangles sum to 360. Consider the pentagon below. It is cut into three triangle, each with internal angles that sum to 360, so the internal angles of a pentagon sum to 3*180 =540.
]]>Area and Circumference of Circle
Area of Segment
{jatex options:inline}\theta{/jatex} on this page is in radians and your calculator should be in radians mode.]]>Suppose then that you put £100 in a bank at 10% interest. This means that the amount of money in the bank increases by 10% each year.
The amount at the end of the 1 ^{st } year is £100 +10% of £100 =£100 +£10=£110
The amount at the end of the 2 ^{nd } year is £110 +10% of £110 =£110 +£11=£121
The amount at the end of the 3 ^{rd } year is £121 +10% of £121 =£121 +£12.1=£133.1
The amount of money in the account each year is shown on the graph below.
There is a formula to find the amount of money,in the bank at the end of each year, after interest has been added:
In this formula,is the original amount invested – the Principle.
is the rate of interest
is the number of years since the investment was made.
For this example, the amount of money in the account at the end of the 20 ^{th } year is
]]>To bisect the angle ABC draw arcs of equal length centred at B. Draw arcs of equal length centred at E and F to cross at G. The line BG bisects the angle.
Constructing the Perpendicular Bisector to a Line
To bisect the above draw arcs centred of equal radius at each end to cross at B and C. Draw arcs centred at B and C to cross at E and F. The line EF bisects the original line.
Constructing an Angle of Sixty Degrees to a Line
Given the line AB, from A draw an arc to cross AB at D, and then draw an arc of equal radius so that the two arcs cross at a point – call this C and draw the line AC. The angle CAD is 60 degrees.]]>There is a simple to solve the inequality approximately. You can sketch the curve and find thosevalues for whichThis will be the solution set of the inequality. The points to be plotted are shown in the table below.
3 
2 
1 
0 
1 
2 
3 

27 
8 
1 
0 
1 
8 
27 

9 
4 
1 
0 
1 
4 
9 

15 
10 
5 
0 
5 
10 
15 

21 
2 
3 
0 
6 
6 
3 
The curve is sketched below.
To solve the inequality we find the intersection of the curve with the lineand read off the – values for those points of intersection.
The lineintersects the curve at the points whereand
The curve is less than, or below the lineforand
]]>If two quantitiesandare in direct proportion then they increase together by the same ratio. If one doubles so does the other, and if one increases by a factor of 10, so does the other. We can write down the relationship between these two quantities in the form of an equation,where is the constant of proportionality. We may have to findbut once we have found it, then we can findfor any given value oforfor any given value of
Direct proportion is illustrated on the graph above. For this direct proportion relationthe gradient of the graph. We can also findif we are told specific values ofandIf we are told thatwhenthen
Now we know thatwe can:
Findif
Findif
Inverse Proportion
Two quantitiesandare inversely proportional if their product is a constant– we can also write this equation asThe graph of two quantities in inverse proportion is given below.
We can find k if we are told specific values ofandIf we are told thatwhen then
Now we know thatwe can:
Findif
Findif
]]>So for example suppose we have the velocity and we want to find the acceleration. We differentiate. If we have the velocity and we want to find the displacement, we integrate.
Example:
a) Find the acceleration when
b) Find the displacement s as a function of time if
a) so when
b}
We are told thatwhenso
]]>Volume of Cone
Curved Surface Area of Conewhere l is the slant height.
Volume of Sphere
Surface Area of Sphere
Probability
If two eventsandare independent with probabilitiesandrespectively then the probability ofandboth happening,
If the probability of an eventhappening isand then areoccasions in whichmay happen then the expected number of occurrences ofis
Integration and Differentiation
Trigonometry
Geometry
Ifat a pointthem the equation of the tangent atisand the equation of the normal is
A circle with equationhas centreand radius
]]>