M May

You must learn the formulae and rules thoroughly.

Some will be given to you on your examination paper, but you should only use this list to check that you have remembered the formula correctly.

M May

M May

56 000 000

2 significant figures

45 0.0062

3 significant figures

378 000 27.3 0.005 32

to nearest 1000 to nearest tenth to nearesthundred thousandthto 1 dec pl

to 5 dec pl

to nearestmillion

to nearestwhole

to 4 dec pl

56.0

to nearest tenth

3050

to nearest ten

0.004 07

to 5 dec pl

7002

4 significant figures

5.308

to 3 dec pl

M May

Percentage used to indicate the RATE at which something is paid.

Simple Interest

Compound Interest / Appreciation / Depreciation

eg 4.5% pa on £400 for 5 months.

Int for 1 yr = 4.5% of 400 = 0.045 x 400

= 18

So for 5 months Interest = 18 ÷ 12 x 5 = 7.50

Each time the interest is added there is a new balance for the interest calculationEach time the interest is added there is a new balance for the interest calculation

eg compound interest on 5000 at 3%pa

After 1 year Int = 3% of 500 = 15, so new Balance = 515

After 1 year Balance = 1.03 x 500 = 515

After 2nd year Balance = 1.03 x 515 = 530.45 and so on ...

M May

Appreciation / Depreciation Appreciation ~ value has increased

Calculate the new value ~ remember to add for appreciation / subtract for depreciation

The value of a painting appreciated each year by 10% In 1990 it was valued at £500 000. What was its value in 1995?

After each year Value= 110% of its value in the previous year.

After 1 year (91) Value = 1.10 x 500 000 = 550 000

After 2nd year (92) Balance = 1.10 x 1.10 x 500 000

Depreciation ~ value has decreased

After 3rd year (93) Balance = 1.10 x 1.10 x 1.10 x 500 000

After 4 th year (94) Balance = 1.104 x 500 000

After 5 th year (95) Balance = 1.105 x 500 000 = 1.61051 x 500 000

= 805 255

M May

Appreciation / Depreciation Appreciation ~ value has increased

Original value = £400Appreciated! By £28

Depreciation ~ value has decreased

Value now = £428

28400 = 0.07 = 7%original

So the £400 item has appreciated by 7%

Car has depreciated by 8% p a Was 100%Now 100 - 8 = 92%

Car was valued at £12 000

After 1 year Value = 0.92 x 12 000

After 2 year Value = 0.92 x 0.92 x 12 000

After 3 year Value = 0.923 x 12 000

M May

Areas

A = 1/2 x b x h

A = π r 2

A = l x b

A= 1/2 x d1 x d2

A = 1/2 a.b.sin C

M May

V = l x b x h

b

h

lV = A x hPrisms

V = π r 2 h

Where A is the area of the cross section

of the prism

Volume of cylinder

M May

Other special objects

Volume of cone V = 1/3 π r 2 h

Volume of sphere V = 4/3 π r 3

M May

Multiplying out brackets Factorising

3(y - 4)

= 3y - 12

7x - 21

= 7(x - 3)

( x + 5 ) ( x + 6 )

= x ( x + 6 ) + 5 ( x + 6 )

= x 2 + 6 x + 5 x + 30

= x 2 + 11 x + 30

= x 2 + 8 x + 15

= ( ) ( )x x5 3+ +

M May

Multiplying out brackets Factorising

( x - y ) ( x + y )

= x2 - y2

[difference of 2 squares]

v2 - 49

= v2 - 72

= (v + 7)(v - 7)

( x - 7 ) ( x + 6 )

= x2 - 7x + 6x - 42

= x2 - x - 42

NB - 7 + 6 = -1

x2 - 3x - 108 -108subtract

1 x 1082 x 543 x 364 x 276 x 189 x 12

+9 - 12 = -3

( ) ( )x x+ 9 - 12

M May

Question 1: Is there a common factor?

Question 2: Is it a difference of 2 squares?

Question 3: Is there still 2 ?: brackets and find factors!

6x - 9

x2 - 81

4 ( x2 - 18x + 81)

M May

•

radius

x˚

sector

arc

x

360

What fraction of …….. ?

arc

C

sector

A= =

M May

• x˚

Isosceles triangles

symmetry

x˚

Diameter / Line of symm

Bisects the chord at right angles

M May

•

90˚

Tangent meets radius

at right angles

90˚

angle in semicircle

is 90˚

M May

Straight Lines

Gradient =vertical

horizontalvertical

horizontal

(x1,y1)

(x2,y2)

gradient m =(y2 - y1)

(x2 - x1)

Gradient is positive

Gradient is negative

M May

Need gradient m Need point on line (a, b)

y - b = m ( x - a )

Line through (2, 6) with gradient 4

y - 6 = 4 ( x - 2 )

y - 6 = 4 x - 8

y = 4 x - 2

•

4

1

(0, -2)

Points (2, 6) and (3, 10)

m = (10 - 6)(3 - 2)

= 41

= 4

M May

Triangle Measure : sides and angles!

c

b

a

Pythagoras Theorem

c2 = a2 + b2

Soh Cah Toa

sin A =

cos A =

tan A =

opposite

hypotenuse

adjacent

hypotenuse

opposite

adjacent

S o h

C a h

T o a

If the triangle has a right angle

sides

sides and angles

M May

More triangle Measure : sides and angles!

A B

C

ab

c

If not right-angled then

Using the Sine Rule

a

sin A=

b

sin B=

c

sin C

a

sin A=

b

sin B=

c

sin CTo find an angle

M May

More triangle Measure : sides and angles!

A B

C

ab

c

If do not know an angle and side opposite then

Using the Cosine Rule

a2 = b2 + c2 - 2.b.c cos A

To find an angle cos A = b 2 + c 2 - a 2

2.b.c

•

M May

A = 1/2 base x height A = 1/2 a b sinC

h

C

A

Ba

b

•

M May

Two equations in two variables are true at the same time:

y = 2x + 5

y = 5x - 1

Graphical Solution

X

Y

(2, 9)

x = 2

y = 9

y = 2 x 2 + 5y = 4 + 5y = 9

Checking!

y = 5 x 2 - 1y = 10 - 1y = 9

5

-1

M May

y = 2x + 5

y = 5x - 1

By substitution

5x - 1 = 2x + 5

3x - 1 = 5

3x = 6

x = 2

y = 2x2 + 5

y = 9

2x + 4y = 24

3x + 2y = 8

…. x -1…. x 2

6x + 4y = 16

- 2x - 4y = - 24

4x = - 8

x = - 2

3x(-2) + 2y = 8

2y = 14

y = 7

Add

x = - 2 y = 7