1. 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.

2. M May

3. 56 000 000 2 significant figures 450.0062 3 significant figures 378 000 27.30.005 32 to nearest 1000to nearest tenthto nearest hundred thousandth to 1 dec pl to 5 dec pl to nearest million to nearest whole 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

4. 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 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...

5. 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 = 1.10 4 x 500 000 After 5 th year (95) Balance = 1.10 5 x 500 000 = 1.61051 x 500 000 = 805 255

6. M May Appreciation / Depreciation Appreciation ~ value has increased Original value = £400 Appreciated! By £28 Depreciation ~ value has decreased Value now = £428 28 400 = 0.07 = 7% original So the £400 item has appreciated by 7% Car has depreciated by 8% p aWas 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 = 0.92 3 x 12 000

7. M May Areas A = 1 / 2 x b x h A = π r 2 A = l x b A= 1 / 2 x d 1 x d 2 A = 1 / 2 a.b.sin C

8. M May V = l x b x h b h l V = A x h Prisms V = π r 2 h Where A is the area of the cross section of the prism Volume of cylinder

9. M May Other special objects Volume of cone V = 1 / 3 π r 2 h Volume of sphere V = 4 / 3 π r 3

10. M May Multiplying out bracketsFactorising 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 = ( ) ( )xx 53++

11. M May Multiplying out bracketsFactorising ( x - y ) ( x + y ) = x 2 - y 2 [difference of 2 squares] v 2 - 49 = v 2 - 7 2 = (v + 7)(v - 7) ( x - 7 ) ( x + 6 ) = x 2 - 7x + 6x - 42 = x 2 - x - 42 NB - 7 + 6 = -1 x 2 - 3x - 108 -108 subtract 1 x 108 2 x 54 3 x 36 4 x 27 6 x 18 9 x 12 +9 - 12 = -3 ( ) xx+ 9- 12

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 x 2 - 81 4 ( x 2 - 18x + 81)

13. M May radius x˚ sector arc x 360 What fraction of …….. ? arc C sector A ==

14. M May x˚ Isosceles triangles symmetry x˚ Diameter / Line of symm Bisects the chord at right angles

15. M May 90˚ Tangent meets radius at right angles 90˚ angle in semicircle is 90˚

16. M May Straight Lines Gradient = vertical horizontal vertical horizontal (x 1,y 1 ) (x 2,y 2 ) gradient m = (y 2 - y 1 ) (x 2 - x 1 ) Gradient is positive Gradient is negative

17. M May Need gradient mNeed 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) = 4 1 = 4

18. M May Triangle Measure : sides and angles! c b a Pythagoras Theorem c 2 = a 2 + b 2 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

19. M May More triangle Measure : sides and angles! A B C a b 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 C To find an angle

20. M May More triangle Measure : sides and angles! A B C a b c If do not know an angle and side opposite then Using the Cosine Rule a 2 = b 2 + c 2 - 2.b.c cos A To find an angle cos A = b 2 + c 2 - a 2 2.b.c

21. M May A = 1 / 2 base x height A = 1 / 2 a b sinC h C A B a b

22. 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 + 5 y = 4 + 5 y = 9 Checking! y = 5 x 2 - 1 y = 10 - 1 y = 9 5

23. M May y = 2x + 5 y = 5x - 1 By substitution 5x - 1 = 2x + 5 3x - 1 = 5 3x = 6 x = 2 y = 2 x 2 + 5 y = 9 2x + 4y = 24 3x + 2y = 8 …. x -1 …. x 2 6x + 4y = 16 - 2x - 4y = - 24 4x = - 8 x = - 2 3 x (-2) + 2y = 8 2y = 14 y = 7 Add x = - 2 y = 7