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CH 1- Law of Indices

CH 1- Law of Indices. +. . =. m. n. m. n. a. a. a. m. a. -. =. m. n. a. n. a. =. 0. a. 1. =. 1. a. a. =. m. n. mn. (. a. ). a. Solve for x. 3(2 k )-2 k-1 =40 3(2 k )-2 -1 (2 k )=40 2 k (3-1/2)=40 2 k =16 2 k =2 4 k=4. =. n. n. n. (. ab. ). a.

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CH 1- Law of Indices

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  1. CH 1- Law of Indices +  = m n m n a a a m a - = m n a n a = 0 a 1 = 1 a a = m n mn ( a ) a Solve for x 3(2k)-2 k-1=40 3(2k)-2 -1(2 k)=40 2k(3-1/2)=40 2k=16 2k=24 k=4 = n n n ( ab ) a b n a a = n ( ) n b b Simplify Simplify Solve for x

  2. Scientific Notation =   < n N d 10 where 1 d 10 , d is a decimal integer n is an Express the following in scientific notation without calculator (a) 2 750 000 000 (b) 0.000 000 137c) 4.7  1012 – 5.2  109

  3. Denary System (base 10) (0,1,2,3,4,5,6,7,8,9) =  +  +  +  +  4 3 2 1 0 abcde a 10 b 10 c 10 d 10 e 10 Binary System (base 2) (0,1) to Denary =  +  +  +  +  4 3 2 1 0 abcde a 2 b 2 c 2 d 2 e 2 Hexadecimal System (base 16)(0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F) to Denary =  +  +  +  +  4 3 2 1 0 abcde a 16 b 16 c 16 d 16 e 16

  4. Denary to Binary System 2 2 2 2 2 27 ........ 1 13 ........ 6 1 ........ 0 3 ........ 1 1 16 16 16 16 16 958521 ........ 1 0 ........ 9 59907 ........ 3 3744 ........ 0 234 ........ A 14 ........ E 0 Denary to Hexadecimal System

  5. CH 2- Percentages - New value Original Value % change =  100 % Original Value M % N % = +  - k ( 1 ) ( 1 ) 100 100 A value V increases by M% and then decrease by N%, find the % increase/decrease in V Example 1 Salary increases by 10% and then decrease by 10%, find % increase/decrease in salary Example 2 Salary increases by 10% and then decrease by 5%, find % increase/decrease in salary

  6. Simple Interest I = PRT A=P(1+RT) P P(1+R1) P(1+R2) P(1+RT) PR PR PR I=PRT For example : Principal = 1000, interest rate p.a. = 10%, years = 3 1000(1+0.11) =1100 1000(1+0.12) =1200 1000(1+0.13) =1300 1000 1000 0.1=100 100 100 I=10000.1 3 =300

  7. Compound Interest I = P(1+R)T-P A=P(1+R)T P P(1+R)1 P(1+R)2 P(1+R)T PR P(1+R)1 R P(1+R)2 R I=P(1+R)T-P 1000(1+0.1)1 =1100 1000(1 + 0.1)2 =1210 1000(1 + 0.1)3 =1331 1000 1000 0.1=100 1210 0.1=121 1100 0.1=110 I=1000(1+0.1)3-1000 =331

  8. CH 3 Factorization +  + + 2 2 2 ( x y ) x 2 xy y + +  + 2 2 2 x 2 xy y ( x y ) -  - + 2 2 2 ( x y ) x 2 xy y - +  - 2 2 2 x 2 xy y ( x y ) + -  - 2 2 -  + - ( x y )( x y ) x y 2 2 x y ( x y )( x y ) -  - + + 3 3 2 2 x y ( x y )( x xy y ) +  + - + 3 3 2 2 x y ( x y )( x xy y ) + +  + + 2 2 Ax Bx C ( dx p )( fx q ) + = = = where pf dq B , df A , pq C + +  + + 2 2 Ax Bxy Cy ( dx py )( fx qy ) + = = = where pf dq B , df A , pq C Expansion Factorization of quadratic equation in 1 variable Factorization of quadratic equation in 2 variables

  9. Factorize x3+64 15x2-41x+28 5+2x-3x2 x4-16y4 x2-6x+5 –3xy+28x2-y2 (a+3) 2+13(a+3)(2a-1)+40(2a-1) 2 3ab-18xy-6ax+9by (x-y)2+2(x-y)+1 8x3+729y3 (x+9)2-25 (x3-y3)+(x2y-y2x) x2-y2+4x+8y-12 If x2+y2=m, x-y=n, find xy Simplify

  10. CH 6- Central Tendency Sum of all the data = Mean Number of Data Arrange n data in ascending / descending order = term if n is odd , Median middle = if n is even , Median the mean of 2 middle terms The mod e of a set of data is the item with the highest frequency

  11. Length of arc = x r h r h h A A h r l r r Area of arc = Surface area of the cylinder = volume of the cylinder = volume of a prism or a cylinder = volume of a pyramid = volume of a cone = Curved surface area of a right circular cone = Volume of a sphere = Surface area of a sphere =

  12. Example 1 = + = + = + = = 2 2 2 2 AG AO OG 12 5 144 25 169 13 = + = + = + = = 2 2 2 2 AF AO OF 12 9 144 81 225 15 A Surface of Area =Area of base + the sum of areas of all sides =Area BCDE + 2 (area of ABC+area of ACD)   BC AF CD AG =  +  + BC CD 2 ( ) 12 2 2   10 15 8 13 =  +  + 10 8 2 ( ) E D 2 2 = + + 80 75 72 O G 18 = 2 F 227 cm B C 10 Volume Pyramid =

  13. Example 2 E 6 A 4 C O 600 4 4 D B Base triangle area = Volume area = Surface of Area =Area of base + the sum of areas of all sides =Area of ABC + 3 (area of BCE)

  14. Example 3 A F 6 12 H 4 12 E D 18 O C B G OF=4 cm. BCDE is a rectangle. Find the volume of the frustum.

  15. Example 3 A F 6 12 H 4 12 E D 18 O C B G OF=4 cm. BCDE is a rectangle. Find the surface area of the frustum.

  16. For 2 similar objects 1 and 2 Ex 1. Radius of a sphere doubled, find the % increase in Area. Ex 2. Original area is 40 cm2. If radius of a sphere doubled, find the new area.

  17. CH 9- Trigonometry BC AC BC B = = = sin cos tan q q q BA AB AC 300 600 450 q A C sin cos tan

  18. B Gradient q sin = BC AC q tan = = gradient 1 : q cos AC BC + = 2 2 q q sin cos 1 BC q tan q = = - 2 2 sin 1 cos q A q C AC = - 2 2 cos 1 sin q q A angle of depression of B from A - = 0 cos( ) sin 90 q q - = 0 sin( 90 q ) cos q 1 - = 0 tan( ) q angle of elevation of A from B 90 B q tan Angle of inclination  Angles of elevation and depression

  19. Compass Bearing True Bearing N N P P   W O O E   Q Q S Measured from N or S towardsE or W. Measured clockwise from N only e.g. P from O =  e.g. Q from O =  e.g. P from O = N  E e.g. Q from O = S W

  20. CH 5 –DEDUCTIVE GEOMETRY x x y y A A B B x + y = 1800 AB is a straight line (adj.  s supp.) AB is a straight line x + y = 1800 (adj. s at st. line) ISOSCELES TRIANGLE A A x y x y B C B C x=y (given) AB=AC AB=AC (given) x=y (sides opp.eq. s) (base s, isos. )

  21. a+ b +c = 1800 ( sum of ) a c b c b a a b a c b a + b + c = 3600 (s at a pt.) (vert. opp. s) a = b a + b = c (ext.  of )

  22. CH 10 –DEDUCTIVE GEOMETRY a a A B A B A B a b b b C C C D D D A B a b C D a = b (given) AB // CD (alt. s eq.) AB // CD (given)a = b (alt. s, AB // CD) a = b (given) AB // CD (corr. s eq.) AB // CD (given) a = b (corr. s, AB // CD)

  23. A B A B a a b b C C D D AB // CD EF // CD AB // EF A B D C F E a + b = 1800AB // CD (int. s supp.) AB // CD (given) a + b = 1800 (int. s, AB // CD) (transitive property of // lines)

  24. Prove the BE is the angle bisector of ABC . A E D  B C A E D B C Proof: Draw BEBD=DE (given)DBE =DEB(base s, isos. )DEB =EBC(alt s, DE//BC)DBE =EBC BE is the angle bisector of ABC .  

  25. Congruent Triangles D D A A D A A D C F C C F F C F E E B B B E E B (given) ABC  DEF  (corr. sides, s) AB = DE AC = DF BC = EF (given) ABC  DEF  (corr. s,   s) ABC = DEF BAC = EDF ACB = DFE AB = DE (given) AC = DF (given) BC = EF (given)  (S.S.S.) ABC  DEF AB=DE CA=FD BAC= EDF  ABC  DEF  (S.A.S.)

  26. Congruent Triangles A D A D D F C F C F E E B B E D A C F B E BAC= EDF (given) AB=DE (given) ABC= DEF (given)  ABC  DEF(S.A.S.)  BAC= EDF (given) ABC= DEF (given) AC=DF (given)  ABC  DEF(A.A.S.) A C  B BAC=EDF= 900 (given) AC= DF (given) BC=EF (given) ABC  DEF (R.H.S)   ABC not necessary  DEF

  27. Similar Triangles A A D D C C F F E E B B AB BC CA = = AB BC CA DE EF FD = = = k k , k DE EF FD  (given) ABC DEF  (corr. sides,  s)   (3 sides prop.) ABC DEF

  28. (given) ABC DEF A A A D D D C C C F F F E E E B B B  ABC=DEF BCA=EFD (corr. s,  s) CAB=FDE  ABC=DEF (given) BAC=EDF (given)  (AA) ABC DEF  ABC=DEF (given) BCA=EFD (given) CAB=FDE (given)  (AAA) ABC DEF

  29. A A ABC DEF (given) D D C C F F AB BC = DE EF E E B B AB BC = DE EF  ABC=DEF   ABC=DEF (ratio of 2 sides, inc. ) ABC DEF

  30. CHAPTER 11 - QUADRILATERALS Kite 2 pairs of eq adj sides Definition 1 diagional is axis of symmetry 2 diagionals are  Properties Trapeziums 1 pair of // sides Definition a c a+b=1800 ,c+d=1800 Properties b d

  31. Parallelogram 2 pairs of opp. // sides Definition Properties (opp. sides of // gram) (opp. s of // gram) Properties Properties (diag. of // gram)

  32. Conditions for Parallelogram B C D A B A B D C O D C A B C D A AB = DC (given) AD = BC (given) (opp. sides eq.) ABCD is a // gram ABC = ADC (given) DAB = DCB (given) (opp. s eq.) ABCD is a // gram AO = OC (given) DO = OB (given) ABCD is a // gram (diag. bisect each other) AB = DC (given) AB // DC (given) (opp. sides eq. and //) ABCD is a // gram

  33. In quadrilateral ABCD, AB=DC, ABC and BCD are equal but not equal to 900. Prove that AD//BC . D A B C A D ( opp . . eq . and //) sides B E C Method 1 AE // DC ( constructi on ) Ð Ð = Ð ( corr . s , AE // DC ) AEB DCE In ABE , D Ð = Ð ABE DCE ( given ) \ Ð = Ð ABE AEB \ = Ð AB AE ( sides . opp . eq . s ) = Q AB DC ( given ) \ = AE DC AE // DC and AE = DC Q \ AECD is a // gram n of // gram ) ( definitio \ AD // EC

  34. In quadrilateral ABCD, AB=DC, ABC and BCD are equal but not equal to 900. Prove that AD//BC . D A Draw AE  BC and DF  BC In ABE and DCF, AEB = DFC =900 (given) ABC =BCD (given) AB=DC (given) ABE  DCF (A.A.S) AE=DF (corr. sides,  s) AEB = DFE =900 AE//DF (corr. s eq) AE=DF and AE//DF ADEF is a // gram (opp. sides eq and //) AD//EF (definition of parallogram) AD//BC Q B C \ Q \ Q \ \ \ Method 2 A D F B C E

  35. A D Given ABCD is a // gram.Prove BPDQ is a // gram Q P B C Method 1 ∵ BQP=DPQ=900(given) ∴BQ//DP (alt. s eq.) In  ABQ and  CDP, AB=CD (opp.sides of // gram) BAQ=DCP (alt s ,AB//DC) BQA=DPC=900(given) ∴  ABQ  CDP (A.A.S.) ∴ BQ=DP (corr.sides, s) ∴BPDQ is a parallelogram (opp sides eq. And //)

  36. Given ABCD is a // gram.Prove BPDQ is a // gram A D Q P B C Method 2 In  ABQ and  CDP, ∵ AB=CD (opp.sides of // gram) BAQ=DCP (alt s ,AB//DC) BQA=DPC=900(given) ∴  ABQ  CDP (A.A.S.) ∴ BQ=DP (corr.sides, s) ∴ AQ=CP (corr.sides, s) In  ADQ and  CBP, ∵ AD=BC (opp.sides of // gram) DAQ=BCP (alt s ,AD//BC) AQ=CP(proof) ∴  ABQ  CDP (S.A.S.) ∴ DQ=BP (corr.sides, s) ∵ BQ=DP and DQ=BP (proof) ∴BPDQ is a parallelogram (opp. sides eq.)

  37. Given ABCD is a // gram.Prove BPDQ is a // gram A D Q O P B C Method 3 ∵ BQO=DPO=900(given) ∴BQ//DP (alt. s eq.) In  BQO and  DPO,∵ BQP=DPQ=900(given) QOB=POD (vert. opp. s) BO=DO (diag. of // gram ABCD)∴  BQO  DPO (A.A.S.) ∴ QO=PO (corr.sides, s) ∵ BO=DO and QO=PO ∴BPDQ is a parallelogram (diag. bisect each other)

  38. Given ABCD is a // gram. Prove BPDQ is a // gram A D Q O P B C Method 4 In  ABQ and  CDP, ∵ AB=CD (opp.sides of // gram) BAQ=DCP (alt s ,AB//DC) BQA=DPC=900(given) ∴  ABQ  CDP (A.A.S.) ∴ BQ=DP (corr.sides, s) In  BQP and  DPQ,∵ QP=QP (common) QOB=POD (vert. opp. s) BQ=DP (proved)∴  BQP  DPQ (S.A.S.) ∴ QBP=PDQ (corr .s, s) PQD =QPB(corr .s, s)∵ BQP =DPQ=900(given ) BQD= BQP+ PQD = QPB+ DPQ= DPB ∵ QBP=PDQ and BQD= DPB ∴BPDQ is a parallelogram (opp. s eq.)

  39. Rhombus A B D C A B D C A B 450 a // gram/a kite of 4 equal sides Definition Diagonals bisects each interior angle Properties Diagonals are  each other Rectangle a // gram of 4 right angles Definition Diagonals are equal AC = BD Properties Square a // gram of 4 right anglesand 4 equal sides Definition Diagonals are  Properties Angles between one diagonal and each side is 450 C D

  40. MID-POINT THEOREM A M N 1 = MN BC 2 B C A M N B C AM = MB (given) AN = NC (given) MN//BC (mid-point theorem) INTERCEPT THEOREM MN//BC (given) AN AM (intercept theorem) = NC MB

  41. Prove PBQR is a // gram CR=RA (given) CQ=QB (given) 1 ∴ = ( RQ AB mid pt theorem ) 2 = PB C = ( given ) AP PB = ( given ) AR RC R Q 1 ∴ ( mid pt theorem ) = RP CB 2 = BQ A P B  PBQR is a // gram (opp. sides eq.)

  42. Prove that AN= NC A AM = MB (given) M N (given) MN//BC AN AM (intercept theorem) = B C NC MB AN MB = NC MB AN 1 = NC 1 AN = NC

  43. A B Prove that AB+EF=2CD. C D AS BD AC = = = 1 (intercept theorem) SF DF CE E F A B S C D 1 1 = = SD CS EF AB 2 2 E F Draw a line AF, cutting CD at S. AC  CE (given) AB // CD // EF (given) ∴ ∴AS  SF and BD  DF In AEF AS  SF(Proved) AC=CE (given) ∴ (mid-point theorem) ∴ EF = 2CS In ABF AS  SF(Proved) BD=DF (proved) ∴ (mid-point theorem) ∴ AB = 2SD ∴AB+EF =2SD +2CS =2(SD+CS) =2 CD

  44. CHAPTER 12 – coordinate geometry = - + - 2 2 AB ( h f ) ( g e ) or = - + - 2 2 AB ( f h ) ( e g ) - - h f f h = = m or m AB AB - - g e e g Distance Formula B(g,h) A(e,f) Slope of a line B(g,h) A(e,f) Parallel Lines D(k,l) mAB =m CD AB // CD C(i,j) AB // CD mAB =m CD

  45. AB CD Perpendicular Lines = - m m 1 C(i,j) x B(g,h) AB CD = - m m 1 x A(e,f) AB CD AB CD D(k,l) Point of Division – coordinates of mid point + + e g f h B(g,h) C ( , ) A(e,f) C(i,j) 2 2 n m + + ne mg nf mh C ( , ) + + m n m n A(e,f) C(i,j) B(g,h) Point of Division – Internet point of division

  46. LINEAR EQUATION L: x=D Y Y mL=0 L: y=C C X X D mL=undefined L: y=mx+C Y mL>0 C X L: y=mx+C Y mL<0 C X

  47. Two Point Form – 2 points are given. - - - - y d b d y b b d = = - - - - x c a c x a a c or (a,b) Y or or (c,d) X Line Equation : Ax+By=E Example (4,5) Y (-2,3) X

  48. Point Slope Form – 1 point and a slope are given. - = - y b k ( x a ) m= Find x intercept Let the point be (0,d) d-b=k(0-a) d =bak (a,b) Y m=k X Line Equation : Ax+By=E Example (4,5) Y X

  49. Y intercept Form – slope (m) and y intercept (c) are given. = + y mx c Find x intercept Let the point be (0,d) y=mx+c 0=md+c Y m c X Example Y m= X

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