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Grade 11 Mathematics: Memorandum Paper 2 - …

mathematics (NSC)/ Grade 11/ P2 86 ExemplarMEMORANDUM Grade 11 mathematics : Memorandum Paper 2 22)04()25( ABD25 = Both points have the same x-value therefore x = m = 0425 = 43D?243 43tan TD?DD87,36 m = 0,81 -1,92 AAcossin DD= - tan tan 2x=31D?Reference angle: 18,43 D?2x= 18,43 + 180 nD?x = 9,22 + 90 nD n Z D?x = 9,22 or 99,22 or 189,22 DD60sin540sin KTDD? KT = 3,71 cm PT2= 72+ 52- 2(7)(5)cos30 DD? PT = 3,66 cm Basic shape DM inimum = 10 DM edian and lower quartile DUpper quartile and maximum DScale shown h = 12 D (Pythagoras) V = 31 r2h=31 (5) (12) D= Diagonals are equal DAdjacent sides are perpendicular 22)520()021( ACD666 D22)025()1011( BDD4626 1150112025 ABmDD21 00020 NoAC and BD are not equal diagonals.

Mathematics(NSC)/Grade 11/ P2 87 Exemplar MEMORANDUM 4.2 PQ 4 1 2 4 2 2 13 D 9 13 3 13 3 12 2 6 12 2 117 u P’Q’ D Area PQRS 13 u 13 = 13D Area P·Q·R·S· 3 13 u ...

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Transcription of Grade 11 Mathematics: Memorandum Paper 2 - …

1 mathematics (NSC)/ Grade 11/ P2 86 ExemplarMEMORANDUM Grade 11 mathematics : Memorandum Paper 2 22)04()25( ABD25 = Both points have the same x-value therefore x = m = 0425 = 43D?243 43tan TD?DD87,36 m = 0,81 -1,92 AAcossin DD= - tan tan 2x=31D?Reference angle: 18,43 D?2x= 18,43 + 180 nD?x = 9,22 + 90 nD n Z D?x = 9,22 or 99,22 or 189,22 DD60sin540sin KTDD? KT = 3,71 cm PT2= 72+ 52- 2(7)(5)cos30 DD? PT = 3,66 cm Basic shape DM inimum = 10 DM edian and lower quartile DUpper quartile and maximum DScale shown h = 12 D (Pythagoras) V = 31 r2h=31 (5) (12) D= Diagonals are equal DAdjacent sides are perpendicular 22)520()021( ACD666 D22)025()1011( BDD4626 1150112025 ABmDD21 00020 NoAC and BD are not equal diagonals.

2 D1 zuBCABmm? AB and BC are not perpendicular to each other. )3;5(' ADD)8;4(' BDD)2;2(' (-y ; x) M idpoint of BB is (8-42 ;4+82 ) = (2;6) DD31' BBmDEquation of perpendicular: y = 3x+cD?6 = 6 + cD?0 = c ?y = Ay point of intersection -4x = 3xD? 7x = 0 ? x = 0D? y = 0D? (0 ; 0) is the point of intersection of AA and BB . )5;3('' AD)4;8('' BD)2;2('' )6;3('PD)12;12('QD)3;18('RD)3;9(' SDLines of enlargement DDP Q R S on graph D7 Copyright reserved mathematics (NSC)/ Grade 11/ P2 87 ExemplarMEMORANDUM 13241422 PQD 13313911712612322 u ''QPDArea PQRS1313u = 13 DArea P Q R S 133133u D= 9 13 = 117 The length of the sides of PQRS increase by a factor of 3 to give the length of the sides of P Q R S.

3 DThe area of PQRS increased by a factor of 9 to give the area of P Q R S . This is 3 the square of the increase of the length of the Dxcos11 or xxcos1cos DD45tan60cos DD121 DD21 cosx (2 cos x 1) cosx = 0 D?x = 90 + 360 n or 270 + 360 n D n Z (add on the period of the cos graph n to get general solution) ORcosx = 12D?x = 60 + 360 n or 300 + 360 nD,n sin (180 +58 ) = - sin 58 D = - sin2 58 + cos2 58 = 1 D? cos2 58 = 1 k22158cosk ? 0,5 or Sipho, Ray and Vishnu get - 0,17 DDLorraine gets 0, TTTT2222cossin1cossin1 DTTTT2222sincossincos D7TT22sincos DTTT222sin21sinsin1 DDor 1cos2cos1cos222 39,69 5318sin xDReference angle is 36,87 DnxDD36087,21618 ?

4 NxDD2012 ?DORnxDD36013,32318 ?nxDD2018 ?D12 ?x, 18, 32 or In PAB: )sin(590sinxyxPB DDD)sin(cos5xyxPB ? In PBT:PBPTy sinD)sin(sincos5xyyxPT ? xKAD DDD9090360 x D180D)180sin(21xbcDAK '?DDxbcsin21 DABC' Sum of lengths is 42,4 DMean length is 4, Length (cm)xxi 2xxi 3,2 -1,04 1,0816 3,6 -0,64 0,4096 50,76 0,5776 4,1 -0,14 0,0196 4,3 0,06 0,0036 4,7 0,46 0,2116 3,4 -0,84 0,7056 5,2 0,96 0,9216 4,6 0,36 0,1296 4,3 0,06DD0,0036DD4,064 DStandard deviation = 67,09064,4 D6 Copyright reserved mathematics (NSC)/ Grade 11/ P2 88 ExemplarMEMORANDUM Length to width comparison of 10 shells012340123456 Length (mm)W idth (mm)

5 2121 xyDDLine on graph 90, 330, 740, 940, 1000 D Values plotted at ends of intervals DD Accurate points D Accurate curve D Labels (Length of shell, cumulative frequency, title) Cumulative frequencyLength of pebble/cumulative frequency graphLength of pebble Median: 49 (47 51) DUpper quartile: 61 (59 63) DLower quartile: 35 (33 37) D3 Copyright reserved


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