Grade12 Mathematics: Memorandum Paper 2

1 mathematics (NCS)/ grade 12/ P2 92 ExemplarMEMORANDUM Grade12 mathematics : Memorandum Paper 2 MBC=2 - (-3)8 -(-2) = 510 = MAD = 12D (lines parallel) AD:y = 12x + cSub in (0;1) 1 = cDAD:y = 12x + 1 Sub(d;6): 6 = 12d + 1Dd = MBC = 12? MAE = -2D( lines A)A = (0;1) which is the y-intercept of AE. AE:y = -2x + F = 221;280 = (4;112 ) 4cos - 7sin = -8D ( 7) 4sin + 7cos = 1D ( -4) 28cos 49sin = -56 -28cos 16sin = -4D(d;6)? -65sin = -60 D? sin = 6065D? = 67,38 D or 180 67,38 = 112,62 D? = 112,62 OROA = 2247 = 65 DOA' = 22( 8)1 = 65 DAA' = 22(4( 8))(71) =180= 65 DUsing the cos rule: (65) = (65) + (65) 2. D180 = 130(1 cos )?180130 1 = cos ? cos = 513D? = 112,62 DC(8;2) B' = (8cos 112,62 14sin 112,62 ; 8sin 112,62 + 14cos 112,62) DD= (-16;2) Tanx = -0,3421D?

2 X = 18,89 Sinx = 0,500D?x = 30 y = < 16% 590 runners = 94 runnersDDD(can accept answers in the range of 94 to 106 runners) A(0;1)B(-2;-3) FE28445772112 Copyright reserved mathematics (NCS)/ grade 12/ P2 93 ExemplarMEMORANDUM Appropriate scale DCorrect drawingDCorrect min and max values for the range DCorrect values for the median etc DDThe values can be out by 1unit on either Let the centre = (a;b)a = 2 D 2222rbyx Sub in (2;0) 222022rb D?22rb DSub in (4; -6) 222624bb D?4 + 36 + 12b +D22bb ?40 + 12b = 0 ? b = 103 DCentre = 310;2 4310222 mMB= 426313 = - 43D?Mtangent = 34 DTangent: y = 34x + c Sub in (4;-6) -6 = 34 (4) + cD-9 = cTangent: y = 43x y = 5 2x 020256122522 xxxxD0201230124202522 xxxxx0152052 xxD0342 xx? 13 xx = 0D?

3 13 xxD 325 y D 125 y?31 yyD?Points of intersection are: (3;-1) and (1:3) AB = 223113 DD3AB = 614 AB = 20AB = mBC=-1 - 33 -1 = -42 = -2 Dmperp = 12 DMidpoint of AB = 213;231 = (2;1) DPerpendicular bisector: y = 12x + c Sub (2;1): 1= 12 (2) + cD?c = 0 ?y = Thex-intercepts of the circle are found by: 020122 xxD? 210 xx= 0D?x = 10 or x = 2D? x = 6 is the perpendicular bisector of thex-interceptsD? the x value of the centre = 6D? y = 3 DThe centre of the circle = (6;3) (5;1) DDp is 1 unit from C to the line x = 4, so the point C' will be 1 unit from x = 4 on the other side 5. The y-value (q) remains the same. (12;4) DD r is 3 units from C' to the line x= 9, so the point C will be 3 units from x = 9 on the other side 12. The y-value (s)remains the same.

4 A translation 10 units right. D Triangle ABC has remained in the same horizontal plane but has moved 10 units along. If point A (1;3) is reflected about the x = 9, it will become A' = (17;3). D If A is then reflected about the x = 4 line, it will become A = (-9;3). D This is not the same result as If A = (4;3) then A' = (3;4) DDand A =(-3;4) Rotation of 90 : A = (4cos 90 3sin 90 ; 4sin 90 + 3cos 90 )DDA = (4 0 3 1; 4 1 + 3 0) DDA = (-3;4) 1cos221sin2coscos2sin2 TTTTTDD1cos4sin)1cos2(coscossin222 TTTTTTDD1cos4)1cos2cos2(sin222 TTTTD6 Copyright reserved mathematics (NCS)/ grade 12/ P2 94 ExemplarMEMORANDUM = If = 60 then T2cos21 = 0 Dand the denominator will be zero which makes the identity undefined. 120 or 240 In TAB:'BT A= 180 ( + )D))(180sin(sinAT0 ETT xD?

5 AT = )(sinsinETT xDIn TAC:'TC = AT sin D?TC = )sin(sinsinETDT 66,60cos2,11800 fD= 7,86 0,86 60minutes = 51,6minutesDTime for sunrise = 7:52 which is the time recorded in the table. 66,6)18060cos(2,16000 fD= 6,06 0,06 60minutes = 3,6minutesDTime for sunrise = 06:04. Actual sunrise is at 06:33. Difference is about Earliest = 17:44 = 17,733 DLatest = 20:01 = 20,016D?20,016 17,733 = 2, a is the amplitude of the cos graph which will be half of the time between the earliest and the latest sunset 2,283 2 = 1,142 DDp represents a horizontal shift which has not occurred therefore p = 0 Dq is the amount that the graph has been shifted upwards and is calculated by : the minimum value + the amplitude of the graph = 17,733 + 1,142 = 18,874. 875,18)180285cos(142,128500 gD= 18,58 0,58 60minutes = 34,8minutesDTime for sunset = 18:39 Actual sunset is at 18:57 Difference is about 875,18cos142,1xxh 66,6180cos2,10 xD 215,12cos2,1cos142,1 xxxh 215,12cos342,2 a1st and 360th dayDD2b180th Predicted: 215,1275cos342,275 hD= 12,82 hoursD= 12hours 49min 4 Actual:19:04 06:46 = 12hrs18minDDiffers by about If the house price was R175 000 then the percentile rank would be = (0,1 + 0,5 + 1,7 + 4,4 + 9,2 + 15 + 19,1 + 19,1 + 15)% 84%DDThis means that 84% of the houses were sold for less than R175 000 and 16% of the houses were sold for more than R175 00.

6 The difference between one standard deviations on either side of the mean = (15 + 19,1 + 19,1 + 15)% = 68,2%DDThis means that 68,2% of the houses were in the price range of R125 000 and R175 000. DDMrs Hlope is therefore correct in saying that most of the house were sold between R125 000 and R175 000. SD = 12 nxx= 16, 46,62 62,7 78,78 DD?1720 = 85% scored within one standard deviation. It is not a normal distribution as we would expect only 68,2% of the students to fall within one standard deviation. , the line is above all the points. , the line goes above the points for lighter eggs and below the point for heavier eggs. , the line goes through the majority of the points. DD284%68,2%125150175 Copyright reserved