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�ظ� # FORM 4 �߼���ѧ (Additional Mathematics) NOTES ���� ������ 9-10-2008 10:46 PM | ��ʾȫ��¥��
 2x(2x+1)^4 + 8x^2(2x+1)^3 = 2x(2x+1)^3��(2x+1) + 2x(2x+1)^3��4x = 2x(2x+1)^3 [(2x+1) + 4x] ���Ǹ��ݳ˷��Լӷ��ķ����ɡ�
�ظ� ������ 10-10-2008 11:34 PM | ��ʾȫ��¥��

## �ظ� 101# mathlim ������

 o�����������ˡ�����ԭ���ǽ�factorize�����ġ�����
�ظ� ������ 11-10-2008 09:48 PM | ��ʾȫ��¥��     additional mathematics �Ǹ߼���ѧ�㣿 �������addition�Ǹ��ӣ� mathematics����ѧ��additional mathematicsӦ�ýи�����������ѧ���ɣ� �������߼���ѧ���ĳ̶�Ӧ��û�н��������� ע:�߼���ѧ��Ӣ����advanced mathematics����
�ظ� ������ 11-10-2008 09:53 PM | ��ʾȫ��¥��
 ˳����һ�� the sum of first nth terms in an  arithmetic progression is said to be   Sn = 6n^2-5n.Find the value of a6 S(6)=6*6^2-5*6        =186 S(5)=6*5^2-5*5        =125 a6 = S(6)-S(5)      =186-125      =61 ���������𣿻᲻�ᱻ�۷֣�
�ظ� ������ 12-10-2008 01:59 AM | ��ʾȫ��¥��
 �������������ʦ�Ļ����һ�����֡�  �ظ� ������ 12-10-2008 01:29 PM | ��ʾȫ��¥��

## �ظ� 105# mathlim ������

 л��     �ظ� ������ 17-1-2009 03:56 PM | ��ʾȫ��¥��
 1.given that function f(x)= 2x+1 , find f^n (x),where n is a positive integer. 2.given that 2^a=5^b=10^c ,express a in term of b and c . 3.Solve the equation 4^x -7 .  2^x - 8 = 0 4.Solve the equation  x - 4��x - 5 = 0 5.Solve the equation x^4 = 7x^2 +18. ���æ.лл. �ظ� ������ 17-1-2009 11:45 PM | ��ʾȫ��¥��
 1. Given that function f(x) = 2x+1, find f^n (x), where n is a positive integer. f^2(x) = f[f(x)] = f(2x+1) = 2(2x+1)+1 = 4x+3 f^3(x) = f[f^2(x)] = f(4x+3) = 2(4x+3)+1 = 8x+7 f^4(x) = f[f^3(x)] = f(8x+7) = 2(8x+7)+1 = 16x+15 �� �� f^n(x) = (2^n)x+(2^n)-1 2. Given that 2^a = 5^b = 10^c, express a in term of b and c. 2^a = 10^c 2^ab = 10^bc ���� �� 5^b = 10^c 5^ab = 10^ca ���� �� �š��� 10^ab = 10^(bc+ca) ab = bc + ca ab �C ca = bc �� a = bc/(b-c) 3. Solve the equation 4^x - 7��2^x - 8 = 0. 4^x - 7��2^x - 8 = 0 (2^x)^2 - 7��2^x - 8 = 0 (2^x + 1)(2^x - 8) = 0 2^x = -1 �� 2^x = 8   ����       x = 3 �� x = 3 4. Solve the equation x - 4��x - 5 = 0. x - 4��x - 5 = 0 (��x + 1)(��x - 5) = 0 ��x = -1 �� ��x = 5   ����       x = 25 �� x = 25 5. Solve the equation x^4 = 7x^2 + 18. x^4 = 7x^2 + 18 (x^2)^2 - 7x^2 - 18 = 0 (x^2 + 2)(x^2 - 9) = 0 x^2 = -2 �� x^2 = 9   ����       x = ��3 �� x = ��3
�ظ� ������ 17-1-2009 11:51 PM | ��ʾȫ��¥��
 1. f^1(x)=2x+1     f^2(x)=2(2x+1)+1=4x+3     f^3(x)=4(2x+1)+1=8x+7    ..... induction  f^n(x)=(2^n)x+(2^n)-1 f^(n+1)=(2^n)(2x+1)+(2^n)-1=(2^(n+1))x+(2^(n+1))-1 so f^n(x)=(2^n)x+(2^n)-1 2. 2^a=5^b=10^c (2^a)^2=5^b*10^c 2^(2a)=5^b*10^c (2a)ln2=(b)ln5+(c)ln10 a=.......... 3. tak tahu 4. x - 4 ��x- 5 = 0 x-5=4��x (x-5)^2=16x ............. warning:x>0 5. x^4 = 7x^2 +18. let y=x^2 y^2=7y+18 y=............ x=��y=........... warning y>=0
�ظ� ������ 17-1-2009 11:55 PM | ��ʾȫ��¥��
 ����������ԭ����7*2^2,�ҵ�����7.2^2��ԩ������ �ڶ���ԭ����������ģ������ֻֽⷨ��Ͷ��ȡ�ɶ��ѣ�
�ظ� ������ 18-1-2009 06:52 AM | ��ʾȫ��¥��
 2^a = 10^c 2^ab = 10^bc ���� �� 5^b = 10^c 5^ab = 10^ca ���� �� �š��� �Ҳ�������2��equation������. ��ȫ������,Ϊʲô2^ab = 10^bc? ������2^ac = 10^ba ? (2^x)^2 - 7��2^x - 8 = 0 (2^x + 1)(2^x - 8) = 0 (��x + 1)(��x - 5) = 0 (x^2 + 2)(x^2 - 9) = 0 ԭ����ѧ�������·���,��ֻѧ�� (ax+c)(bx+d)=0 ��quadratic equation����,����û���뵽��������...ԭ����ѧ����ô�Ĺ�� ���ǵ���ѧ������,������������? ԭ���� puangenlun �� 17-1-2009 11:55 PM ���� ����������ԭ����7*2^2,�ҵ�����7.2^2��ԩ������ �ڶ���ԭ����������ģ������ֻֽⷨ��Ͷ��ȡ�ɶ��ѣ� ʲô��In?��û��ѧ��,���ǿ�����calculator ����. form 5Ҳ���ָ��Ŷ ȥ��ѧ�� [ ��������� ������aries �� 18-1-2009 06:54 AM �༭ ]
�ظ� ������ 18-1-2009 11:51 AM | ��ʾȫ��¥��
 2^a = 10^c ����ȡb�η� (2^a)^b = (10^c)^b 2^(ab) = 10^(cb)
�ظ� ������ 18-1-2009 03:53 PM | ��ʾȫ��¥��
 ԭ���� mathlim �� 18-1-2009 11:51 AM ���� 2^a = 10^c ����ȡb�η� (2^a)^b = (10^c)^b 2^(ab) = 10^(cb) ԭ�����,������ �ظ� ������ 19-1-2009 12:45 AM | ��ʾȫ��¥��
 ln �� log ���� ֻ����һ��ȡ2            һ��ȡe ��������ȫ���ܺ�mathlim�Ľ�����Ƚ� �׳��ˣ�
�ظ� ������ 19-1-2009 09:32 PM | ��ʾȫ��¥��
 ԭ���� puangenlun �� 19-1-2009 12:45 AM ���� ln �� log ���� ֻ����һ��ȡ2            һ��ȡe ��������ȫ���ܺ�mathlim�Ľ�����Ƚ� �׳��ˣ� in calculator, log is base on 10 not 2 ln is log which base on e, e = ??? i forget le
�ظ� ������ 19-1-2009 11:26 PM | ��ʾȫ��¥��
 ��ο��� ����ln��log��ʲô���?
�ظ� ������ 20-1-2009 09:46 AM | ��ʾȫ��¥��
 in calculator, log a=log10 a the base of log is 10 or you can change ln a = loge a e = 2.7.... ln is natural log i forget how to get e le....
�ظ� ������ 20-1-2009 04:51 PM | ��ʾȫ��¥��
 ԭ���� TKCboy �� 20-1-2009 09:46 AM ���� in calculator, log a=log10 a the base of log is 10 or you can change ln a = loge a e = 2.7.... ln is natural log i forget how to get e le.... �������Ļظ� e ��defination�Ǵ�intergration�Ǳ������� int a^x=a^x+c a=e=2.718281828459...
�ظ� ������ 20-1-2009 05:58 PM | ��ʾȫ��¥��
 ��Դ�� ��һ���ᵽ����e���Ǽs�����{Ƥ���1618�����Č�����������е�һ�����������]��ӛ��@������ֻ���������Ӌ�����һ����Ȼ�����б���ͨ���J�������������W���׵�(William Oughtred)�u������һ�ΰ�e���鳣�������Ÿ�����Ŭ��(Jacob Bernoulli) The compound-interest problem Jacob Bernoulli discovered this constant by studying a question about compound interest. One simple example is an account that starts with \$1.00 and pays 100% interest per year. If the interest is credited once, at the end of the year, the value is \$2.00; but if the interest is computed and added twice in the year, the \$1 is multiplied by 1.5 twice, yielding \$1.00��1.5² = \$2.25. Compounding quarterly yields \$1.00��1.254 = \$2.4414��, and compounding monthly yields \$1.00��(1.0833��)12 = \$2.613035��. Bernoulli noticed that this sequence approaches a limit (the force of interest) for more and smaller compounding intervals. Compounding weekly yields \$2.692597��, while compounding daily yields \$2.714567��, just two cents more. Using n as the number of compounding intervals, with interest of 1/n in each interval, the limit for large n is the number that came to be known as e; with continuous compounding, the account value will reach \$2.7182818��. More generally, an account that starts at \$1, and yields (1+R) dollars at simple interest, will yield eR dollars with continuous compounding. ���� wikipedia..
�ظ� ������ 20-1-2009 08:14 PM | ��ʾȫ��¥��
 ԭ���� ����һ�� �� 20-1-2009 04:51 PM ���� �������Ļظ� e ��defination�Ǵ�intergration�Ǳ������� int a^x=a^x+c a=e=2.718281828459... ԭ��������ѧ��intergration��Ȼ����ô�ؼ� ,��ֻѧ ��  integrate ax^n dx = ax^n+1/n+1 +c . ��Ȼ�Ҳ�֪���Ǹ�a ��x ����ʲô,���������� ʲôʱ��ѧ��In ��sign
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