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1) z is partly varies directly as x and partly inversely as y. When x=y=2, z=43: and when x=y=6,z=121.A) express z in terms of x and y.2)For an arithmetic progression the sum of the first term T(1) and fourth term T(4) is 23, the ninth term T(9) is 18a) Find the general TERM OF AP.b) Find the sum of the... 顯示更多 1) z is partly varies directly as x and partly inversely as y. When x=y=2, z=43: and when x=y=6,z=121. A) express z in terms of x and y. 2)For an arithmetic progression the sum of the first term T(1) and fourth term T(4) is 23, the ninth term T(9) is 18 a) Find the general TERM OF AP. b) Find the sum of the first terms. THANKS. 之前補考答錯左,現家大考前想問返係點?? 更新: 吾明 a = 20, b = 6.係第一題我答 1) Let z= ax+ b/y when x=2, y=2, z= 43 43=2a+b/2 86=2a+b------(1) when x=6,y=6 , z=121 121=6a + b/6 726= 6a + b ------(2) (2)- (1) 726- 86= 6a+b-(2a-b) 640 = 4a a=160 更新 2: 第二題我答 T(1)+T (4) = a=(1+1)d + a (4-1)d = a+d +a + 3d =2a +4d = 23 ------(1) T(9)= a+ (9-1)d= 18 = a+ 8d =18 ---------(2) (2)X2 =2a +16d= 36 ----------(3) (3)-(1) 2a+16d- (2a +4d)=36-23 12d= 13 d= 13/12 a=28/3 咁樣話我錯左....
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1. Let z = ax + b/y, where a, b are constants. Put x = 2, y = 2, z = 43, we have 43 = 2a + b/2--------------(1) Put x = 6, y = 6, z = 121, we have 121 = 6a + b/6-------------(2) Solving (1), (2), a = 20, b = 6. Hence z = 20x + 6/y 2(a) Let T(1) = a, d be the common difference. Then T(4) = a + (4 - 1)d = a + 3d T(9) = a + (9 - 1)d = a + 8d T(1) + T(4) = 23 a + a + 3d = 23 2a + 3d = 23----------------(3) T(9) = 18 a + 8d = 18------------------(4) Solving (3), (4), we have a = 10, d = 1. Hence general term = 10 + (n - 1)(1) = n + 9 (b) 打漏字?補充番之後先答你 2011-05-31 18:36:38 補充： 1. 43 = 2a + b/2 86 = 4a + b,唔係86 = 2a+b 同時, 726 = 36a + b 先岩,唔係726 = 6a+b 2. 因為設 a 係首項, 即 T(1)已經等於a. T(1) + T(4) = a + a + (4 - 1)d = 2a + 3d,唔係2a+4d
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F.5 maths..考試前問一問嫁發問:
1) z is partly varies directly as x and partly inversely as y. When x=y=2, z=43: and when x=y=6,z=121.A) express z in terms of x and y.2)For an arithmetic progression the sum of the first term T(1) and fourth term T(4) is 23, the ninth term T(9) is 18a) Find the general TERM OF AP.b) Find the sum of the... 顯示更多 1) z is partly varies directly as x and partly inversely as y. When x=y=2, z=43: and when x=y=6,z=121. A) express z in terms of x and y. 2)For an arithmetic progression the sum of the first term T(1) and fourth term T(4) is 23, the ninth term T(9) is 18 a) Find the general TERM OF AP. b) Find the sum of the first terms. THANKS. 之前補考答錯左,現家大考前想問返係點?? 更新: 吾明 a = 20, b = 6.係第一題我答 1) Let z= ax+ b/y when x=2, y=2, z= 43 43=2a+b/2 86=2a+b------(1) when x=6,y=6 , z=121 121=6a + b/6 726= 6a + b ------(2) (2)- (1) 726- 86= 6a+b-(2a-b) 640 = 4a a=160 更新 2: 第二題我答 T(1)+T (4) = a=(1+1)d + a (4-1)d = a+d +a + 3d =2a +4d = 23 ------(1) T(9)= a+ (9-1)d= 18 = a+ 8d =18 ---------(2) (2)X2 =2a +16d= 36 ----------(3) (3)-(1) 2a+16d- (2a +4d)=36-23 12d= 13 d= 13/12 a=28/3 咁樣話我錯左....
最佳解答:
1. Let z = ax + b/y, where a, b are constants. Put x = 2, y = 2, z = 43, we have 43 = 2a + b/2--------------(1) Put x = 6, y = 6, z = 121, we have 121 = 6a + b/6-------------(2) Solving (1), (2), a = 20, b = 6. Hence z = 20x + 6/y 2(a) Let T(1) = a, d be the common difference. Then T(4) = a + (4 - 1)d = a + 3d T(9) = a + (9 - 1)d = a + 8d T(1) + T(4) = 23 a + a + 3d = 23 2a + 3d = 23----------------(3) T(9) = 18 a + 8d = 18------------------(4) Solving (3), (4), we have a = 10, d = 1. Hence general term = 10 + (n - 1)(1) = n + 9 (b) 打漏字?補充番之後先答你 2011-05-31 18:36:38 補充： 1. 43 = 2a + b/2 86 = 4a + b,唔係86 = 2a+b 同時, 726 = 36a + b 先岩,唔係726 = 6a+b 2. 因為設 a 係首項, 即 T(1)已經等於a. T(1) + T(4) = a + a + (4 - 1)d = 2a + 3d,唔係2a+4d
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