標題:
maths ce 1990 mc
發問:
how to solve the following questions 1990 maths ce mc no. 4, 31,43, 49 https://docs.google.com/file/d/0By_fBXid_HcUYmlGbl9SZFdxWWc/edit?pli=1 plz show the steps clearly! thanks
最佳解答:
4. 1/2 (n+1)(n+1-1)-1/2n(n-1) = 1/2n(n+1)-1/2n(n-1) =1/2(n^2+n-n^2+n) =1/2(2n) =n E 31. I is correct. Since the graph is concave downwards. II is incorrect. Sum of root =-b/a, which is larger than 0 a is smaller than 0, -1/a is larger than 0. b is then larger than 0. III is correct. Since c is equal to y intercept, which is smaller than zero. C 43 y= k1+ k2/x When x is very small, tends to 0, y should be very large or small, tends to infinity depending on whether k2 is positive or negative. D or E should be the answer. When x is very large, the term k2/x tends to 0. y tends to equal to k1, a constant, so it is the shape of curve indicated by E. 49 let angle ACB AND ECD be x. Sin 30/(sinACB)= AC/AB. 1 SIN 120 /(SIN ECD)= CD/ED. 2 combine the two equations together, (1/2) Sin 30/sin 120= AC(ED) /AB /CD. (Angle ACB = ECD) 1/ root 3= ED/AB. (AC= CD) AB/ ED = root 3 D
其他解答:
maths ce 1990 mc
發問:
how to solve the following questions 1990 maths ce mc no. 4, 31,43, 49 https://docs.google.com/file/d/0By_fBXid_HcUYmlGbl9SZFdxWWc/edit?pli=1 plz show the steps clearly! thanks
最佳解答:
4. 1/2 (n+1)(n+1-1)-1/2n(n-1) = 1/2n(n+1)-1/2n(n-1) =1/2(n^2+n-n^2+n) =1/2(2n) =n E 31. I is correct. Since the graph is concave downwards. II is incorrect. Sum of root =-b/a, which is larger than 0 a is smaller than 0, -1/a is larger than 0. b is then larger than 0. III is correct. Since c is equal to y intercept, which is smaller than zero. C 43 y= k1+ k2/x When x is very small, tends to 0, y should be very large or small, tends to infinity depending on whether k2 is positive or negative. D or E should be the answer. When x is very large, the term k2/x tends to 0. y tends to equal to k1, a constant, so it is the shape of curve indicated by E. 49 let angle ACB AND ECD be x. Sin 30/(sinACB)= AC/AB. 1 SIN 120 /(SIN ECD)= CD/ED. 2 combine the two equations together, (1/2) Sin 30/sin 120= AC(ED) /AB /CD. (Angle ACB = ECD) 1/ root 3= ED/AB. (AC= CD) AB/ ED = root 3 D
其他解答:
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