(２０分) (２０分) F.2 數學問題超easy (２０分) (２０分) @ gpjqem1的部落格

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(２０分) (２０分) F.2 數學問題超easy (２０分) (２０分)

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ＳＴＥＰＡＮＤＡＮＳＷＥＲＰＬＺ答得幾多得幾多！！！答得最多＋對＝最佳！！！1. express the following no. in scientific notation (A) (5.03x10^-2) X (7.3x10^-2)2. Linear equations in 2 unknowns (A) the age of a boy exceeds one-third of his father's by 1. five years later his age will be 4 less than half of his father's.... 顯示更多ＳＴＥＰＡＮＤＡＮＳＷＥＲＰＬＺ答得幾多得幾多！！！答得最多＋對＝最佳！！！ 1. express the following no. in scientific notation (A) (5.03x10^-2) X (7.3x10^-2) 2. Linear equations in 2 unknowns (A) the age of a boy exceeds one-third of his father's by 1. five years later his age will be 4 less than half of his father's. what is the persent age of his father? 3. identities and factorization ( no step is ok) (a) factorize x ^4-1 (b) factorize (2a-b)^2 - (a-7b)^2 更新: 4.Percentages (a). what is the increase in interest for saving $10000 in a bank at a compound interest of 2% p.a. for 2 years, if the interest is compounded half-yearly instead of annually? 更新 2: (b) what is the difference between simple interest and compound interest (compounded annually) at 4% p.a. for 2 years with a principal of $20000? (c) at what simple interest rate will a sum of money triple itself in 20 years? 更新 3: (d) the population of a city increases steadily at 4% every year. if the present population is 250000, find its population after 3years. Give your answer correct to the nearest thousand. 更新 4: (e) a tree is now 6.76m high. 2 years ago it was 4m high. if the tree grows at a constant rate each year, what is the annual growth rate? 更新 5: (f) a man put $10 into a bank at the beginning of 1900 at a compound interest of 8% per annum, compounded annually. how much can he get back at the end of 2003? ( correct the answer to nearest cent) 更新 6: (g) the value of a washing machine is now $5000. if its value depreciates by 12% each year, find (i) its value after 4 years (ii) its value 2 year ago ( correct the answer to nearest cent)

最佳解答:

1(A) (5.03x10^-2) X (7.3x10^-2) = (5.03X7.3X10^(-2+-2)) = (36.719X10^-4) =(3.6719X10^-3) 2(A) Let the age of a boy be x and the age of his father be y x-(y/3)=1......................(1) (x+5)+4=(y+5)/2..........(2) From (1), 3x-y=3.......(3) From (2), 2(x+9)=y+5..(4) (3) X 2, 6x-2y=6.......(5) (4) X 3, 6x-3y=3X(5-18)...(6) (5)-(6), y= 6+39 =45 3. (a) x ^4-1 = (x^2+1)(x^2-1) = (x^2+1)(x+1)(x-1) (b) (2a-b)^2 - (a-7b)^2 = [(2a-b)+(a-7b)][(2a-b)-(a-7b)] = (3a-8b)(a+6b) 4. (a). At a compound interest of 2% p.a. for 2 years, Interest = 10000*(1+2%)^2 - 10000 = 404 If the interest is compounded half-yearly, Interest = 10000*(1+1%)^4 -10000 = 406.04 Interest increased = 406.04-404=2.04 (b) Simple Interest = 20000*4%*2 - 20000 = 1600 Compound Interest = 20000*(1+4%)^2 -20000 = 1632 Difference between simple interest and compound interest =1632-1600=32 (c) y * (20i)=3y i=3/20 i=15% (d) the population of a city will increase to 250000 * (1+4%)^3 = 281216 The population is 281000 (e) Let annual growth rate be r. 4*(1+r)^2 = 6.76 (1+r)^2=1.69 (1+r) = 1.3 r=30% (f) No. of year = 104 At the end of 2003, total deposit = $10*(1+8%)^104 = $29927.51 (g) the value of a washing machine is now $5000. if its value depreciates by 12% each year, find (i) After 4 years, the value = $5000*(1-12%)^4 = $2998.48 (ii) Let its value 2 year ago be y, y * (1-12%)^2=5000 y= $6456.61