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发表于 29-5-2011 05:51 PM
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sold price - cost price = profit? |
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发表于 30-5-2011 08:14 PM
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发表于 2-6-2011 01:18 PM
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Proof (A - B) ∪ (B - A) = (A ∪ B) - (A ∩ B)
我做的
(A - B) ∪ (B - A)
= (A ∩ B') ∪ (B ∩ A')
然后这样,
= (A ∪ B) ∩ (A' ∪ B')
= (A ∪ B) ∩ (A ∩ B)'
= (A ∪ B) - (A ∩ B)
对吗?
做proving时需不需要在每个步骤旁边写是什么law? |
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发表于 2-6-2011 02:09 PM
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步驟寫得詳細一點哦
書的example很少導致有些不知道如何開始做
1.Given that x*y=x^y,whre x and y are positive integers,state whether the binary operation * is
a) commutative
b)associative
2. x*y=(1/2)(x+y) >>題目也是和上面一樣<<
a)commutative
b)associative |
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发表于 2-6-2011 05:53 PM
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发表于 2-6-2011 11:56 PM
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回复 2603# 轻风
写比较好,好像某一年的stpm需要写 |
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发表于 2-6-2011 11:57 PM
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回复 2604# woeifang
FMT的wor  |
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发表于 3-6-2011 12:02 AM
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回复 2605# Minute
1/(√2+√3+√5)= (√2+√3-√5) /(2+3+2√6 - 5) = (√2+√3-√5) /2√6
= (1/2)[√1/3 + √1/2 - √5/6] |
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发表于 5-6-2011 08:51 AM
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Points A,B and C have coordinates (1,1) , (5,1) and (5,5) respectively.
(a)find the equation of the line that passes through A and the midpoint M of BC
(b)Find the coordinates of the point P where these two lines intersect |
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发表于 5-6-2011 12:24 PM
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回复 2609# hongji
如果有出差錯,請多多包涵
a) Midpoint of BC=[(5+5)/2 , (1+5)/2]
=(5,3)
equation of the line that passes through A and the midpoint M of BC:
(y-1)/(x-1)=(3-1)/(5-1)
2y-2=x-1
2y=x+1
b)
The equation of bc:
x=5
sub x=5 into 2y=x+1
2y=5+1
y=3
Hence, P(5,3) |
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发表于 5-6-2011 04:40 PM
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回复 hongji
如果有出差錯,請多多包涵
a) Midpoint of BC=[(5+5)/2 , (1+5)/2]
...
woeifang 发表于 5-6-2011 12:24 PM 
a的对了
b的对了
谢谢 |
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发表于 5-6-2011 04:50 PM
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a)find the equation of locus of moving point such that the sum of the squares of its distances from points(2,0) and (-2,0) is equal to 72 units.
b)a straight line y=m(x-2) which passes through a fixed point (2,0) meets the lines x=1 and y=k at A and B respectively. if O is the origin and P is the points (1,0), find the equations of lines OA and PB. if m varies, find the equations of the locus of the point of intersection between lines OA and PB. |
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发表于 6-6-2011 10:16 PM
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请问 1) 3 log8 X = 2 logx 8 + 5 ----------0.5,64
2) 2 log4 X + 3 logx 4 =7-----------2,64
3)log5 X + logx 25 =3--------------5,25
4)12^x^2 > 10^2x----------------x>1.853
5) e^x -3e^-x <2-----------------x<1.099
怎样做?我做来做去都跟书上的答案不一样。帮帮我....pls~ |
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发表于 6-6-2011 11:20 PM
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本帖最后由 blazex 于 6-6-2011 11:21 PM 编辑
回复 2612# hongji
我不知道是否正确.....
a)let the moving point be P(x,y), A(2,0) and B(-2,0) distance of PA,d1=((x-2)^2+(y-0)^2)^(1/2)
=(x^2+y^2-4x+4)^(1/2)
distance of PB,d2=((x+2)^2+(y-0)^2)^(1/2)
=(x^2+y^2+4x+4)^(1/2)
(d1)^2+(d2)^2=72
x^2+y^2-4x+4+x^2+y^2+4x+4=72
2x^2+2y^2-64=0
x^2+y^2-32=0
b) when x=1, y=m(1-2)
y=-m coordinate of A(1,-m)
when y=k , k=m(x-2)
x= k/m+2 coordinate of B(k/m+2,k)
gradient of OA= (-m-0)/(1-0)=-m
equation of OA, y=-mx -----(1)
gradient of PB=(k-0)/(k/m+2-1)=km/(k+m)
equation of PB,y=(km/(k+m))(x-1)
=(kmx-km)/(k+m) -----------(2)
(1)=(2) mx=(kmx-km)/(k+m)
x=(kx-k)/(k+m)
kx+mx=kx-k
x=-k/m ---------(3)
如果答案没有m的话,把(4)subtitute 进(3)
y=m(x-2)
m=y/(x-2) -------(4) |
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发表于 7-6-2011 12:04 AM
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本帖最后由 blazex 于 7-6-2011 12:05 AM 编辑
回复 2613# Stabilo_2B
1)3log8(x)=2logx(8)+5 注意:log8(x)是指 没有挂号()的8是base 3log8(x)=2(log8(8)/log8(x))+5 --换base
3log8(x)=2/(log8(x))+5 let y=log8(x)
3y=2/y+5
3y^2-5y-2=0
(3y+1)(y-2)=0
y=-1/3 , log8(x)=-1/3 y=2 , log8(x)=2
x=8^(-1/3) x=8^2
=0.5 =64
2 和3用同样的方法换base就可以得到答案。
4) 12^x^2>10^2x
log10(12^x^2)>log10(10^2x)
x^2(log10(12))>2x(log10(10))
1.079x^2-2x>0
x(1.079x-2)>0
let y=x(1.079x-2),if y=0
x=0,x=1.853
这题很难用字面来讲解,form 4 有学过。这题 还有一个答案是不是?
5)e^x-3e^-x<2
e^2x-3<2e^x let e^x=z
z^2-2z-3<0
(z-3)(z+1)<0
let y=(z-3)(z+1),if y=0
z=3, e^x=3 z=-1
ln(e^x)=ln3 e^x=-1(invalid,e^x cannot be less or equal to zero)
x=1.099 (接下来用第四题的方法) |
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发表于 9-6-2011 11:54 AM
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请问大家一个问题:
Without evaluating the square root or using the calculator, show that
[(18)^1/2]+[(3)^1/2] is less than 6.
做了好多次还是想不通,请大家帮帮忙~
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发表于 9-6-2011 01:42 PM
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本帖最后由 peaceboy 于 9-6-2011 04:50 PM 编辑
回复 2616# genji
let [(18)^1/2]+[(3)^1/2] < 6
[18+3 + 2 (54)^(1/2) ]^(1/2) < (36)^(1/2)
[21+ (216)^(1/2) ]^(1/2) < (36)^(1/2)
21+ (216)^(1/2) < 36
(216)^(1/2) < 15
(216)^(1/2) < (225)^(1/2)
216< 225 shown |
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发表于 9-6-2011 10:06 PM
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回复 2617# peaceboy
谢谢帮忙!我明白了~ |
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发表于 11-6-2011 11:05 AM
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這題我不會做,教我
1.
我用其他符號來代替 *
a◆b=a+b+ab
a◇b=ab+1
a)for each operation,determine if there are exist an identity element for the operation
b)for each operation,determine if ◇ is distribitive on ◆
2.
a*b=a(b-1)+b(a-1)
determine if the operation is closed |
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发表于 13-6-2011 09:12 PM
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這題我不會做,教我
1.
我用其他符號來代替 *
a◆b=a+b+ab
a◇b=ab+1
a)for each operation,det ...
woeifang 发表于 11-6-2011 11:05 AM 
想说的是。。这些题目算超出STPM的水准了吧。。。 |
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