数学Paper 1讨论专区

walrein_lim88

1.solve the equation 4^(2x)-8(2^(2x))-9=0
   giving your answer up to three significant figures.
2.solve the simultaneous equation
   log9(xy)=1/2
   log3(x)log3(y)=-2
天空之道 发表于 14-6-2010 03:20 PM

Allmaths

需要各位帮忙.

Leo has RM60. With the amount of money, he can buy 25 more Type B pens than Type A pens. If a Type A pen is RM0.20 more expensive than a Type B pen, what are the prices of the Type A pen and the Type B pen respectively?

题目虽看简单,但是我还是不会做啊!!!

乙劍真人

需要各位帮忙.

Leo has RM60. With the amount of money, he can buy 25 more Type B pensthan Type A pens. If a Type A pen is RM0.20 more expensive than a TypeB pen, what are the prices of the Type A pen and the Type B penrespectively?

题目虽看简单,但是我还是不会做啊!!!
Allmaths 发表于 15-6-2010 08:52 AM

A + 25B = 60 ------(1)
A - 0.20 = B -------(2)

From (2), A = B + 0.20 ---(3)
Sub. (3) into (1);
B + 0.20 + 25B = 60
                26B = 59.80
                   B = 2.30 ---(4)
Sub. (4) into (3);
                   A = 2.30 + 0.20
                      = 2.50

Therefore, Type A pen = RM2.50; Type B pen = RM2.30

Allmaths

回复 1183# 乙劍真人


答案不对啊....
Type A: RM0.80
Type B: RM0.60

乙劍真人

回复  乙劍真人
答案不对啊....
Type A: RM0.80
Type B: RM0.60
Allmaths 发表于 15-6-2010 09:12 AM

抱歉，应该是我的 1st equation 错了，

再等各路高手相助吧！

你给的答案好像也是不对哦！

32(RM0.80) + 57(RM0.60) = RM59.80 ??

Lov瑜瑜4ever

抱歉，应该是我的 1st equation 错了，

再等各路高手相助吧！

你给的答案好像也是不对哦 ...
乙劍真人 发表于 15-6-2010 09:37 AM

也许那个60块是不用用到完的吧？

Lov瑜瑜4ever

Let Type A pen : RM(x+0.2)
      Type B pen : RM x

then the equation can be written as below
a(A)+(a+25)B=RM 60 ,
a(x+0.2)+(a+25)x=60
ax+0.2a+ax+25x=60
2ax+0.2a+25x=60
a(2x+0.2)=60-25x
            a=(60-25x)/(2x+0.2)

a是买多少枝pen...所以a一定是natural number
If a=1, then 2x+0.2=60-25x and x=RM 2.21
If a=2, then 2(2x+0.2)=60-25x and x=RM 2.06
If a=3, then 3(2x+0.2)=60-25x and x=RM 1.92
......
If a=299, then 299(2x+0.2)=60-25x and x=RM 3.21x10^-4
a不能等于300,否则x会变成RM 0...
这样不是很多答案
我也不知道这样做有没有错
欢迎指正错误><

Lov瑜瑜4ever

需要各位帮忙.

Leo has RM60. With the amount of money, he can buy 25 more Type B pens than Type A  ...
Allmaths 发表于 15-6-2010 08:52 AM

可以问下你这题是从哪里看到的呀？

walrein_lim88

需要各位帮忙.

Leo has RM60. With the amount of money, he can buy 25 more Type B pens than Type A  ...
Allmaths 发表于 15-6-2010 08:52 AM

    PEN A的数量 和 PEN B的数量 是 UNKNOWN。
    PEN A的价钱 和 PEN B的价钱 也是 UNKNOWN.
   有4个UNKNOWN了。。。要找出unique solution ,一定要4个EQUATION。。。
   而你的问题最多也是只能有3个EQUATION :
   Let A - quantity of pen A, B-quantity of pen B, x-price of pen A, y-price of pen B
    B - A = 25
    x - y =0.2
    Ax + By = 60
   这种情况下，答案会很多～～～

砂砾

请教教我这题。。。
if loga(x/a^2)=3loga2-loga(x-2a),express x in terms of a.
谢谢前辈的请教

walrein_lim88

请教教我这题。。。
if loga(x/a^2)=3loga2-loga(x-2a),express x in terms of a.
谢谢前辈的请教
砂砾 发表于 15-6-2010 02:24 PM

ELFofWAR

话说
fx-570ES和ms有什么不同?
读math t一定要换去es吗?

Lov瑜瑜4ever

话说
fx-570ES和ms有什么不同?
读math t一定要换去es吗?
ELFofWAR 发表于 15-6-2010 03:49 PM

其实要不要都可以的啦
我懂有其中一个不同之处
如果是MS啦
你打sin 60就会=0.866...
如果是ES的话
那就会=(3^1/2)/2

Allmaths

回复 1188# Lov瑜瑜4ever


说到这个还真的不好意思...是F4 F5 Add maths 的题目来的...在pelangi的书找到的...

Allmaths

回复 1185# 乙劍真人


我也是觉得怪怪的.....这是书给的答案...

HanQing

find the modulus and argument of the 2 complex numbers that satisfy the equation
(1+Z^2)/(1-Z^2)=i. ...
qwer0909 发表于 15-6-2010 07:38 PM

From

(1+z^2)/(1-z^2)=i
( 1+ z^2)= i ( 1-z^2)
1 + z^2 = i - i z^2
(1+i )z^2 = -1 +i
z^2 = (-1+i)/(1+i)
Taking conjugate of denominator

z^2 = {(-1+i)(1-i)}/2
z^2 = i

Consider  z = x +i y, where x , y ∈R

z^2 = x^2-y^2 +2xyi = i

By comparing both side,
x^2-y^2 = 0 ----(1)
2xy = 1 ----------(2)

after solving the 2 equations,
x = 1/√2 , -1/√2
y = 1/√2 , -1/√2.

so
z = -1/√2 -1/√2 i  
or z= 1/√2 +1/√2 i

|z|= 1
arg z = pi/4 or -3pi/4

walrein_lim88

find the modulus and argument of the 2 complex numbers that satisfy the equation
(1+Z^2)/(1-Z^2)=i. ...
qwer0909 发表于 15-6-2010 07:38 PM

    (1+Z^2)/(1-Z^2)=i
    Let z = x + iy
     z^2 = x^2 - y^2 + 2xyi
   
   then 1 + z^2 = i(1-z^2)
          1 + x^2 - y^2 +2xyi = i ( 1 - x^2 + y^2 -2xyi)
          1 + x^2 - y^2 +2xyi = 2xy + ( 1 - x^2 + y^2)i
By doing comparison:
1 + x^2 - y^2 =2xy
1 - x^2 +y^2 = 2xy

therefore:  1 + x^2 - y^2 = 1 -x^2 + y^2
                2x^2  - 2y^2 =0
                 x^2 - y^2 =0
                ( x+y)(x-y)=0
               x=-y, or x = y
if x = -y:
1  + x^2 - x^2 = - 2x^2
1 + 2x^2 =0
which is impossible:

If x = y:
1 = 2x^2
x^2 = 1/2
x = +- 1/surd 2
y = +- 1/surd 2
z =+- 1/surd 2 ( 1 + i )

modulus = 1
arg = 45 degree or 225 degree

Lov瑜瑜4ever

回复  Lov瑜瑜4ever


说到这个还真的不好意思...是F4 F5 Add maths 的题目来的...在pelangi的书找到的 ...
Allmaths 发表于 15-6-2010 10:34 PM

好奇怪哦
哈哈

Lov瑜瑜4ever

i ( 1 - x^2 + y^2 -2xyi)
= 2xy + ( 1 - x^2 + y^2)i这个是怎样变的呢？
qwer0909 发表于 15-6-2010 10:58 PM

先把i乘进去
然后再把i抽出来><

HanQing

i ( 1-z^2)
= i - i z^2想问这个是怎样变的？
qwer0909 发表于 15-6-2010 10:57 PM

multiply i into it