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【纪念当年的帖子(2010)】Add Maths功课讨论区
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发表于 27-2-2011 09:53 AM
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if f(x)=10+4x-x^2 can be written as c-(x+b)^2,
(a)find the values of b and c.
我做到的答案是b=2,c= ...
×寂寞°心 发表于 26-2-2011 09:03 AM
f(x)=-x^2+4x+10
f(x)=-(x^2-4x)+10
f(x)=-[(x-2)^2-4]+10
f(x)=-(x-2)^2+14
所以b=-2 |
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发表于 27-2-2011 11:14 AM
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看不到图。。。
Allmaths 发表于 27-2-2011 09:52 AM
你点击第二个网址看看~~看看能不能看到~~ |
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发表于 27-2-2011 11:17 AM
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用b^2-4ac来算咯。。。如果b^2-4ac>0就有两个real roots, b^2-4ac=0就有一个real roots。。
Allmaths 发表于 27-2-2011 09:51 AM
那如果是b^2-4ac<0呢?就连一个real root都没有吗? 0 ? |
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发表于 27-2-2011 08:20 PM
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那如果是b^2-4ac
×寂寞°心 发表于 27-2-2011 11:17 AM
对一个都没有~~ 但不是0 |
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发表于 3-3-2011 12:03 PM
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In an arithmetic progression, the sum of first four term is 28 and the sum of the first 8 term is 48. Find the sum of first 6 terms.
遇到这种题目就难倒我~
有谁愿意帮忙吗?
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发表于 3-3-2011 12:41 PM
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本帖最后由 c3harry 于 4-3-2011 07:19 PM 编辑
S4= 4/2 (2a + (4-1)d) S8 = 8/2 (2a + (8-1)d)
28 = 4a + 6d ---(1) 48 = 8a +28d ---(2)
(1)-->(2)
d = -1/2 a = 31/4
S6 = 6/2 ( 2(31/4)+ (6-1)(-1/2))
=39
sorry... 多谢Allmaths。。。
p/s:但是S6=S8 是可能的
如:
当TERMS =6 5 4 3 2 1 0 -1的时候 |
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发表于 3-3-2011 01:11 PM
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本帖最后由 Allmaths 于 3-3-2011 01:13 PM 编辑
S4= 4/2 (2a + (4-1)d) S8 = 8/2 (2a + (8-1)d)
28 = 4a + 6d ---(1) ...
c3harry 发表于 3-3-2011 12:41 PM
simultaneous做错了。。。
我得到a=31/4 和 d=-1/2
再说你的S6怎么等于S8? |
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发表于 9-3-2011 10:10 AM
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发表于 9-3-2011 10:17 AM
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If the number sequence 2, p, q, 25/2, r is an arithmetic progression, find the value of p, q, r.
Answer: p=11/2, q=9, r=16
有劳。
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发表于 9-3-2011 10:04 PM
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If the number sequence 2, p, q, 25/2, r is an arithmetic progression, find the value of p, q, r.
...
寂寞乄好了 发表于 9-3-2011 10:17 AM
p-2=q-p
q=2p-2 --->eq 1
25/2-q=p-2
p+q=29/2 --->eq 2
Sub eq 1 into eq 2,
p+(2p-2)=29/2
3p=33/2
p=11/2
q=2(11/2)-2
q=9
r-25/2=p-2
r=(11/2)-2+25/2
r=16 |
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发表于 26-3-2011 10:08 AM
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我有问题想问大家
Find the volume of revolution, in terms of π, when the region bounded by the curve x=y²-4y, the y-axis and the line x=3 is rotated completely about the y-axis. |
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发表于 30-3-2011 07:14 PM
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我有问题想问大家
Find the volume of revolution, in terms of π, when the region bounded by the curve x=y²-4y, the y-axis and the line x=3 is rotated completely about the y-axis.
jiayi_94 发表于 26-3-2011 10:08 AM
y-axis为轴。
当x=3,y=3²-4(3)=-3
∫πx²dy from -3 to 0
=π∫(y²-4y)²dy from -3 to 0
看可以吗? |
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发表于 17-5-2011 12:58 AM
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有一题问题想请教大家。。。
if log a P= m and log b P = n
show that log ab P = (mn)/(m+n)
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发表于 17-5-2011 11:42 PM
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有一题问题想请教大家。。。
if log P= m and log P = n
show that log P = (mn)/(m+n)
:hands ...
Saturn20 发表于 17-5-2011 12:58 AM
log_a P=m
log_p a=1/m --->eq 1
log_b P=n
long_p b=1/n ---eq 2
eq 1+eq 2,
log_p a+log_p b=(1/m)+(1/n)
log_p (ab)=(m+n)/mn
log_(ab) P=mn/(m+n) |
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发表于 18-5-2011 01:15 PM
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我的答案好长啊。。。 |
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发表于 26-5-2011 01:04 PM
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本帖最后由 gerrykhang 于 26-5-2011 01:05 PM 编辑
有一题问题想请教大家。。。
if log P= m and log P = n
show that log P = (mn)/(m+n)
:hands ...
Saturn20 发表于 17-5-2011 12:58 AM
log(ab)p=1/[log (p)ab]
=1/[log(p)a+log(p)b]
change log(p)a to 1/[log (a)p] and subsitute log(a)p=m
change log(p)b to 1/[log (b)p] and subsitute log(b)p=n
=1/[1/m+1/n] , then simplify it
=mn/(m+n) (QED) |
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发表于 29-5-2011 12:00 PM
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发表于 4-6-2011 10:03 PM
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有一题问题。。。
find two numbers in the ratio 7:12 so that the greater exceeds the less by 257 . |
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发表于 5-6-2011 12:36 AM
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有一题问题。。。
find two numbers in the ratio 7:12 so that the greater exceeds the less by 257 .
josser 发表于 4-6-2011 10:03 PM
答案是616.8和359.8吗? |
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发表于 5-6-2011 12:56 PM
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回复 839# 睡星
我没有答案 ><
请问你怎样找到这答案的呢 ? |
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