1.(logab+logb^2)/(loga^2b^3+logb^3)
a.2,
b.1/2,
c.1/3,
d.logab^3
2.cos^4(x)-sin^4(x)=
a.1
b.-1
c.(cos(x)-sin(x))(cos(x)+sin(x))\
d.(cos(x)-sin(x))^2
3.Let 101010(2)=P(10) and 1010000(2)=Q(10).How many prime
numbers are there between P and Q ?
a.6
b.7
c.8
d.9
4.A paper sector with central angle x is folded up to form a cone with
semi-vertical angle y .The relationship between x and y is
a.x/360=siny
b.x/360=1/siny
c.x/360=cosy
d.x/180=siny
5.2^m*4^n=16^n,find M:N
math mc+LQ
2009-07-06 4:28 am
回答 (1)
2009-07-06 5:10 am
✔ 最佳答案
1.The answer is b.
(logab + logb2)/(loga2b3 + logb3)
= log(abb2)/log(a2b3b3)
= log(ab3)/log(a2b6)
= log(ab3)/log(ab3)2
= log(ab3)/[2log(ab3)]
= 1/2
**********
2.
The answer is a.
cos4(x) - sin4(x)
= [cos2(x)]2 - [sin2(x)]2
= [cos2(x) + sin2(x)][cos2(x) - sin2(x)]
= 1[cos2(x) - sin2(x)]
= [cos(x) + sin(x)][cos(x) - sin(x)]
**********
3.
答案是 d。
1010102 = 1x(2)5 + 1x(2)3 + 1x(2) = 4210
10100002 = 1x(2)6 + 1x(2)4 = 8010
在 42 與 80 間的質數有：43, 47, 53, 59, 61, 67, 71. 73, 79
共有質數 9 個。
**********
4.
The answer is a.
Let R be the radius of the paper sector.
Arc length of the sector
= 2πR•(x/360)
Let r be the base radius of the cone.
Circumference of the base of the cone
= 2πr
Arc length of the sector = Circumference of the base of the cone
2πR•(x/360) = 2πr
r/R = x/360
Consider the cone.
The height, r and R form a right-angled triangle with hypotenuse R.
siny = r/R
Hence, siny = x/360
**********
5.
2m x 4n = 16n
2m x (22)n = (24)n
2m x 22n = 24n
2m = 24n / 22n
2m = 24n-2n
2m = 22n
m = 2n
m/n = 2/1
Hence, m : n = 2 : 1
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