费费数学第七章 几题不明！

hedonism555

38、K is the square of an integal N,which of the choice below could be the number of factors of K?
Ⅰ奇数  Ⅱ偶数  Ⅲ质数

该题并没问有多少“不同”的因子个数，所以答案是否应是2。 偶数。


44、The possibility that the value of stock A will increase is 0.34 and the possibility that stock B will increase is 0.68.What is the biggest possibility that neither will happen？

Why the answer is 0.32?  My way is 1-0.34*0.68

48、一位男士有5套衣服，7条领带；其中有一套衣服只能配一条领带，其余衣服和领带都可搭配，问一共有几种搭配方法。

题目说“有一套衣服”是否是指具体的某一套， 如果按C1,5+4*6的算法是不是违背题意？

49、一个平行四边形的边长分别为6、7，问以下那个可能是平行四边形的面积？
(1)21  (2)42  (3)84

Why 21 ?


64、x,y为整数，y能否被3整除？
(1)y=2x^3+9x^2-8x   (2)x=3


No clue why A.

72、四封信装入四个已写好正确地址的信封(每封信只装一个信封)，仅有一封信装对地址的概率是多少？

The question is similar to 48. Since the question refers to only one single letter can be correctly enveloped, then why C1,4*2. Its hidden meaning is that each letter can be fitted into a correct envelop in each one time. This is not what the question indicates.  My answer is  3,3-1(the remaining 3 all correct)-3(2 out of 3 wrong)/p4/4


Any strong mind can kindly pierce a few ray of inspirational light into my confusing mental fog ? Thank you very much !

hedonism555

Nobody can do me a favor?   help!!!!!!!!!!!!!!!!!!!!!!

sftd

49、一个平行四边形的边长分别为6、7，问以下那个可能是平行四边形的面积？
(1)21  (2)42  (3)84

Why 21 ?


64、x,y为整数，y能否被3整除？
(1)y=2x^3+9x^2-8x   (2)x=3


No clue why A.


49. s=absinA=42sinA, A为a,b夹角

64. y=2x^3+9x^2-8x,    9x^2能被3整除, 只要考虑2x^3-8x能否被3整除,
2x^3-8x=2x(x+2)(x-2)能被3整除

hedonism555

Thank you!   49 题还是不太明白。 能详细讲一下吗？

sftd

49. 平行四边形, 如果a为底边, bsinA为底边上的高啊, 画个图就知道了.
同理,三角形面积为1/2absinA

hedonism555

Okay, i roughly grasp the idea. This area is my weak point. Thanks again!