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let a and b be any positive integers and x=0 or 1,then

a)a^xb(1-x)=x*a+b(1-x)
b)a^xb(1-x)=a(1-x)+x*b
c)a^x*b(1-x)=a*(1-x)*b*x
d)None of these necessarily true

2006-07-19 04:43:27 · 7 answers · asked by sanko 1 in Science & Mathematics Mathematics

is this question not clear ?

my book has given only the answer but no explanation....the book has suggested to use "method of substitution"

but as you see, its difficult to use what to substitute .

BTW, any have a nice answer ?

2006-07-19 04:55:11 · update #1

------------------------
hmm...none of the answer so far matches with my book.....

2006-07-19 05:22:54 · update #2

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anser given is (a)

2006-07-19 06:03:28 · update #3

7 answers

_________a)a^xb(1-x)=x*a+b(1-x)
4 x=1
lhs = 0; rhs = 0

4 x=0
lhs = b; rhs = 0

__________b)a^xb(1-x)=a(1-x)+x*b
4 x = 1
lhs = 0; rhs = b

4 x = 0
lhs = b; rhs = a

____________c)a^x*b(1-x)=a*(1-x)*b*x
4 x = 1
lhs = 0; rhs = 0

4 x = 0
lhs = b; rhs = 0






none is true. ur ans is D

2006-07-19 05:13:22 · answer #1 · answered by Sean 3 · 0 0

options a) and b) are incorrect since the left hand side of both equations will always be 1 either x=0 or 1,since it will always equal a^0, while the right hand side may equal either a or b.
option c) is also incorrect because, as i assumed that the right hand side is a^(1-x)*bx, when x=0, then b=0, while when x=1, then also b=0. And as you said, b is a positive integer.
so none of the above options is necessarily true, which is option d).

2006-07-19 12:10:08 · answer #2 · answered by Turkleton 3 · 0 0

Why not go to a homework help site?

P.S. If you had listened in class, you would not have had to go to summer school and you would also know the answer!

2006-07-19 11:49:25 · answer #3 · answered by Rosie 2 · 0 0

hu what in the world

2006-07-19 17:38:08 · answer #4 · answered by Anonymous · 0 0

Option (e)

2006-07-19 11:48:44 · answer #5 · answered by Anonymous · 0 0

ans:(D)

2006-07-19 11:52:42 · answer #6 · answered by Anonymous · 0 0

wtf?

2006-07-19 11:45:35 · answer #7 · answered by Anonymous · 0 0

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