The first thief takes (1/2 x + 1) . What remains ? x - (1/2x + 1)
So the 2nd thief takes 2/3 of [ x - (1/2x + 1) ]
What remains ? x - 2/3 [ x - (1/2x + 1) ]
So the 3rd thief takes 2/3 of { x - 2/3 [ x - (1/2x + 1) ] } and he takes 1 more .
What remains ? x - ( 2/3 { x - 2/3 [ x - (1/2x + 1) ] } + 1 )
And that whole ugly thing is equal to ' 1 ', so you can solve it for 'x'..
The whole problem from here on is an exercise in simplifying
an expression with a bunch of 'nested' parentheses in it.
===============================================
This is a lot harder than just solving the problem with logic and
waving your hands in the air. Here's how you would do that:
Start from the end and work backwards:
-- One diamond is left.
-- Before the 3rd thief took 1 more, there were 2.
-- That was 1/3 of what was there before the 3rd man took 2/3.
So he found 6 when he arrived.
-- 6 was 1/3 of what was there before the second thief helped himself.
So there were 18 when the 2nd man arrived.
-- 18 was 1 less than what was there before the first thief took 1 extra.
So he took his 1 extra from 19.
-- 19 was the remaining after the first man took 1/2 of all on the table.
So there were 38 on the table when he arrived.
Thank you for your generous 5 points.
Answer:
I will say C, because tossing a coin twice has two outcomes. H is for heads while T is for heads. Tell me if I'm right.
Step-by-step explanation:
Answer:
c = 
Step-by-step explanation:
Given
R= 
Clear the radical by squaring both sides
R² = b² - 4ac ( subtract b² from both sides )
R² - b² = - 4ac ( multiply all terms by - 1 )
b² - R² = 4ac ( divide both sides by 4a )
= c
Answer:
D
Step-by-step explanation:
