Quick Question... (2 Viewers)

[aussiechick007]

Hi, I can't work out the answers for 2004 q10 b iv, the last bit... it just doesnt make sense to me.

how does...
(a - b + c)(a + b - c)
go to
(a + b + c - 2b)(a + b + c - 2c)

???

 

[tommykins]

a-b+c = a+b-2b+c

Logic is kinda like, 3 = 2+1 or 4-1 or 5-2

 

[H4rdc0r3]

Haaaaaaaahahaha!

 

[Timothy.Siu]

lol, silly question...

 

[trailblazer]

lol, silly question...

Nah it's harder than you think.

 

[Timothy.Siu]

lol not that part i know -b=b-2b lol

 

[trailblazer]

Haaaaaaaahahaha!

That's not very nice, just because someone didnt get that doesnt mean you can laugh at them. It's not like you get all the maths questions either. You some kind of toolfag?

 

[trailblazer]

lol not that part i know -b=b-2b lol

The question won't be that simple. As if it would be

Prove that:
(a - b + c)(a + b - c) = (a + b + c - 2b)(a + b + c - 2c)

 

[Graceofgod]

The question won't be that simple. As if it would be

Prove that:
(a - b + c)(a + b - c) = (a + b + c - 2b)(a + b + c - 2c)

Yeah I think that is what he is saying. He was just laughing that he didn't get that part.

But mm, reasonable question. I often look for the harder solution first.

 

[trailblazer]

Yeh if only we had the actual question. Ill go find it.

 

[trailblazer]

Thought so....




The diagram shows a triangular piece of land ABC with dimensions
AB = c metres, AC = b metres and BC = a metres, where a ≤ b ≤ c.
The owner of the land wants to build a straight fence to divide the land into two
pieces of equal area. Let S and T be points on AB and AC respectively so that ST
divides the land into two pieces of equal area.
Let AS = x metres, AT = y metres and ST = z metres.

 

[tommykins]

lol, silly question...

don't laugh, you start to use that technique alot in 4unit.

 

[aussiechick007]

ok, thanks for the first person for explaining, and thanks for the others for laughing at me....

like someone said, yes that wasn't just the whole question, it was the last bit in that triangle question that someone posted.