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VCE Methods Question Thread!

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TrueTears:
probably not, but just remember the binomial theorem is more of a tool that is used in the process of solving some problem rather than simply applying the theorem. It's a very versatile tool that can be used in different kinds of situations ;)

Dominatorrr:
OK yeah, that's kind of what I was thinking, thanks.

tqn:
Hi I have a question from the Essentials book:

The dimensions of an enclosure are shown. The perimeter of the enclosure is 160 m.
Find a rule for the area, A m^2, of the enclosure in terms of x.

Can someone please help me out? I can't seem to get it through my head :S

(the Image is in the form of an attachment as i don't know how to embed it into a post -_-")

b^3:

--- Quote from: tqn on December 08, 2011, 08:00:07 pm ---Hi I have a question from the Essentials book:

The dimensions of an enclosure are shown. The perimeter of the enclosure is 160 m.
Find a rule for the area, A m^2, of the enclosure in terms of x.

Can someone please help me out? I can't seem to get it through my head :S

(the Image is in the form of an attachment as i don't know how to embed it into a post -_-")

--- End quote ---
Firstly find the perimeter in terms of x and y
perimeter = y+x+20+12+(y-20)+(x-12)
now that equals 160 m
2y+2x=160
so y=80-x

Now to find the area split it up into two shapes, lets split it the vertical way.
Area=(20)(x)+(y-20)(x-12)
Area=20x+(y-20)(x-12)
Now we know that y=80-x, so sub that in
Area=20x+(80-x-20)(x-12)
=-x2+92x-720 m2

EDIT: missed the negative - fixed

tqn:

--- Quote from: b^3 on December 08, 2011, 08:23:34 pm ---
--- Quote from: tqn on December 08, 2011, 08:00:07 pm ---Hi I have a question from the Essentials book:

The dimensions of an enclosure are shown. The perimeter of the enclosure is 160 m.
Find a rule for the area, A m^2, of the enclosure in terms of x.

Can someone please help me out? I can't seem to get it through my head :S

(the Image is in the form of an attachment as i don't know how to embed it into a post -_-")

--- End quote ---
Firstly find the perimeter in terms of x and y
perimeter = y+x+20+12+(y-20)+(x-12)
now that equals 160 m
2y+2x=160
so y=80-x

Now to find the area split it up into two shapes, lets split it the vertical way.
Area=(20)(x)+(y-20)(x-12)
Area=20x+(y-20)(x-12)
Now we know that y=80-x, so sub that in
Area=20x+(80-x-20)(x-12)
=x2+92x-720 m2

--- End quote ---

Ohhh I get it, I forgot to find y first then to sub it in ^^" Although, the answer I got was -x2 + 92x - 720 m2 so is there any difference with the x2 or... :S

Also thanks for the help as well :)

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