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Author Topic: VCE Methods Question Thread!  (Read 2365078 times)  Share 

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TrueTears

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Re: VCE Methods Question Thread!
« Reply #15 on: December 02, 2011, 08:23:13 pm »
+3
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

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Re: VCE Methods Question Thread!
« Reply #16 on: December 03, 2011, 10:49:49 am »
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OK yeah, that's kind of what I was thinking, thanks.

tqn

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Re: VCE Methods Question Thread!
« Reply #17 on: December 08, 2011, 08:00:07 pm »
0
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 -_-")
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Re: VCE Methods Question Thread!
« Reply #18 on: December 08, 2011, 08:23:34 pm »
+4
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 -_-")
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
« Last Edit: December 08, 2011, 08:54:58 pm by b^3 »
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tqn

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Re: VCE Methods Question Thread!
« Reply #19 on: December 08, 2011, 08:48:16 pm »
0
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 -_-")
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

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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b^3

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Re: VCE Methods Question Thread!
« Reply #20 on: December 08, 2011, 08:52:08 pm »
+3
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 -_-")
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

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 :)
Yeh thats right, just missed the negative when I typed it out, had it written down right and copied it wrong.
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tqn

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Re: VCE Methods Question Thread!
« Reply #21 on: December 08, 2011, 09:04:01 pm »
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O ok then. Once again thanks very much for the assistance :)
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Dominatorrr

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Re: VCE Methods Question Thread!
« Reply #22 on: December 15, 2011, 08:30:58 pm »
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QUESTION ATTACHED

How do I find the values of a and b?

b^3

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Re: VCE Methods Question Thread!
« Reply #23 on: December 15, 2011, 08:34:38 pm »
+5
QUESTION ATTACHED

How do I find the values of a and b?
This exact question has been on one of my SACS before. Anyway since the diameter at the top of the structure is 20m, you have the point 20/2=10 away from the origin and up 40m (height) i.e. the point (10,40). Now at the base you have a height of 0 and 50m from the origina so you have the point (50,0). Then plug those two points back into the equation to find a and b.
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brightsky

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Re: VCE Methods Question Thread!
« Reply #24 on: December 15, 2011, 08:36:51 pm »
+6
we know that the graph passes through these points:
(50,0), (-50,0), (10, 40), (-10,40)

so...subbing x and y coordinates:
0 = a/50^2 + b
40 = a/10^2 + b

0 = a + 2500b..[1]
4000= a +100b..[2]

[2]-[1]:
4000 = -2400 b
b = -5/3
so a = 2500*5/3 = 12500/3

hope these correlate to the answers, these numbers look weird.
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Dominatorrr

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Re: VCE Methods Question Thread!
« Reply #25 on: December 15, 2011, 11:11:21 pm »
0
QUESTION ATTACHED

How do I find the values of a and b?
This exact question has been on one of my SACS before. Anyway since the diameter at the top of the structure is 20m, you have the point 20/2=10 away from the origin and up 40m (height) i.e. the point (10,40). Now at the base you have a height of 0 and 50m from the origina so you have the point (50,0). Then plug those two points back into the equation to find a and b.
we know that the graph passes through these points:
(50,0), (-50,0), (10, 40), (-10,40)

so...subbing x and y coordinates:
0 = a/50^2 + b
40 = a/10^2 + b

0 = a + 2500b..[1]
4000= a +100b..[2]

[2]-[1]:
4000 = -2400 b
b = -5/3
so a = 2500*5/3 = 12500/3

hope these correlate to the answers, these numbers look weird.

Ah I see, thanks to the both of you and yes, those answers are right.

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Re: VCE Methods Question Thread!
« Reply #26 on: December 16, 2011, 11:12:54 am »
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QUESTION ATTACHED

How do I find the values of a and b?

Another part of this question is "A junior version of the ride is proposed, with the sides of the central structure dilated by a factor of 2 from the y-axis. Find the new equation."

The answer for this question is y=(16666.33/x^2)-1.67. I'm a little confused as to how the 16666.33 was obtained.

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Re: VCE Methods Question Thread!
« Reply #27 on: December 16, 2011, 11:36:49 am »
+2
To dilate by factor of 2 from y-axis means you need to replace x with x/2. This should give you the answer.
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Re: VCE Methods Question Thread!
« Reply #28 on: December 16, 2011, 05:56:22 pm »
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To dilate by factor of 2 from y-axis means you need to replace x with x/2. This should give you the answer.

Yeah I think it makes sense now.

y=(4167/x^2)-1.67
y=(4167/0.5x*0.5x)-1.67
y=(4167/0.25x^2)-1.67
y=(16668/x^2)-1.67 (CLOSE ENOUGH)

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Re: VCE Methods Question Thread!
« Reply #29 on: December 16, 2011, 06:02:12 pm »
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To dilate by factor of 2 from y-axis means you need to replace x with x/2. This should give you the answer.

Yeah I think it makes sense now.

y=(4167/x^2)-1.67
y=(4167/0.5x*0.5x)-1.67
y=(4167/0.25x^2)-1.67
y=(16668/x^2)-1.67 (CLOSE ENOUGH)

Yep, and the reason your answer was a little off is because you rounded 12500/3 up to 4167, when it should be 4166.6(recurring)
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