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Mth163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
http://vhcc2.vhcc.edu/pc1fall9
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question form
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Mth163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
missing qa's
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qa_16,qa_17,Qa-18,qa-19
are there qa_'s for these assignments because I clicked on them and they said path not specified ?????
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There are no more qa's after Assignment 15, just the worksheets and the Queries.
I don't see the qa's on either the Assignments Page or the Brief Assignments Page.
I want to be sure you're viewing the correct page. Can you send me a copy of the Address Box of the page where you see links to these qa's?
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http://vhcc2.vhcc.edu/pc1fall9/frames%20pages/assignments_grid_full.htm#Assignments_15-19_Below
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Thanks.
That page has been changed, and includes a link to the correct version of the Assignments Page.
The correct version is the one accessible directly from your course homepage.
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Mth163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
TheReturnOf Proportionality
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y = k x^2
When you double the square of 2 you get 4
4= k 2^2
so when x doubles 4 quads
...I still feel like i am doing something wrong
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First let me address your use of fractions (and order of operations) when you solved the equation
4= k / 1/3^3
for k.
In the first place you meant
4 = k / (1/3)^3.
k divided by 1/3 would not be k / 1 / 3, which by order of operations tells you to divide k by 1 then divide the result by 3.
k divided by 1/3 would be expressed as k / (1/3).
Of course in your equation you are cubing the 1/3. I didn't include the exponent above so we could focus on just the order of operations. Including the cube you would have to write this as
k / (1/3)^3.
This is then equal to k / (1/27), so your equation becomes
4 = k / (1/27).
Multiplying 27 by both sides of this equation doesn't help, because k / (1/27) * 27 means to divide k by 1/27, then multiply the result by 27. But k divided by 1/27 is k multiplied by 27/1, so
k / (1/27) * 27 = k * 27 * 27 = 729 k, not just k.
To solve the equation
4 = k / (1/27)
you would multiply both sides by 1/27, not by 27.
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This is then equal to k / (1/27), so your equation becomes
4 = k / (1/27).
Multiplying 27 by both sides of this equation doesn't help, because k / (1/27) * 27 means to divide k by 1/27, then multiply the result by 27. But k divided by 1/27 is k multiplied by 27/1, so
k / (1/27) * 27 = k * 27 * 27 = 729 k, not just k.
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what happened to the 4??? and why would it not be 1/729(k)=4
(1/729*729)(k)= 4*729?? or would be 1/27*27=1 so k=4, no k=4*1/27 is that what you are saying ???
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The correct solution would not involve 729 k, but
k / (1/27) * 27
would give you 729 k, by the order of operations and the behavior of fractions.
The statement you're referring to reads
"Multiplying 27 by both sides of this equation doesn't help, because k / (1/27) * 27 means to divide k by 1/27, then multiply the result by 27. But k divided by 1/27 is k multiplied by 27/1, so
k / (1/27) * 27 = k * 27 * 27 = 729 k, not just k. "
That statement tells you why your algebra step, where you multiplied by 27, does not lead you to a correct result.
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#$&*
Mth163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
the return of the cube
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** Question Form_labelMessages **
cubes
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Question: `q012. A sculpture is made of small cubes, about the size of playing dice.
Suppose each small cube could be cut into smaller cubes, each having edges 1/3 as long as the edges of the original. Suppose that somehow each of the smaller cubes is then expanded until it is the size of one of the original cubes, forming a larger sculpture. The shape of the sculpture would not change, only its size would be different. Don't worry about how this might be accomplished or even whether this is possible.
If this could be done, how many times higher would the larger sculpture be?
How many times as much volume would the larger sculpture occupy?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
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The same size of the original larger sculpture
The same
Because if you make it small and then expanded it again wouldn’t it be the same size since you are saying that the cube basically gets as big as it was before. But are the 1/3 edges included on how big it was before if not then we could take the orginal size*1/3 to get the new size to get volume
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The figure isn't shrunk in the prcess of subdividing the original cubes. Those cubes are cut into smaller cubes without changing the total volume.
The smaller cubes are now inflated to the size of the original cubes, so the sculpture now gets bigger.
By what factor does its volume change?
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so does the cube get 3 times as big as the original since you have a bunch of cubes cut into 1/3's then each of those small cubes are re-expanded to the size of a original cube???
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See if you can answer the questions below. You should either sketch this or cut up an actual cube to be sure you see what's going on here (you could cut up something like a potato or a piece of fruit).
What we're trying to visualize here is a cube being cut into smaller cubes, then each of the smaller cubes being expanded to the size of the original (of course you have to imagine the expansion).
Suppose you have a cube 3 feet on a side.
Into how many 1-foot cubes can it be cut?
Now if a 1-foot cube is expanded to form a 3-foot cube, by how many times does its volume increase?
If each of the 1-foot cubes into which you divided the original 3-foot cube was expanded back into a 3-foot cube, then how many times as much volume would you have?
So in general if a cube is cut into smaller cubes, each having edges 1/3 as long as the edges of the original, and each of the smaller cubes is then expanded until it is the size of the original cube, by how many times does its volume increase?
What are the implications for this sculpture?
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question form
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Suppose you have a cube 3 feet on a side.
Into how many 1-foot cubes can it be cut? 9 cubes
Now if a 1-foot cube is expanded to form a 3-foot cube, by how many times does its volume increase? the volume increases by 3*3 so the volume increases by 9 times the original one foot cube
If each of the 1-foot cubes into which you divided the original 3-foot cube was expanded back into a 3-foot cube, then how many times as much volume would you have? 3*9= (or increasing for the third time so 3^3)=27
So in general if a cube is cut into smaller cubes, each having edges 1/3 as long as the edges of the original, and each of the smaller cubes is then expanded until it is the size of the original cube, by how many times does its volume increase? the volume increases by 27 cubic feet
What are the implications for this sculpture?
When the one foot sided cube is divided into 3ft per side then expanded until it reached the size of the original cube you would have to expanded it 3*3*3 times to fill in each cubic foot which equals 27 cubic feet
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You are correct that the volume of the original 1-foot cube would expand to 27 cubic feet.
The 'it' in your sentence would refer to either the one foot sided cube or the original cube. However the 1 foot cube expanded 3 * 3 * 3 times would not fill a cubic foot, nor would the original cube. So your statement doesn't quite say what you want it to. It's worth thinking about how you would word a statement to say exactly what you're trying to say here.
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In any case the expansion would result in each of the original 1-foot cubes expanding to a volume of 27 cubic feet, as you say.
What would therefore happen to the volume of the original statue?
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#$&*
Mth163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
y2 / y1 = (x2 / x1)^2
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y = k x^2
When you double the square of 2 you get 4
4= k 2^2
so when x doubles 4 quads
...I still feel like i am doing something wrong
@&
First let me address your use of fractions (and order of operations) when you solved the equation
4= k / 1/3^3
for k.
In the first place you meant
4 = k / (1/3)^3.
k divided by 1/3 would not be k / 1 / 3, which by order of operations tells you to divide k by 1 then divide the result by 3.
k divided by 1/3 would be expressed as k / (1/3).
Of course in your equation you are cubing the 1/3. I didn't include the exponent above so we could focus on just the order of operations. Including the cube you would have to write this as
k / (1/3)^3.
This is then equal to k / (1/27), so your equation becomes
4 = k / (1/27).
Multiplying 27 by both sides of this equation doesn't help, because k / (1/27) * 27 means to divide k by 1/27, then multiply the result by 27. But k divided by 1/27 is k multiplied by 27/1, so
k / (1/27) * 27 = k * 27 * 27 = 729 k, not just k.
To solve the equation
4 = k / (1/27)
you would multiply both sides by 1/27, not by 27.
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See my notes directly above.
First be sure you understand the fractions and order of operations addressed in my first note.
Then I suggest you solve the problem in three different ways, as indicated by my second note.
If you can do this successfully you'll be well on your way to a solid mastery of the extremely important concept of proportionality.
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Now you can use the correct proportionality
y = k x^2.
There are many ways to set this up.
You set this up by saying that
y = 4 m^2 when x = 1/3
and find y when x = 1. This would be along the lines of your original thinking, but you have to specify both values of x. I'm not sure you saw that if x for the first figure is 1/3, then x for the second has to be 1.
You could alternatively set this up as
y = 4 m^2 when x = 1
and find y when x = 3. You would get a different value of k, but the final result for the volume would be the same.
I recommend that you set this up in both ways and verify that is doesn't matter whether x goes from 1 to 3, or from 1/3 to 1. You get the same result both ways.
Another way to solve the problem is to recall that
y = k x^2
implies that
y2 / y1 = (x2 / x1)^2.
For this question, what is the ratio x2 / x1, and what therefore is the ratio y2 / y1? What do you therefore conclude is the area of the larger figure?
This reasoning will lead to the same result as before.
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please ignore my last proportionality question form labeled TheReturnOf Proportionality
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for y=4(1)^2 i got 4
y=4(3)^2 =36
which combined is 4/36 or 1/9
(y2/y1)=(x2/x1)^2
1/9=(1/3)^2
you were right both are equal ways to finding 1/9, but the second way is a little faster
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Very good.
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