open qa 12

#$&*

course MTH 163

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

012.

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Question: `q001. Note that this assignment has 3 questions

If we know that y = k x^2, then if (x2/x1) = 7, what is (y2/y1)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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Given Solution:

If y2 = k x2^2 and y1 = k x1^2, then y2 / y1 = (k x2^2) / ( k x1^2). Since k / k = 1 this is the same as

y2 / y1 = x2^2 / x1^2, which is the same as

y2 / y1 = (x2 / x1)^2.

In words this tells us if y to is proportional to the square of x, then the ratio of y2 to y1 is the same as the square of the ratio of x2 to x1.

Now if (x2 / x1) = 7, we see that y2 / y1 = (x2 / x1)^2 = 7^2 = 49.

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Self-critique (if necessary):

I am totally and completely lost on this one. I don’t understand how to get the first problem broken down or anything. I don’t understand all of the extra 1s and 2s and such.

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Self-critique rating:1

@& x1 and x2 just stand for two different values of x.

y1 and y2 stand for the corresponding values of y.

Written out x1, for example, would be written as x with a subscript of 1.*@

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Question: `q002. If we know that y = k x^3, then if (x2/x1) = 7, what is (y2/y1)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

If y2 = k x2^3 and y1 = k x1^3, then y2 / y1 = (k x2^3) / ( k x1^3). Since k / k = 1 this is the same as

y2 / y1 = x2^3 / x1^3, which is the same as

y2 / y1 = (x2 / x1)^3.

In words this tells us if y to is proportional to the cube of x, then the ratio of y2 to y1 is the same as the cube of the ratio of x2 to x1.

Now if (x2 / x1) = 7, we see that y2 / y1 = (x2 / x1)^3 = 7^3 = 343.

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Self-critique (if necessary):

Again, I’m very confused. I understand that k is a constant., but again I don’t understand the random 1s and 2s.

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Self-critique rating:

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Question: `q003. If we know that y = k x^-2, then if (x2/x1) = 64, what is (y2/y1)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

If y2 = k x2^-2 and y1 = k x1^-2, then y2 / y1 = (k x2^-2) / ( k x1^-2). Since k / k = 1 this is the same as

y2 / y1 = x2^-2 / x1^-2, which is the same as

y2 / y1 = (x2 / x1)^-2, which is the same as

1 / (x2 / x1)^2, which gives us

(x1 / x2)^2.

So if y = k x^-2, then (y2 / y1) = (x1 / x2)^2.(

In words this tells us if y to is inversely proportional to the square of x, then the ratio of y2 to y1 is the same as the square of the ratio of x1 to x2 (note that this is a reciprocal ratio).

Now if (x2 / x1) = 64, we see that y2 / y1 = (x1 / x2)^2 = (1/64)^2 = 1/ 4096.

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Self-critique (if necessary):

Its definitely going to take me a while to get this. I just don’t know what to do at all.

Self-critique

"

Self-critique (if necessary):

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Self-critique rating:

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Question: `q003. If we know that y = k x^-2, then if (x2/x1) = 64, what is (y2/y1)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

If y2 = k x2^-2 and y1 = k x1^-2, then y2 / y1 = (k x2^-2) / ( k x1^-2). Since k / k = 1 this is the same as

y2 / y1 = x2^-2 / x1^-2, which is the same as

y2 / y1 = (x2 / x1)^-2, which is the same as

1 / (x2 / x1)^2, which gives us

(x1 / x2)^2.

So if y = k x^-2, then (y2 / y1) = (x1 / x2)^2.(

In words this tells us if y to is inversely proportional to the square of x, then the ratio of y2 to y1 is the same as the square of the ratio of x1 to x2 (note that this is a reciprocal ratio).

Now if (x2 / x1) = 64, we see that y2 / y1 = (x1 / x2)^2 = (1/64)^2 = 1/ 4096.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

Its definitely going to take me a while to get this. I just don’t know what to do at all.

Self-critique

"

Self-critique (if necessary):

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Self-critique rating:

#*&!

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Question: `q003. If we know that y = k x^-2, then if (x2/x1) = 64, what is (y2/y1)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

If y2 = k x2^-2 and y1 = k x1^-2, then y2 / y1 = (k x2^-2) / ( k x1^-2). Since k / k = 1 this is the same as

y2 / y1 = x2^-2 / x1^-2, which is the same as

y2 / y1 = (x2 / x1)^-2, which is the same as

1 / (x2 / x1)^2, which gives us

(x1 / x2)^2.

So if y = k x^-2, then (y2 / y1) = (x1 / x2)^2.(

In words this tells us if y to is inversely proportional to the square of x, then the ratio of y2 to y1 is the same as the square of the ratio of x1 to x2 (note that this is a reciprocal ratio).

Now if (x2 / x1) = 64, we see that y2 / y1 = (x1 / x2)^2 = (1/64)^2 = 1/ 4096.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

Its definitely going to take me a while to get this. I just don’t know what to do at all.

@& x1 and x2 stand for two different values of x.

y1 and y2 are the corresponding values of y.

So y1 = k x1^-2, and y2 = k x2^-2.*@

Self-critique

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Self-critique (if necessary):

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Self-critique rating:

#*&!#*&!

@& Check my notes and let me know if they help. This really isn't too difficult to understand once you understand the notation.*@