open qa 12

course MTH 163

10/31/09 10:42 p.m.

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

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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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Self-critique (if necessary):

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Self-critique Rating:

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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)?

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Your solution:

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

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

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

(y2/y1)=(7)^2

(y2/y1)= 49

confidence rating: 3

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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):

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Self-critique Rating:

*********************************************

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

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Your solution:

y2/y1=(K(X2)^3)/(k(X1)^3)

y2/y1=(X2/X1)^3

y2/y1=7^3

y2/y1= 343

confidence rating: 3

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

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

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):

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Self-critique Rating:

*********************************************

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

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Your solution:

Y2/Y1=(k(x2)^-2)/(k(x1)^-2)

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

y2/y1=64^-2

y2/y1=.000244

confidence rating: 3

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

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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, 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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&#Good work. See my notes and let me know if you have questions. &#