Assignment 12

course Mth 163

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012.

Precalculus I

03-02-2008

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18:14:12

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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RESPONSE -->

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.

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

self critique assessment: 3

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18:18:29

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

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RESPONSE -->

Like the previous question. If y2/y1 = kx2^3/kx1^3. K divided by k is equal to 1. so y2/y1 = (x2/x1)^3. This then gives us the equation y2/y1 = 7^3 which then equals 343.

confidence assessment: 2

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18:27:52

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

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RESPONSE -->

If y = kx^-2 and (x2/x1) = 64 then (y2/y1) is equal to kx2^-2/kx1^-2. Since k/k is equal to 1 we can eliminate. Now (y2/y1) = (x2/x1)^-2 thus y2/y1 = 64^-2 which is equal to 1/4096 or 2.44x10^-4

confidence assessment: 3

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18:28:04

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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RESPONSE -->

self critique assessment: 3

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Assignment 12

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