course Math 163 012.*********************************************
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating #$&*: OK ********************************************* Question: `q002. If we know that y = k x^3, then if (x2/x1) = 7, what is (y2/y1)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = k x^3 (x2/x1) = 7 (y2/y1) = ? y2/y1 = (k x2^3)/(k x1^3) y2/y1 = x2^3/x1^3 y2/y1 = (x2/x1)^3 (x2/x1) = 7 y2/y1 = (x2/x1)^3 = 7^3 = 343 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*: OK ********************************************* Question: `q003. If we know that y = k x^-2, then if (x2/x1) = 64, what is (y2/y1)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = k x^-2 (x2/x1) = 64 (y2/y1) = ? y2/y1 = (k x2^-2)/(k x1^-2) y2/y1 = (x2^-2)/(x1^-1)this is the same as below, because it is a negative power so we have to use inverse proportions. 1/(x2/x1)^2 (x2/x1) = 64 (y2/y1) = 1/(x2/x1)^2 = 64^2 = 4096 confidence rating #$&*: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.