course Mth 271 10/05 7:40pm 008. `query 8*********************************************
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Given Solution: `aCORRECT STUDENT RESPONSE: f(z)=2^z and g(t)=3t-5, so that f(g(t)) = 2^g(t) = 2^(3t-5). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q1.3.81 (was 1.3.66 temperature conversion. What linear equation relates Celsius to Fahrenheit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y=mt + b or F=1.8C+32 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aCORRECT STUDENT RESPONSE: degrees Fahrenheit=1.8(degrees Celsius)+32, or F = 1.8 C + 32. INSTRUCTOR COMMENT: Since each Fahrenheit degree is 1.8 Celsius degrees a graph of F vs. C will have slope 1.8. Since F = 32 when C = 0 the graph will have vertical intercept at (0, 32) so the y = m x + b form will be F = 1.8 C + 32. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qHow did you use the boiling and freezing point temperatures to get your relationship? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since we are taking Celsius vs. Fahrenheit, we can say at 0 degrees Celsius, it is 32 degrees Fahrenheit which is the freezing point of water. Similarly, we can also say that the boiling point of water is 100 degrees Celsius or 212 degrees Fahrenheit. This can be plotted on a graph as (0, 32) and (100, 212). If you take the slope of the line between the two points, you get 180/100 or 1.8. You can take the y intercept of the freezing point and plug it in as b since it crosses the y axis at that point so your equation will be F=1.8C+32. Plug 0 in at C and all that remains is 32 so that has to be right. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a A graph of Fahrenheit vs. Celsius temperatures gives us the two (x,y) points (o,32) and (100,212). We use these two points to find the slope m=(y2-y1)/(x2-x1) = (212 - 32) / 100 = 1.8. Now we insert the coordinates of the point (0,32) and into the point-slope form y = 1.8 x + b of a line to get 32 = 1.8 * 0 + b. We easily solve to get b = 32. {So the equation is y = 1.8 x + 32, or F = 1.8 C + 32. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK "