course Mth 271
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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): Self-critique Rating:ok ********************************************* Question: `q1.3.81 (was 1.3.66 temperature conversion. What linear equation relates Celsius to Fahrenheit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F= 1.8(Celsius) +32
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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): Self-critique Rating: ok ********************************************* Question: `qHow did you use the boiling and freezing point temperatures to get your relationship? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Freezing for Celsius is 0 degrees F= 1.8(0) +32 F= 32 degrees is the freezing temperature for Far. (0,32) Boiling point for Celsius is 100 degrees F= 1.8(100) +32 F= 212 degrees is the boiling temperature for Far. (100, 212)
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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): Self-critique Rating:ok "