Q8

#$&*

course MTH271

8-5-10

008. `query 8

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

Question: `qWhat functions f(z) and g(t) express the function 2^(3t-5) as a composite f(g(t))?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

f(z)=2^z and g(t)=3t-5

f(g(t)) = 2^g(t) = 2^(3t-5)

confidence rating #$&*

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

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

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):okay

------------------------------------------------

Self-critique rating #$&*3

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

Question: `q1.3.81 (was 1.3.66 temperature conversion. What linear equation relates Celsius to Fahrenheit?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Degrees Fahrenheit=1.8(degrees Celsius)+32, or F = 1.8 C + 32

confidence rating #$&*

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

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

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):okay

------------------------------------------------

Self-critique rating #$&*3

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

Question: `qHow did you use the boiling and freezing point temperatures to get your relationship?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

M = (y2-y1)/(x2-x1) = (212 - 32) / 100 = 1.8.

y = 1.8 x + b

32 = 1.8 * 0 + b.

b = 32.

y = 1.8 x + 32, or F = 1.8 C + 32.

confidence rating #$&*

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

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

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):okay

------------------------------------------------

Self-critique rating #$&*3

"

&#Good work. Let me know if you have questions. &#

#$&*