For my depth vs. time model I was asked the question: On the average how many seconds did it take for the depth to change by one cm. between t=23sec. and t=34sec., I came up with the answer -1/2. Then I was asked, On the average by how many cm. did the depth change per second between t=23sec. and t=34sec., would the answer simply be -2cm.?
You don't include the function so I can't check your accuracy, and you don't say how you got your results so I can't check your procedure.
However your two answers are consistent, and I suspect that you are right.
If the answer to the first question is -1/2 second, then the answer to the second is -2 cm.
Look at what you got for y when t = 23 and when t = 34. If your values that differ by -22 cm from t = 23 sec to t = 34 sec, then the average rate of change of depth with respect to clock time will be -22 cm / (11 sec) = -2 cm/s, and depth will change by an average of -2 cm every second.
How do you determine the average deviation for a model you made after choosing the points, (20,68);(40,49);(60,21)?
See also Completion of the Initial Flow Model under the heading Assessing the Function Model.
Briefly, assuming your data points are given for t = 10, 20, ..., 80, you take the y = a t^2 + b t + c model and do the following:
Evaluate your function at each of these t values.
Subtract the result you get from each of the given y values, and take the absolute value of each. These are the deviations of your model from the data.
Add your deviations and divide by the number of data points to get the average deviation.
I don't understand the following question: Using x=-5,-2,-1,-1/2,
-1/10,1/10,1/2,1,2,5 for positive powers and excluding x=0 for the negative powers make tables and sketch graphs for the basic power functions y=x^2, y=x^3, y=x^-2, y=x^-3. Are there parts of the questions missing or am I reading to much into it?
That question could be better stated. It's saying to use x values -5,-2,-1,-1/2,-1/10,1/10,1/2,1,2,5 to make tables for each of the functions y=x^2, y=x^3, y=x^-2, y=x^-3. You have four functions, you will have four tables, and you will get four graphs. For y = x^2, the table would be
x y
-5 25
-2 4
-1 1
-1/2 1/4
0 0
1/2 1/4
1 1
2 4
5 25
and the graph would be a parabola.
When substituting 3f(x)into f(x)=2x+3 do you get an answer of 3(2x+3)which equals 6x+9, and how do you find f(3x) for the same function?
Since
f(x) = 2x + 3 it follows that
3 f(x) = 3 ( 2x + 3) = 6x + 9
so your are correct.
To find f(3x), since
f(x) = 2 x + 3 it follows that
f(3x) = 2 ( 3x) + 3 = 9x + 3.
Can you simplify 3(2^x)more than it already is after you substitute into f(x)+2^x?
The expression 3 * 2^x cannot be simplified further.
2 is raised to the x power, and 3 is not, so there is no way to combine these numbers to simplify the expression.