A person goes from a sauna at 115 degrees to an outside temp of -15 degrees. What is the change in temperature?
1 answer:
You might be interested in
Question #1
Part A:
The y-intercept can be found when x = 0. If you look at your table, when x = 0, y = 5. So the y-intercept is 5.
Part B:
The slope is 22.
Part C:
y = mx + b
y = 22x + 5
We are given 225 as the range, or in place of y.
225 = 22x + 5
220 = 22x
x = 10
The domain is 10.
Question #2
Part A:
(2,255)
(5,480)
Standard form is Ax + By = C
Let's plug this into this form first:
Now, let's make it into Standard Form.
What, which is in the box, is your final answer. :)
Part B:
Function notation simply means replacing y with f(x).
We had y = 85x + 55
So your answer is:
Part C:
Using the final answer which we got in Part A, we would know that the y-intercept is (0,55) and the x-intercept is (-55/85, 0). We would plot these 2 points, and then draw a line between them. :)
Part A:
The y-intercept can be found when x = 0. If you look at your table, when x = 0, y = 5. So the y-intercept is 5.
Part B:
The slope is 22.
Part C:
y = mx + b
y = 22x + 5
We are given 225 as the range, or in place of y.
225 = 22x + 5
220 = 22x
x = 10
The domain is 10.
Question #2
Part A:
(2,255)
(5,480)
Standard form is Ax + By = C
Let's plug this into this form first:
Now, let's make it into Standard Form.
What, which is in the box, is your final answer. :)
Part B:
Function notation simply means replacing y with f(x).
We had y = 85x + 55
So your answer is:
Part C:
Using the final answer which we got in Part A, we would know that the y-intercept is (0,55) and the x-intercept is (-55/85, 0). We would plot these 2 points, and then draw a line between them. :)
(a) See the first attachment for a graph. This graphing calculator displays roots to 3 decimal places. (The third attachment shows a different graphing calculator and 10 significant digits.)
(b) In the table of the first attachment, the column headed by g(x) gives iterations of Newton's Method. (For Newton's method, it is convenient to let the calculator's derivative function compute the derivative f'(x) of the function f(x). We have defined g(x) = x - f(x)/f'(x).) The result of the 3rd iteration is ...
... x ≈ 3.0473167
(c) The function h(x₁, x₂) computes iterations using the secant method. The results for three iterations of that method are shown below the table in the attachment. The result of the 3rd iteration is ...
... x ≈ 3.2291234
(d) The function h(x, x+0.01) computes the modified secant method as required by the problem statement. The result of the 3rd iteration is ...
... x ≈ 3.0477377
(e) Using <em>Mathematica</em>, the roots are found to be as shown in the second attachment. The highest root is about ...
... x ≈ 3.0466805180
_____
<em>Comment on these methods</em>
Newton's method can have convergence problems if the starting point is not sufficiently close to the root. A graphing calculator that gives a 3-digit approximation (or better) can help avoid this issue. For the calculator used here, the output of "g(x)" is computed even as the input is typed, so one can simply copy the function output to the input to get a 12-significant digit approximation of the root as fast as you can type it.
The "modified" secant method is a variation of the secant method that does not require two values of the function to start with. Instead, it uses a value of x that is "close" to the one given. For our purpose here, we can use the same h(x1, x2) for both methods, with a different x2 for the modified method.
We have defined h(x1, x2) = x1 - f(x1)(f(x1)-f(x2))/(x1 -x2).
