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goldenfox [79]
3 years ago
8

A linear relationship has a constant rate of change. agree? or disagree? and why?

Mathematics
2 answers:
mestny [16]3 years ago
7 0
Yes, a linear relationship (mathematical) is a constant change happening to x, for example, if in the first step it adds 2 to the table chart or graph, it will always be adding 2 every step of the equason
slega [8]3 years ago
5 0
I agree based on the question
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The estimated probability of a football player scoring a touchdown during a particular game is 26%. If several simulations of th
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I believe the answer is C
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3 years ago
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3 determine the highest real root of f (x) = x3− 6x2 + 11x − 6.1: (a) graphically. (b) using the newton-raphson method (three it
Juliette [100K]

(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).

6 0
2 years ago
PLEASE HELP FAST!!!
grigory [225]

Answer:

x = 9

Step-by-step explanation:

Since the figures are similar then the ratios of corresponding sides are equal, that is

\frac{x-1}{6} = \frac{16}{12} ( cross- multiply )

12(x - 1) = 96 ( divide both sides by 12 )

x - 1 - 8 ( add 1 to both sides )

x = 9

7 0
3 years ago
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{6 3/6 + 10 4/6) + 9 2/6
n200080 [17]

Answer:

26.5

Step-by-step explanation:

used calculator

4 0
3 years ago
Which is bigger 0.21, 21%
andrew11 [14]
21% is bigger than 0.21
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