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3 years ago

A. Is this mapping diagram a function? Why or why not?

Mathematics

1 answer:

nikitadnepr [17]3 years ago

5 0

Answer:

40--C A ---is not[figure it out in your own words,your teacher might suspect something] and B---times 40

Step-by-step explanation:

b--40+40=80 80+40=120 and 120+40=160 A=subtract 40 from all the other answers such as 40-160=120 40-120=80 and 40-80=40

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3+ jk + k*3 when j=2 and k=6

Licemer1 [7]

The answer is 39, jk=18 because j multiplied by k is 18 and then i added 3 and i got 21, then k (6) multiplied by 3 is 18, so then i added 21 and 18 and i got 39

5 0

3 years ago

Read 2 more answers

If y=e5t is a solution to the differential equation

a_sh-v [17]

Answer:

k = 30, $y(t) = C_1e^{5t}+C_2e^{6t}$

Step-by-step explanation:

Since $y=e^{5t}$ is a solution, then it must satisfy the differential equation. So, we calculate the derivatives and replace the value in the equation. We have that

$\frac{d^2y}{dt^2} = 25 e^{5t},\frac{dy}{dt} = 5e^{5t}$

Then, replacing the derivatives in the equation we have:

$25e^{5t}-11(5)e^{5t}+ke^{5t}=0 e^{5t}(25-55+k) =0$

Since $e^{5t}$ is a positive function, we have that

$25-55+k = 0 \rightarrow k = 30$ .

Now, consider a general solution $y(t) = Ae^{rt}, A \in \mathbb{R}$ , then, by calculating the derivatives and replacing them in the equation, we get

$Ae^{rt}(r^2-11r+30)=0$

We already know that r=5 is a solution of the equation, then we can divide the polynomial by the factor (r-5) to the get the other solution. If we do so, we get that (r-6)=0. So the other solution is r=6.

Therefore, the general solution is

$y(t) = C_1e^{5t}+C_2e^{6t}$

8 0

3 years ago

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).