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Travka [436]
3 years ago
15

Please help. A B C D

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
6 0

Answer:

A. t(x) = 450 - 0.06x

Step-by-step explanation:

The layer which is originally 450 m thick is decreasing by 0.06 m.

450 - 0.06

... per day. x represents days.

0.06x

So,

450 - 0.06x

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Read 2 more answers
In how many distinct ways can the letters of the word mathematics be arranged? (first, does the order matter?)
TEA [102]
Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).

Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
4 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).

6 0
2 years ago
What is the measure of X! Please help if you know how to do this!
deff fn [24]

Answer:

28°

Step-by-step explanation:

You're given that line DE and line FG are parallel and KL and FG are perpendicular. Then you can find out angle ∠BAC by using the vertical angles property: ∠BAC=62°. Then since KL and FG are perpendicular ∠ABC = 90°. So you find the angle ∠BCA by finding the sum of interior angles: 62+90+∠BCA=180, therefore ∠BCA is 28°. Finally, ∠x or ∠JCG = 28 because ∠JCG and ∠BCA are vertical angles and congruent.

4 0
3 years ago
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