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kobusy [5.1K]
3 years ago
9

Lucio quiere repartir su colección de estampillas entre la mayor cantidad de amigos pero de manera que todos reciban la misma ca

ntidad tiene parar repartir 30 estampillas de animales 75 estampillas de flores y 160 de ciudades del mundo cuantos grupos iguales puede armar
Mathematics
1 answer:
Daniel [21]3 years ago
4 0

Answer:

Lucio podrá armar 5 grupos iguales de estampillas.

Step-by-step explanation:

El problema plantea un caso de distribución mediante la aplicación del divisor común mayor a todos los números involucrados. Para ello, primero debemos analizar los divisores de cada uno de estos números:

-30: divisible por 2, 3, 5, 6, 10, 15 y 30.

-75: divisible por 3, 5, 15, 25 y 75.

-160: divisible por 2, 4, 5, 8, 10, 16, 20, 32, 40, 80 y 160.

Así, podemos ver que el divisor común mayor a los tres números es 5, ya que 30, 75 y 160 dan como resultado un número entero tras ser divididos por 5.

Entonces, para poder repartir equitativamente sus estampillas entre la mayor cantidad de amigos posibles, deberá repartir 6 estampillas de animales, 15 estampillas de flores y 32 estampillas de ciudades a 5 amigos. De esta manera, Lucio podrá armar 5 grupos iguales de estampillas.

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5 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

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<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 multiplies to 378 and adds to -51
Oliga [24]
Your answer would be -9 & -42. 

Here's why ...
(-9)*(-42) A negative times a negative equals a positive (378).
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Hope this makes sense & helps you ! (:


6 0
3 years ago
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