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nikitadnepr [17]
3 years ago
13

Write a method doubleUp that doubles the size of a list of integers by doubling-up each element in the list. Assume there's enou

gh space in the array to double the size. Suppose a list stores the values [1, 3, 2, 7]. After calling list.doubleUp(), the list should store the values [1, 1, 3, 3, 2, 2, 7, 7].
Computers and Technology
1 answer:
madreJ [45]3 years ago
6 0

Answer:

def double_up(self, mylist):

   doubled = list()

   for item in mylist:

       for i in range(2):

           doubled.append(item)

   return doubled

Explanation:

The program method double_up gets the list argument, iterates over the list and double of each item is appended to a new list called doubled.

Methods are functions defined in classes. A class is a blueprint of a data structure. Each instance of the same class holds the same attributes and features.

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A continuous and differentiable function f(x) with the following properties: f(x) is decreasing at x=−5 f(x) has a local minimum
butalik [34]

The continuous and differentiable function where f(x) is decreasing at x = −5 f(x) has a local minimum at x = −2 f(x) has a local maximum at x = 2 is given as: y = 9x - (1/3)x³ + 3.

<h3>What is a continuous and differentiable function?</h3>

The continuous function differs from the differentiable function in that the curve obtained is a single unbroken curve in the continuous function.

In contrast, if a function has a derivative, it is said to be differentiable.

<h3>What is the solution to the problem above?</h3>

It is important to note that a function is differentiable when x is set to a if the function is continuous when x = a.

Given the parameters, we state that

f'(5) < 0; and

x = -5

The local minimum is given as:
x = -3;

the local maximum is given as

x = 3

Thus, x = -3 ; alternatively,

x = 3.  With this scenario, we can equate both to zero.

Hence,

x + 3 = 0;

3-x = 0.

To get y' we must multiply both equations to get:

y' = (3-x)(x + 3)

y'   = 3x + 9 - x² - 3x

Collect like terms to derive:

y' = 3x - 3x + 9 - x²; thus

y' = 9-x²

When y' is integrated, the result is

y = 9x - (x³/3) + c

Recall that

F (-5) < 0

This means that:

9 x -5 - (-5³/3) + c < 0
⇒ -45 + 125/3 + c <0
⇒ -10/3 + c < 0

Collecting like terms we have:
c < 10/3; and

c < 3.33


Substituting C into

f(x) = 9x - x³/3 + c; we have

f(x) = 9x - x³/3 + 3, which is the same as  y = 9x - (1/3)x³ + 3.

Learn more about differentiable functions at:
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#SPJ1

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