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natta225 [31]
3 years ago
8

Write the function lettersOnly(s) that takes in a string called s, and returns a string containing only the alphabetic character

s of s.
Computers and Technology
1 answer:
Phantasy [73]3 years ago
4 0

Answer:

The program to this question as follows:

Program:

def lettersOnly(s): #defining method lettersOnly

   val="" #defining string variable that return value

   for i in s: #defining loop to calculate value

       if(i.isalpha()): #check condition that value is string

           val=val+ i #add value

   return val #return value

print(lettersOnly("data3base_ro1c3k5s")) #call method and print value

Output:

databaserocks

Explanation:

In the above python code, a method lettersOnly is declared that accepts a string variable  "s" in its parameter. Inside the method, a string variable "val", and loop is declared, in which the "val" variable holds method return value.

  • In the loop and if block is used that uses "isalpha" string method, which checks the check alphabetic character in the given value. if this is true it will calculate all value in "val" variable and return its value.    
  • At the last, the print method is used, which calls the lettersOnly method and prints its return value.
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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
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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.

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