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alukav5142 [94]
3 years ago
15

Write a C program that includes a function of type double called divemaster accepts two double arguments (you must write the div

emaster function). When called, your function will return the first argument divided by the second argument. Make sure your function includes a decision statement so that it can't divide by 0--if the caller attempts to divide by 0 the function should return a value of -1.0.
Computers and Technology
1 answer:
Firlakuza [10]3 years ago
6 0

Answer:

Following is the C program to divide two numbers:

#include<stdio.h>

double divemaster (double first, double second)

{

//Check if second argument is zero or not

if(second!=0)

    return first/second;

else

 return -1.0;

}

int main()

{

printf("%f\n", divemaster(100,20));

printf("%f\n", divemaster(100,0));

}

Output:

5.000000

-1.000000

Explanation:

In the above program, a function divemaster is declared which divedes the first argument with second and return the result.

Within the body of divemaster function if else block is used to verify that the  second argument is not zero, and thus there is no chance of divide by zero error.

In main two printf statements are used to call the divemaster function. One to check normal division process and second to check divide by zero condition.  

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Answer:

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Explanation:

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Putting the values of t we can get the range of varaiable resistor as;

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Putting t=2 we get the first value of the range for the variable resistor

2=R*0.500*10^-6

R=2/(0.500*10^-6)

R=4*10^6

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Now putting t=5 we get the final value for the range of variable resistor

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5=R*0.500*10^-6

R=5/(0.500*10^-6)

R=10*10^6

R=10,000k-ohm

So variable resistance must be made to vary in the range from 4000k-ohm to 10,000k-ohm

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An anagram of a string is another string with the same characters in the same frequency, in any order. For example 'ab', 'bca, a
nlexa [21]

Answer:

from collections import Counter

def anagram(dictionary, query):

   newList =[]

   for element in dictionary:          

       for item in query:

           word = 0

           count = 0

           for i in [x for x in item]:

               if i in element:

                   count += 1

                   if count == len(item):

                       newList.append(item)

   ans = list()

   for point in Counter(newList).items():

       ans.append(point)

   print(ans)        

mylist = ['jack', 'run', 'contain', 'reserve','hack','mack', 'cantoneese', 'nurse']

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Explanation:

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3 years ago
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