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Vilka [71]
3 years ago
14

You have activated pop-up blockers in your web browser to prevent pop-up windows from continually interrupting your browsing exp

erience. what could be an unintended consequence of this action?
Computers and Technology
2 answers:
mylen [45]3 years ago
6 0
Content Creators not gaining any money because they are paid by advertised content or sponsorship deals!
bazaltina [42]3 years ago
5 0
Content Creators not gaining any money as they are paid by advertised content or sponsored.
You might be interested in
What is the definition of overflow in binary?
asambeis [7]
Overflow occurs when the magnitude of a number exceeds the range allowed by the size of the bit field. The sum of two identically-signed numbers may very well exceed the range of the bit field of those two numbers, and so in this case overflow is a possibility.
4 0
3 years ago
1. Scrieţi un program care citeşte un număr natural n şi determină produsul cifrelor impare ale lui n. De exemplu, pentru n = 23
saveliy_v [14]

Answer:

1.  

num1 = input("Enter the value of n")  

n = int(num1)  

def calc(n):  

Lst = []  

while n > 0:  

remainder = n % 10  

Lst.append(remainder)  

quotient = int(n / 10)  

n = quotient  

i = len(Lst) - 1  

sum = 1  

for i in range(0, len(Lst)):  

if(i % 2 != 0):  

sum *=Lst[i];  

else:  

continue  

print(sum)  

return(0)  

\ r2

num=input("Enter the value of n")

arr=[12,-12,13,15,-34,-35,35,42]

def div7positive():

  sum = 0

  m = 0

  i = 0

  while m <= 7:

      if(arr[i]>=0 and arr[i]%7 == 0):

          sum +=arr[i]

      i = i + 1

      m += 1

       

   

  print("sum of number divisible by 7 and positive is", +sum);

div7positive()

\ r

\ r3.Write an algorithm that reads a natural number n and calculates the sum:

\ rS = 1 / (1 * 2) + 1 / (2 * 3) + 1 / (3 * 4) +… + 1 / ((n-1) * n)

\ r

num=input("Enter the value of n")

def sumseries():

  sum = 0

  m= 0

  while m <= 7:

      sum +=1/((int(num)-1)*int(num))

      m += 1

  print("sum of series is", +sum)

sumseries()

\ R4. Read a natural number n. Calculate the sum of its own divisors n. For example, for n = 12, the sum of its own divisors is 2 + 3 + 4 + 6 = 15

num=input("Enter the value of n")

def sumdivisors():

  sum = 0

  m= 2

  while m <= int(num):

      if int(num) % m == 0:

          sum += m

      m += 1

  print("sum of divisors is", +sum)

sumdivisors()\ r

\ R5. We read a natural number n and then whole numbers. Calculate and display the sum of the natural numbers between 10 and 100. For example, if n = 5 and then read 30, –2, 14, 200, 122, then the sum will be 44 (that is, 30 + 14).

\ r

Lst = []  

num=input("Enter the value of n")

i = 0

while i<= int(num):

  num1=input("Enter the element of array")

  Lst.append(int(num1));

  i += 1

def numbet0and100():

  sum = 0

  m= 0

  while m <= int(num):

      if Lst[m] <= 100 or Lst[m] >=0:

          sum += Lst[m]

      m += 1

  print("sum of numbers between 0 and 100 is", +sum)

numbet0and100()

\ R6. A natural number n of maximum 4 digits is read. How many digits are in all numbers from 1 to n? For example, for n = 14 there are 19 digits, and for n = 9 there are 9 digits.

\ r

num = input("Enter the value of n")

n = int(num)

print(“sum as required is:” (n -9) + n)  

 

\ R7. Read the natural numbers n and S, where n can be 2, 3, 4 or 5. Show all the numbers of n digits that have the numbers in strictly ascending order, and the sum of the digits is S. For example, for n = 2 and S = 10, 19, 28, 37, 46 will be displayed.

num = input("Enter the value of n")

n = int(num)

Lst = []

def calc(n):

  i = 1

  total = pow(10,n)

  while i <= total:

      j = i

      while j > 0:

          remainder = j % 10  

          Lst.append(remainder)

          quotient = int(j / 10)

          if quotient > 0:

              j = quotient

          else:

              length = len(Lst) - 1

              sum = 0

              while length >= 0:

                  sum += Lst[length]

                  length = length - 1

              if sum == pow(10,n):

                  print(j)

              k = len(Lst)

              del Lst[0:k]

      i = i + 1

  return(0)

calc(n)

\ r

\ R8. We consider the row 1, 1, 2, 3, 5, 8, 13, ... in which the first two terms are 1, and any other term is obtained from the sum of the preceding two. Write an algorithm that reads a natural number n and displays the first n terms of this string. For example, for n = 6, 1, 1, 2, 3, 5, 8 will be displayed.

\ r

nterm = int(input("What number of terms do you want?"))

a, b = 0, 1

totalcount = 0

if nterm <= 0:

 print("Please input a positive number")

elif nterm == 1:

 print("Fibonacci number upto which",nterm,":")

 print(a)

else:

 print("Fibonacci series:")

 print(nterm)

 while totalcount < nterm:

     print(a)

     nth = a + b

     a = b

     b = nth

     totalcount += 1

 

\ 9. Write an algorithm that reads two natural numbers n1 and n2 and displays the message "yes" if the sum of the squares of the digits of n1 is equal to the sum of the numbers of n2 or "no" otherwise. For example, for n1 = 232 and n2 = 881, "yes" will be displayed, and for n1 = 45 and n2 = 12, "no" will be displayed.

num1 = input("Enter the value of n")

num3 = input("Enter the value of n")

n = int(num1)

num2 = int(num3)

def calc(n):

  Lst = []

  while n > 0:

      remainder = n % 10  

      Lst.append(remainder)

      quotient = int(n / 10)

      n = quotient

       

  i = len(Lst) - 1

  sum = 0

  while i >= 0:

      sum += Lst[i]* Lst[i]

      i = i - 1

  return(sum)

def calc1(num2):

  Lst = []

  while num2 > 0:

      remainder = num2 % 10  

      Lst.append(remainder)

      quotient = int(num2 / 10)

      num2 = quotient

       

  i = len(Lst) - 1

  sum = 0

  while i >= 0:

      sum = Lst[i] + Lst[i]

      i = i - 1

  return(sum)

a = calc(n)

b = calc1(num2)

if a == b:

  print("yes")

else:

  print("No")

\ r

\ R10. Write an algorithm that reads a natural number n and displays the message "yes" if all of its n numbers are distinct, or "no" if n does not have all the distinct digits. For example, for n = 37645 it will display "yes" and for 23414 it will show "no".

\ r

num1 = input("Enter the value of n")

n = int(num1)

def calc(n):

  Lst = []

  while n > 0:

      remainder = n % 10  

      Lst.append(remainder)

      print( remainder)

      quotient = int(n / 10)

      n = quotient

       

  flag = 1

  for i in range(0, len(Lst)):

      for j in range(i+1, len(Lst)):    

          if Lst[i] == Lst[j]:    

              flag = 0

              break

           

  if flag == 1:

      print("YES")

  else:

      print("NO")

  return(0)

Explanation:

Please check answer.  

3 0
2 years ago
If you have a list consisting of just numbers, then you can add all of the values in the list using the sum() function.
Vlada [557]

Answer:

See explaination

Explanation:

def sum_lengths(value_list):

total = 0

for x in value_list:

if (type(x)==int) or (type(x)==float):

total += x

return total

6 0
3 years ago
Given positive integer numinsects, write a while loop that prints that number doubled without reaching 100. follow each number w
ELEN [110]
Please provide the language you're using when you ask for programming help, otherwise you aren't going to get the answer that you are looking for.

Here it is in Java, and I'm assuming the number is given via user input? Otherwise, just remove the user input function and replace the integer with a value of your choice. Note, that this isn't the full code; only what is relevant to the question.


public static void main(String[] args) {
    int num = numInput(10);
    printDoubles(num, 100);  // You can create a user input function for
                                             // maxValue if you wanted to.
}

/**
 * Receives user input between 0 and the absolute value of maxInput.
 * @param maxInput The largest absolute value that can be input.
 */
private static int numInput(int maxInput) {
    Scanner sc = new Scanner(System.in);

    maxInput = Math.abs(maxInput);

    int num = 0;
    while (!(num > 0 && num <= maxInput)) {
        num = sc.nextInt();

        if (!(num > 0 && num <= maxInput)) {
            System.out.println("Input too small or too large");
        }
    }

    return num;
}

/**
 * Continues to print out num doubled until maxValue is reached.
 * @param num The number to be printed.
 * @param maxValue The maximum value (not including in which num can                                       be doubled to.
 */
private static void printDoubles(int num, int maxValue) {
    if (num >= maxValue) {
        System.out.println("No output.");
    }

    while (true) {
        if (num >= maxValue) {
            break;
        }

        if (num < maxValue) {
            System.out.print(num + " ");
        }

            num *= 2;
    }

    System.out.println();
}

8 0
3 years ago
Read 2 more answers
Knowing how to use your learning style to study is not important.
qaws [65]

False. It helps u learn better if u use ur learning style to study

3 0
3 years ago
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