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posledela
3 years ago
10

What item on a business card is generally the most prominent?

Computers and Technology
2 answers:
andrew11 [14]3 years ago
4 0
Company name is usually the most prominent
AveGali [126]3 years ago
3 0
The most prominent
Company name that's most important
You might be interested in
You are given a list of n positive integers a1, a2, . . . an and a positive integer t. Use dynamic programming to design an algo
Anna11 [10]

Answer:

See explaination for the program code

Explanation:

The code below

Pseudo-code:

//each item ai is used at most once

isSubsetSum(A[],n,t)//takes array of items of size n, and sum t

{

boolean subset[n+1][t+1];//creating a boolean mtraix

for i=1 to n+1

subset[i][1] = true; //initially setting all first column values as true

for i = 2 to t+1

subset[1][i] = false; //initialy setting all first row values as false

for i=2 to n

{

for j=2 to t

{

if(j<A[i-1])

subset[i][j] = subset[i-1][j];

if (j >= A[i-1])

subset[i][j] = subset[i-1][j] ||

subset[i - 1][j-set[i-1]];

}

}

//returns true if there is a subset with given sum t

//other wise returns false

return subset[n][t];

}

Recurrence relation:

T(n) =T(n-1)+ t//here t is runtime of inner loop, and innner loop will run n times

T(1)=1

solving recurrence:

T(n)=T(n-1)+t

T(n)=T(n-2)+t+t

T(n)=T(n-2)+2t

T(n)=T(n-3)+3t

,,

,

T(n)=T(n-n-1)+(n-1)t

T(n)=T(1)+(n-1)t

T(n)=1+(n-1)t = O(nt)

//so complexity is :O(nt)//where n is number of element, t is given sum

6 0
3 years ago
For those that play pc games how do you go outside in The Sims 2?
yanalaym [24]

Answer:

My sister plays a lot of the sims 4 but I do recall she also played the Sims 2 so when she would go outside she would command them to do that.

Explanation:

If this does not work I would suggest you get a video that can explain it better than I possibly would? Considering the Sims 2 was made a while back you would maybe want somebody with a little more experience.

5 0
3 years ago
Read 2 more answers
Amye is in high school. As part of one of her classes, she had to identify the legality of different situations that involved
adell [148]
I say you should pick answer 1. an HR makes more sense since she is interested in a topic about the legality of different situations that are related to workers and their employers.
8 0
2 years ago
Read 2 more answers
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
Using emails for time-sensitive messages is an example of:
blagie [28]

Answer:time-sensitive email

5 0
3 years ago
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