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Sauron [17]
3 years ago
6

Nathan wants to create multiple worksheet containing common formatting styles for his team members. Which file extension helps h

im to save these worksheets?
Computers and Technology
2 answers:
Luden [163]3 years ago
5 0
Templates help Nathan to create multiple worksheets with common styles. He needs to save them with the xls extension.
Lorico [155]3 years ago
5 0
<h2>Answer:</h2>

<u>He should save it in </u><u>xls format </u>

<h2>Explanation:</h2>

xls format is related to Microsoft Excel files. Excel sheets are automatically saved under the extension xls. Excel sheets allow us to save the sheets having same format and different data. We can usually save two or more sheets with different data in only one file. These sheets may contain common formatting styles which can be taken into consideration for multiple team members.

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what could a company do if it was using a commercial wan and a vulnerability appered thst allowed attackers
Radda [10]

Answer:

The answer is below

Explanation :

If a company was using a commercial WAN and a vulnerability appeared that allowed attackers find routing information and therefore be able to eavesdrop on corporate transmissions "It is expected that such a company should try to formulate and establish Virtual Private Networks between corporate sites.

This will ultimately lead to intercepted transmissions to be unreadable. Though such means have been seen as time-consuming."

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2 years ago
Write a lottery program that will ask the user if they would like to pick 5 numbers (1-30) or if they would like to choose EZ Pi
Llana [10]

Answer:

Explanation:

The following code is written in Python and creates arrays for the user's numbers and the next chance numbers, it also creates a cost variable and a did_you_win variable. The function takes in the winning numbers as an array parameter. It asks all the necessary questions to generate data for all the variables and then compares the user's numbers to the winning numbers in order to detect if the user has won. Finally, it prints all the necessary information.

import random

def lottery(winning_numbers):

   user_numbers = []

   next_chance_drawing = []

   cost = 0

   did_you_win = "No"

   #print("Would you like to choose your own 5 numbers? Y or N")

   answer = input("Would you like to choose your own 5 numbers? Y or N: ")

   if answer.capitalize() == "Y":

       for x in range(5):

           user_numbers.append(input("Enter number " + str(x+1) + ": "))

   else:

       for x in range(5):

           user_numbers.append(random.randint(0,31))

           cost += 1

   next_chance = input("Would you like to enter a Next Chance drawing for $1.00? Y or N: ")

   if next_chance.capitalize() == "Y":

       for x in range(4):

           next_chance_drawing.append(random.randint(0, 31))

           cost += 1

           

   if user_numbers == winning_numbers or next_chance_drawing == winning_numbers:

       did_you_win = "Yes"

   print("User Numbers: " + str(user_numbers))

   print("Next Chance Numbers: " + str(next_chance_drawing))

   print("Cost: $" + str(cost))

   print("Did you win: " + did_you_win)

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2 years ago
Implement the function first chars() that takes a list of strings as a parameter and prints to the screen the first character of
Fiesta28 [93]
Def firstChars( word ):
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a cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits such that the giv
Leokris [45]

Using the knowledge in computational language in C++ it is possible to write a code that  cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits

<h3>Writting the code:</h3>

<em>#include <bits/stdc++.h></em>

<em>using namespace std;</em>

<em>// chracter to digit mapping, and the inverse</em>

<em>// (if you want better performance: use array instead of unordered_map)</em>

<em>unordered_map<char, int> c2i;</em>

<em>unordered_map<int, char> i2c;</em>

<em>int ans = 0;</em>

<em>// limit: length of result</em>

<em>int limit = 0;</em>

<em>// digit: index of digit in a word, widx: index of a word in word list, sum: summation of all word[digit]  </em>

<em>bool helper(vector<string>& words, string& result, int digit, int widx, int sum) { </em>

<em>    if (digit == limit) {</em>

<em>        ans += (sum == 0);</em>

<em>        return sum == 0;</em>

<em>    }</em>

<em>    // if summation at digit position complete, validate it with result[digit].</em>

<em>    if (widx == words.size()) {</em>

<em>        if (c2i.count(result[digit]) == 0 && i2c.count(sum%10) == 0) {</em>

<em>            if (sum%10 == 0 && digit+1 == limit) // Avoid leading zero in result</em>

<em>                return false;</em>

<em>            c2i[result[digit]] = sum % 10;</em>

<em>            i2c[sum%10] = result[digit];</em>

<em>            bool tmp = helper(words, result, digit+1, 0, sum/10);</em>

<em>            c2i.erase(result[digit]);</em>

<em>            i2c.erase(sum%10);</em>

<em>            ans += tmp;</em>

<em>            return tmp;</em>

<em>        } else if (c2i.count(result[digit]) && c2i[result[digit]] == sum % 10){</em>

<em>            if (digit + 1 == limit && 0 == c2i[result[digit]]) {</em>

<em>                return false;</em>

<em>            }</em>

<em>            return helper(words, result, digit+1, 0, sum/10);</em>

<em>        } else {</em>

<em>            return false;</em>

<em>        }</em>

<em>    }</em>

<em>    // if word[widx] length less than digit, ignore and go to next word</em>

<em>    if (digit >= words[widx].length()) {</em>

<em>        return helper(words, result, digit, widx+1, sum);</em>

<em>    }</em>

<em>    // if word[widx][digit] already mapped to a value</em>

<em>    if (c2i.count(words[widx][digit])) {</em>

<em>        if (digit+1 == words[widx].length() && words[widx].length() > 1 && c2i[words[widx][digit]] == 0) </em>

<em>            return false;</em>

<em>        return helper(words, result, digit, widx+1, sum+c2i[words[widx][digit]]);</em>

<em>    }</em>

<em>    // if word[widx][digit] not mapped to a value yet</em>

<em>    for (int i = 0; i < 10; i++) {</em>

<em>        if (digit+1 == words[widx].length() && i == 0 && words[widx].length() > 1) continue;</em>

<em>        if (i2c.count(i)) continue;</em>

<em>        c2i[words[widx][digit]] = i;</em>

<em>        i2c[i] = words[widx][digit];</em>

<em>        bool tmp = helper(words, result, digit, widx+1, sum+i);</em>

<em>        c2i.erase(words[widx][digit]);</em>

<em>        i2c.erase(i);</em>

<em>    }</em>

<em>    return false;</em>

<em>}</em>

<em>void isSolvable(vector<string>& words, string result) {</em>

<em>    limit = result.length();</em>

<em>    for (auto &w: words) </em>

<em>        if (w.length() > limit) </em>

<em>            return;</em>

<em>    for (auto&w:words) </em>

<em>        reverse(w.begin(), w.end());</em>

<em>    reverse(result.begin(), result.end());</em>

<em>    int aa = helper(words, result, 0, 0, 0);</em>

<em>}</em>

<em />

<em>int main()</em>

<em>{</em>

<em>    ans = 0;</em>

<em>    vector<string> words={"GREEN" , "BLUE"} ;</em>

<em>    string result = "BLACK";</em>

<em>    isSolvable(words, result);</em>

<em>    cout << ans << "\n";</em>

<em>    return 0;</em>

<em>}</em>

See more about C++ code at brainly.com/question/19705654

#SPJ1

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2 years ago
A(n) ________ is a wonder of miniaturization combining a CPU, GPU, and sundry other support logic onto a single silicon die, sav
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Answer:

system on a chip (SoC)

Explanation:

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