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castortr0y [4]
2 years ago
10

Donte needs to print specific sections in the workbook, as opposed to the entire workbook. Which feature should he use? Page Set

up Backstage view Print Area Option.
Computers and Technology
1 answer:
siniylev [52]2 years ago
3 0

Donte needs to use the feature Print Area to print specific sections in the workbook.

Before workbooks are being printed out, they are usually being typed with Microsoft Office packages such as:

  • Microsoft word
  • Microsoft Excel.
  • Microsoft Powerpoint

In the Microsoft Office Package, to print a page, you can hover your cursor to the select file button and click on print pages.

But when you want to print some selected sections in the Microsoft Office Package, let's say the Microsoft Excel for example;

  • After using the file button → print pages

You will need to set the print features you need. To do that, under the settings, you will click on the Print selection which will help you to select the sections you want to print from the entire workbook.

Learn more about the use of Microsoft Office Packages here:

brainly.com/question/25203116?referrer=searchResults

You might be interested in
2. (8 points) When creating the Academic Database, there were several instances of data
Bad White [126]

Answer:

See explaination

Explanation:

An Academic Database deals with the information pertaining to the records of students of an institute. The various fields which can be associated with a student are Name, Unique Identification Number, Marks in various subjects, Grades, Courses Taken and so on. Most of the fields mentioned above have some or other form of validation that is required for the DB to be consistent and follow the basic properties of a database.

One field where range validation can be used is MARKS. In the marks field, the range of marks will be say 0 to 100 and anything below 0 or above 100 must be reported to the database administrator. Another field where we can apply a range validation is AGE in which the range of age allowed could greater than say 10, if it is a college and maximum age could be say 60. Thus, the range check on the AGE field is 10 to 60.

One field where choice validation can be used is GENDER. In the gender field, there could be multiple choices like Male, Female and Others. Thus, out of the available choices we have to select only the available choice and only one choice must be selected where we can apply choice validation. Another field is COURSE, where we can have a list of courses a student can opt for and out of the available courses can select one.

Thus, there are multiple fields where we can apply various types of validation and it is important to explore the area for which the DB has been created understanding all the scenarios and attributes of the problems that are associated with that area.

8 0
4 years ago
The groups_per_user function receives a dictionary, which contains group names with the list of users. Users can belong to multi
Akimi4 [234]

The groups_per_user function receives a dictionary, which contains group names with the list of users.

Explanation:

The blanks to return a dictionary with the users as keys and a list of their groups as values is shown below :

def groups_per_user(group_dictionary):

   user_groups = {}

   # Go through group_dictionary

   for group,users in group_dictionary.items():

       # Now go through the users in the group

       for user in users:

       # Now add the group to the the list of

         # groups for this user, creating the entry

         # in the dictionary if necessary

         user_groups[user] = user_groups.get(user,[]) + [group]

   return(user_groups)

print(groups_per_user({"local": ["admin", "userA"],

       "public":  ["admin", "userB"],

       "administrator": ["admin"] }))

3 0
3 years ago
Answer the question ASAP please
RUDIKE [14]

Answer:

Shows the programming checking if num1 is greater than num2

Explanation:

So num1 and num2 are inputs

for you to code this you would need to put

num1=int(input("What is your first number? ))

and the same for num2 except change num1 for num 2 and first for second

When the input is completed, the computer will check if num 1 is greater than num2

it will do this by using a code something like:

if num1>num2:

    Print("Your first input was greater than your second")

But in this example if it greater it just ends

But if it was less than you would put

if num1>num2:

    Print("Your first input was greater than your second")

elif num1<num2:

    Print("Your first input is less than your second")

So basically this code shows the computer checking if one number is greater than the other or not

3 0
3 years ago
Read 2 more answers
Can anyone help? Please.
Shtirlitz [24]
I can’t see it... can you type it out please?
3 0
3 years ago
Read 2 more answers
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

3 0
2 years ago
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