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zepelin [54]
3 years ago
6

Ruby is creating a presentation. She wants each slide displayed at intervals of five seconds. Which feature in the presentation

software should she use to produce this particular effect?
Computers and Technology
2 answers:
gregori [183]3 years ago
7 0

Answer:

Hello! I think its "slide advance"

Explanation:

I've being trying to find this answer myself and this is the closest I feel being correct. Why? <em><u>To make the slides automatically, you need enter the number of minutes or seconds that you want.</u></em> <em>The timer starts when the final animation or other effect on the slide finishes</em>. SO hey I did what I could. However if anyone has a better answer, feel free to type it.

I hope this helps. Thank you!

andrew-mc [135]3 years ago
6 0

i believe its transition

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Which of the following is normally mutually exclusive<br> Read only files<br> Printer<br> WIFI
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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

3 0
2 years ago
Edhesive assignment 7 calendar
ikadub [295]

Answer:

def leap_year(y):

 if y % 4 == 0:

     return 1

 else:

     return 0

def number_of_days(m,y):

 if m == 2:

     return 28 + leap_year(y)

 elif m == 1 or m == 3 or m == 5 or m == 7 or m == 8 or m ==10 or m == 12:

     return 31

 elif m == 4 or m == 6 or m == 9 or m == 11:

     return 30

def days(m,d):

 if m == 1:

     return 0 + d

 if m == 2:

     return 31 + d

 if m == 3:

     return 59 + d

 if m == 4:

     return 90 + d

 if m == 5:

     return 120 + d

 if m == 6:

     return 151 + d

 if m == 7:

     return 181 + d

 if m == 8:

     return 212 + d

 if m == 9:

     return 243 + d

 if m == 10:

     return 273 + d

 if m == 11:

     return 304 + d

 if m == 12:

     return 334 + d

def days_left(d,m,y):

 if days(m,d) <= 60:

     return 365 - days(m,d) + leap_year(y)

 else:

     return 365 - days(m,d)

print("Please enter a date")

day=int(input("Day: "))

month=int(input("Month: "))

year=int(input("Year: "))

choice=int(input("Menu:\n1) Calculate the number of days in the given month.\n2) Calculate the number of days left in the given year.\n"))

if choice == 1:

 print(number_of_days(month, year))

if choice == 2:

 print(days_left(day,month,year))

Explanation:

Hoped this helped

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Answer: (B) Legal hold

Explanation:

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The window of the command will be given the root accessing. Then, it provide the associated by means of a protected shell with the access root in the system.

 

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