What will the following loop display? 0 1 2 3 4 5 0 1 2 3 4 0 1 2 3 4 The loop will display numbers starting at 0, for infinity.

To set up scenarios, you need to first use ______ to set up a list, then ______ to set up the reference cell. Last you need to u

Sholpan [36]

To set up scenarios, you need to first use Goal seek to set up a list, then data table to set up the reference cell. Last you need to use VLOOKUP to set up the cells that display the output results from the scenario.

<h3>What is a Goal Seek in Excel?</h3>

Microsoft Excel is known to have a goal seeking tool that is said to be built-in and it is one that gives room for the user to know the desired input value for any kind of formula if the output value is said to be already known.

Hence, To set up scenarios, you need to first use Goal seek to set up a list, then data table to set up the reference cell. Last you need to use VLOOKUP to set up the cells that display the output results from the scenario.

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To set up scenarios, you need to first use to set up a list, then to set up the reference cell. Last you need to use to set up the cells that display the output results from the scenario. CHOOSE; VLOOKUP; data table Goal seek; data table; VLOOKUP Data validation; VLOOKUP; CHOOSE Data table; VLOOKUP;

Learn more about VLOOKUP from

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6 0

2 years ago

The ability to anticipate and determine upcoming driving hazards and conditions are adversely affected by stress.

dybincka [34]

It can be stress or drinking, drugs, or anything of the sort.

6 0

3 years ago

Plz go sub 2 "Shyy096" he is a 11 y/o boy and makes music fortnite vids, it would rly help him if u subscribe cuz he only has 8.

Pachacha [2.7K]

Answer:

OKK

Explanation:

5 0

3 years ago

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>

#include <bits/stdc++.h>

using namespace std;

// chracter to digit mapping, and the inverse

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

unordered_map<char, int> c2i;

unordered_map<int, char> i2c;

int ans = 0;

// limit: length of result

int limit = 0;

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

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

if (digit == limit) {

ans += (sum == 0);

return sum == 0;

}

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

if (widx == words.size()) {

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

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

return false;

c2i[result[digit]] = sum % 10;

i2c[sum%10] = result[digit];

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

c2i.erase(result[digit]);

i2c.erase(sum%10);

ans += tmp;

return tmp;

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

if (digit + 1 == limit && 0 == c2i[result[digit]]) {

return false;

}

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

} else {

return false;

}

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

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

return helper(words, result, digit, widx+1, sum);

}

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

if (c2i.count(words[widx][digit])) {

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

return false;

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

}

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

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

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

if (i2c.count(i)) continue;

c2i[words[widx][digit]] = i;

i2c[i] = words[widx][digit];

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

c2i.erase(words[widx][digit]);

i2c.erase(i);

}

return false;

}

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

limit = result.length();

for (auto &w: words)

if (w.length() > limit)

return;

for (auto&w:words)

reverse(w.begin(), w.end());

reverse(result.begin(), result.end());

int aa = helper(words, result, 0, 0, 0);

}

int main()

{

ans = 0;

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

string result = "BLACK";

isSolvable(words, result);

cout << ans << "\n";

return 0;

}

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

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3 0

2 years ago

Which of the following is the smallest aperture (f-stop) opening?

Firlakuza [10]

Answer:

E) f/16

Explanation:

6 0

3 years ago