Consider the following code:

failure of the _ cylinder will often result in sudden unexpected loss of the ability to stop the vehicle

grin007 [14]

Failure of the Brake master cylinder will often result in sudden unexpected loss of the ability to stop the vehicle

Explanation:

The ability to stop the vehicle lies with the Brakes. If brakes of a vehicle do not work properly then it might become difficult to stop the vehicle. This happens when a cylinder called brake master cylinder fails.

The brake master cylinder might not work properly with the passage of time or it can form internal leaks. This is the master cylinder and it controls other cylinders in a vehicle. Its failure affect the brakes badly, it will be unsafe to drive such a vehicle.

5 0

3 years ago

Easy coding question, please help.

borishaifa [10]

Answers:

What is the index of the last element in the array? stArr1.length()-1

This prints the names in order. How would I print every other value? Change line 4 to: index = index +2

Change line 7 to: i < names.length

5 0

3 years ago

A driving school uses this rule to estimate how many lessons a learner will require.

HACTEHA [7]

Answer:

This is correct. And to remove the confusion, I am adding the meaning of the Pseudocode. You need to begin with the algo that you are working upon, and then you need it to phrase the algo with the words which are easy to be transcribed in the form of the computer instructions. Now you need to indent the instructions properly inside the loop or within the conditional clauses. And while doing this, you need to not use the words which are used in certain forms of computer language. However, IF and THEN and ELSE and ENDIF are very frequently used as pseudo-code. Hence, your answer is correct.

Explanation:

Please check the answer section.

8 0

3 years ago

Which of the following scenarios might indicate that you have been a victim of identity theft?A)Your credit report displays acco

victus00 [196]

Answer:

Option A is the correct answer for the above question.

Explanation:

A Victim is a person, who faces the problem of a criminal person. Here in the question, the scenario is that a user is a victim of the identity of being theft by some other person. So the user can get known by the help of option a, which suggests that the user's credit card report display something which is not done by him then it can be the scenario, where the user can understand that he is a victim. hence option a is the correct answer where the other option is not because--

Option B suggests the user does not pay the bill on time but it is a general case because the bill is paid by the user is in every month not a single time.
Option C suggests both options (A and B) are correct, but option b is not the correct.
Option D suggests that no option is valid from the above but option A is the correct.

7 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

#SPJ1

3 0

2 years ago