Answer:
This tool lets users change text from one language to another.
translation tool
This tool helps users find definitions for difficult vocabulary.
online dictionary
This tool allows users to listen to passages of text that they select.
text-to-speech tool
This tool helps users memorize vocabulary by giving them representations of words using pictures.
flash cards
Explanation:
just did the assignment on edg 2020
Answer:
#include <stdio.h>
#include <ctype.h>
void printHistogram(int counters[]) {
int largest = 0;
int row,i;
for (i = 0; i < 26; i++) {
if (counters[i] > largest) {
largest = counters[i];
}
}
for (row = largest; row > 0; row--) {
for (i = 0; i < 26; i++) {
if (counters[i] >= row) {
putchar(254);
}
else {
putchar(32);
}
putchar(32);
}
putchar('\n');
}
for (i = 0; i < 26; i++) {
putchar('a' + i);
putchar(32);
}
}
int main() {
int counters[26] = { 0 };
int i;
char c;
FILE* f;
fopen_s(&f, "story.txt", "r");
while (!feof(f)) {
c = tolower(fgetc(f));
if (c >= 'a' && c <= 'z') {
counters[c-'a']++;
}
}
for (i = 0; i < 26; i++) {
printf("%c was used %d times.\n", 'a'+i, counters[i]);
}
printf("\nHere is a histogram:\n");
printHistogram(counters);
}
Network management is a broad range of functions including activities, methods, procedures and the use of tools to administrate,operate,and reliably maintain computer network system. <span />
Answer
Technology can help government handle economic emergencies such as crop and resource shortages.
Explanation
The government can address the concerns of food shortages and water scarcity through embracing new technology interventions to increase farm yields and mitigate the impacts of water shortages. Through crop protection methods, weeds and pest can be controlled. Drip irrigation technology applies water directly to the roots of crops to facilitate high crop production. Other technologies to apply can include organic agriculture and integrated soil fertility management.
Answer:
The program in Python is as follows:
num1 = int(input())
num2 = int(input())
if num1 >=0 and num2 >= 0:
print(num1+num2)
elif num1 <0 and num2 < 0:
print(num1*num2)
else:
if num1>=0:
print(num1**2)
else:
print(num2**2)
Explanation:
This gets input for both numbers
num1 = int(input())
num2 = int(input())
If both are positive, the sum is calculated and printed
<em>if num1 >=0 and num2 >= 0:</em>
<em> print(num1+num2)</em>
If both are negative, the products is calculated and printed
<em>elif num1 <0 and num2 < 0:</em>
<em> print(num1*num2)</em>
If only one of them is positive
else:
Calculate and print the square of num1 if positive
<em> if num1>=0:</em>
<em> print(num1**2)</em>
Calculate and print the square of num2 if positive
<em> else:</em>
<em> print(num2**2)</em>
