Answer:
its okay. i mean like you have to do a bunch off stuff. andsomtimes it gets a little boring.
Explanation:
i am doing online school.
Answer:
Your answer would be more likely.
Explanation:
When someone is upset and they are driving they usually have their music baring and the they can't hear their surroundings and all they can think about is what they are upset about so therefor they don't pay attention. Hope this helped! Have a great day! :)
Complete Question:
Which of the following is a trademark automatically received by an organization when a symbol is being consistently used in the normal course of business?
Group of answer choices
A. Open source trademark.
B. Common law trademark.
C. Registered trademark.
D. Open source trademark.
Answer:
B. Common law trademark.
Explanation:
A common law trademark can be defined as a protection or enforceable mark for a product name, logo, symbol or brand name used to distinguish goods and services prior to its registration with the state or federal government. Common law trademark is a trademark which is automatically received by an organization when a symbol is being consistently used in the normal course of business.
This ultimately implies that, common law trademarks are not governed by any statute and as such are only limited to the geographical location where they are used.
For instance, if a tomato paste is being sold to consumers with the product name "Ginoo" in Florida, the company's trademark applies to Florida only. Thus, another company can use the product name without any trademark infringement in other states of the country such as New York, Washington DC, California etc. except in Florida due to a common law trademark.
Answer:
See Explaination
Explanation:
#include <iostream>
#include <string.h>
using namespace std;
char *mixem(char *s1, char *s2);
int main() {
cout << mixem("abc", "123") << endl;
cout << mixem("def", "456") << endl;
return 0;
}
char *mixem(char *s1, char *s2) {
char *result = new char[1 + strlen(s1) + strlen(s2)];
char *p1 = s1;
char *p2 = s2;
char *p = result;
while (*p1 || *p2) {
if (*p1) {
*p = *p1;
p1++;
p++;
}
if (*p2) {
*p = *p2;
p2++;
p++;
}
}
*p = '\0';
return result;
}
