Conduct online research and provide some guidelines on how to secure a web browser. (CyberSecurity plato)

Nezavi [6.7K]

portable computing tool

7 0

3 years ago

Input 3 positive integers from the terminal, determine if tlrey are valid sides of a triangle. If the sides do form a valid tria

butalik [34]

Answer:

In Python:

side1 = float(input("Side 1: "))

side2 = float(input("Side 2: "))

side3 = float(input("Side 3: "))

if side1>0 and side2>0 and side3>0:

if side1+side2>=side3 and side2+side3>=side1 and side3+side1>=side2:

if side1 == side2 == side3:

print("Equilateral Triangle")

elif side1 == side2 or side1 == side3 or side2 == side3:

print("Isosceles Triangle")

else:

print("Scalene Triangle")

else:

print("Invalid Triangle")

else:

print("Invalid Triangle")

Explanation:

The next three lines get the input of the sides of the triangle

side1 = float(input("Side 1: "))

side2 = float(input("Side 2: "))

side3 = float(input("Side 3: "))

If all sides are positive

if side1>0 and side2>0 and side3>0:

Check if the sides are valid using the following condition

if side1+side2>=side3 and side2+side3>=side1 and side3+side1>=side2:

Check if the triangle is equilateral

if side1 == side2 == side3:

print("Equilateral Triangle")

Check if the triangle is isosceles

elif side1 == side2 or side1 == side3 or side2 == side3:

print("Isosceles Triangle")

Otherwise, it is scalene

else:

print("Scalene Triangle")

Print invalid, if the sides do not make a valid triangle

else:

print("Invalid Triangle")

Print invalid, if the any of the sides are negative

else:

print("Invalid Triangle")

4 0

3 years ago

How has the global marketplace used emerging technologies to expand businesses

madam [21]

Answer:

Information Technology (IT) has revolutionized the way organizations conduct business by enabling small and medium businesses to level the playing field with larger organizations. Small businesses use an array of technology based on everything from computer server stations to portable mobile devices to expand competitive advantages in the global economic marketplace and marketing environment. So, small & medium business organizations owners are considering implementing information technology (IT) in their planning process for streamlined integration. Thus, this process is leading to future business expansion locally as well as globally. This decision allows proprietors to create smooth business operations using the most effective information technology available. On the other hand, the Internet has further linked the overseas suppliers of goods and services and their buyers. During the e-commerce era, internet traffic is increasing day by day, and e-commerce business is in the hyper growth stage. Information technology tremendously helps to boost the expansion of the current and future of marketing environment; communication technology became faster than ever. Though, globalization creates a little challenge for multinational companies to share resources and knowledge across a number of businesses inside and outside the country. Outsourcing and freelancer ease this challenge because multinational companies recruit talent freelancers and outsource their projects to the highly expert vendors and then the works are done through the team of people very effectively and efficiently.

Explanation:

6 0

3 years ago

The angles of a quadrilateral are in the ratio 1:2:3:4 what is the largest angle

zheka24 [161]

Answer:

144°

Explanation:

The sum of the measures of interior angles of a quadrilateral is 360°

The ratio of angle measures is 1:2:3:4

Total ratio = 1+2+3+4=10

The angles will be :

1/10 * 360°= 36°

2/10*360°=72°

3/10*360°=108°

4/10*360°=144°

The largest angle is 144°

7 0

3 years ago

Suppose that you are asked to modify the Stack class to add a new operation max() that returns the current maximum of the stack

Advocard [28]

Answer:

Following are the code to the given points:

Explanation:

For point 8.1:

public String max()//defining a method max

{

String maxVal=null;//defining a string variable that holds a value

for(int x=0;x<n;x++)

{

if(maxVal==null || a[i].compareTo(maxVal)>0)//defining if blok to comare the value

{

maxVal=a[i];//holding value in maxVal variable

}

return maxVal;//return maxVal variable value

}

For point 8.2:

public void push(String item)//defining a method push that accepts item value in a parameter

{

a[n]=item;//defining an array to hold item value

if(n==0 || item.compareTo(maxVals[n-1])>0)//use if to comare item value

{

maxVals[n]=item;//holding item value in maxVals variable

}

else

{

maxVals[n]=maxVals[n-1];//decreasing the maxVals value

}

n++;//incrementing n value

}

public String pop()//defining a method pop

{

return a[--n];//use return value

}

public String max()//defining a method max

{

return maxVals[n-1];//return max value

}

In the first point, the max method is declared that compares the string and returns its max value.
In the second point, the push, pop, and max method are declared that works with their respective names like insert, remove and find max and after that, they return its value.

7 0

3 years ago