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IgorLugansk [536]
3 years ago
10

Which term refers to a type of an attack in which an attacker makes his data look like it is coming from a different source addr

ess, and is able to intercept information transferred between two computers?
Computers and Technology
1 answer:
Andrew [12]3 years ago
8 0

<u>Man-in-the-middle attack</u> refers to a type of an attack in which an attacker makes his data look like it is coming from a different source address, and is able to intercept information transferred between two computers.

<u>Explanation:</u>

A man-in-the-middle attack (MITM) is an assault where the aggressor furtively transfers and potentially changes the correspondences between two gatherings who accept that they are straightforwardly speaking with one another. This happens when the assailant catches a segment of a correspondence between two gatherings and retransmits it sometime in the future. The assailant would then be able to screen and perhaps change the substance of messages. The utilization of such encoded burrows makes extra secure layers when you get to your organization's secret systems over connections like Wi-Fi.

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(15) What is the best definition of a contextual tab?
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  1. What is the best definition of a contextual tab-<u>(c) A tab that appears in context with what you are working on</u>
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Explanation:

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6 0
3 years ago
Given parameters b and h which stand for the base and the height of an isosceles triangle (i.e., a triangle that has two equal s
schepotkina [342]

Answer:

The area of the triangle is calculated as thus:

Area = 0.5 * b * h

To calculate the perimeter of the triangle, the measurement of the slant height has to be derived;

Let s represent the slant height;

Dividing the triangle into 2 gives a right angled triangle;

The slant height, s is calculated using Pythagoras theorem as thus

s = \sqrt{b^2 + h^2}

The perimeter of the triangle is then calculated as thus;

Perimeter = s + s + b

Perimeter = \sqrt{b^2 + h^2} + \sqrt{b^2 + h^2} +b

Perimeter = 2\sqrt{b^2 + h^2} + b

For the volume of the cone,

when the triangle is spin, the base of the triangle forms the diameter of the cone;

Volume = \frac{1}{3} \pi * r^2 * h

Where r = \frac{1}{2} * diameter

So, r = \frac{1}{2}b

So, Volume = \frac{1}{3} \pi * (\frac{b}{2})^2 * h

Base on the above illustrations, the program is as follows;

#include<iostream>

#include<cmath>

using namespace std;

void CalcArea(double b, double h)

{

//Calculate Area

double Area = 0.5 * b * h;

//Print Area

cout<<"Area = "<<Area<<endl;

}

void CalcPerimeter(double b, double h)

{

//Calculate Perimeter

double Perimeter = 2 * sqrt(pow(h,2)+pow((0.5 * b),2)) + b;

//Print Perimeter

cout<<"Perimeter = "<<Perimeter<<endl;

}

void CalcVolume(double b, double h)

{

//Calculate Volume

double Volume = (1.0/3.0) * (22.0/7.0) * pow((0.5 * b),2) * h;

//Print Volume

cout<<"Volume = "<<Volume<<endl;

}

int main()

{

double b, h;

//Prompt User for input

cout<<"Base: ";

cin>>b;

cout<<"Height: ";

cin>>h;

//Call CalcVolume function

CalcVolume(b,h);

//Call CalcArea function

CalcArea(b,h);

//Call CalcPerimeter function

CalcPerimeter(b,h);

 

return 0;

}

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