Answer:
A downloaded executable file may contain harmful software know as malware.
Further details:
Malwares:
Malware (a portmanteau for pernicious programming) is any product purposefully intended to make harm a PC, server, customer, or PC network. Malware does the harm after it is embedded or brought somehow or another into an objective's PC and can appear as legitimately executable code, contents, supposed "dynamic substance" (Microsoft Windows), and different types of data. Some sorts of malware are to a great extent alluded to in the media as PC infections, worms, Trojan steeds, ransomware, spyware, adware, and scareware, among different terms. Malware has a malignant expectation, acting against the enthusiasm of the PC client—thus does exclude programming that causes accidental damage because of some lack, which is regularly portrayed as a product bug.
Uses:
Malware is once in a while utilized comprehensively against government or corporate sites to assemble monitored information, or to upset their activity as a rule. However, malware can be utilized against people to pick up data, for example, individual recognizable proof numbers or subtleties, bank or charge card numbers, and passwords.
Answer details:
Subject: Computer and technology
Level: College
Keywords:
• Harmful software
• Malware
• Malware software
• Uses of malware
• Purpose of malware
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Answer:
Explanation:
#include <iostream>
using namespace std;
int costdays(int);
int costhrs(int,int);
int main()
{
int dd,hh,mm,tmph,tmpd,tmpm=0;
int pcost,mcost=0;
cout<<"Enter Parking time" << endl;
cout<<"Hours: ";
cin>>hh;
cout<<"Minutes: ";
cin>>mm;
if (mm>60)
{
tmph=mm/60;
hh+=tmph;
mm-=(tmph*60);
}
if (hh>24)
{
tmpd=hh/24;
dd+=tmpd;
hh-=(tmpd*24);
}
if ((hh>4)&&(mm>0))
{
pcost+=costdays(1);
}
else
{
mcost=costhrs(hh,mm);
}
cout<<"Total time: ";
if (dd>0)
{
cout<<dd<<"days ";
pcost+=costdays(dd);
}
pcost+=mcost;
cout<<hh<<"h "<<mm<<"mins"<<endl;
cout<<"Total Cost :"<<pcost<<"Won";
return 0;
}
int costdays(int dd)
{
return(dd*25000);
}
int costhrs(int hh,int mm)
{
int tmpm, tmp=0;
tmp=(hh*6)*1000;
tmp+=(mm/10)*1000;
tmpm=mm-((mm/10)*10);
if (tmpm>0)
{
tmp+=1000;
}
return(tmp);
}
Answer:
1 Introduce the problem.
2 Explain your perspective.
3 Explain your opponent's perspective. Refute their points one-by-one as you go.
4 Present your evidence.
5 Conclude your argument.
1 Introduce the problem.
2 Explain your opponent's perspective first. ...
3 Explain your perspective.
Explanation:
Chchvjvcuvggiiog. Correct
