Answer:
telnet
Explanation:
<h2><u>Fill in the blanks </u></h2>
Terminal emulation, especially the unprotected <u>telnet </u>protocol, should be blocked from any access to all internal servers from the public network.
I believe that the answer is A: Extension but I am not 100% sure, hope this helps.
Answer: They must be dated.
Explanation: If pyrotechnic VDSs are used, they must be dated. Expired VDSs may be carried on board, but a minimum of three unexpired VDSs must be carried in the vessel.
Answer:
1. Hydrostatic pressure.
2. Pressure.
3. Equilibrium forces.
Explanation:
1. <u>Hydrostatic pressure</u>: pressure exerted by a stationary liquid. It is simply the pressure in a liquid such as water at a specific depth.
Mathematically, hydrostatic pressure is given by the formula;
Where:
- H.P is the hydrostatic pressure of a liquid.
- rho represents the density of a liquid.
- g is the acceleration due to gravity.
- d is the depth (height) of a liquid.
2. <u>Pressure</u>: it is the force in a direction perpendicular to a given surface area. Thus, it's typically measured in Newton per meters (N/m²).
Mathematically, pressure is given by the formula;
3. <u>Equilibrium forces</u>: a pair of forces whose resultant is zero. It is also referred to as balanced forces because the resultant of all the pair of forces acting on a body or system is at equilibrium (balanced).
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);
}
