Explanation:
idealized model of computation or physical electronic device implementing a Boolean function, a logical operation performed on one or more binary inputs that produces a single binary output.
Answer: What key value does Accenture's myConcerto bring to clients looking for automated solutions? myConcerto functions an out-of-the-box platform solution that Accenture can hand off to clients. myConcerto uses slide presentations to demonstrate the capabilities of Accenture's automated solutions. myConcerto provides clients with a suite of ready-to-use productivity dashboards. myConcerto lets clients work side-by-side with Accenture to tailor and visualize their solutions.
Explanation:
Answer:
2.342m
Explanation:
Given
Time = 0.5 s
Height of Window = 2m
Because the pot was in view for a total of 0.5 seconds, we can assume that it took the cat 0.25 seconds to go from the bottom of the window to the top
Using this equation of motion
S = ut - ½gt²
Where s = 2
u = initial velocity = ?
t = 0.25
g = 9.8
So, we have.
2 = u * 0.25 - ½ * 9.8 * 0.25²
2 = 0.25u - 0.30625
2 + 0.30625 = 0.25u
2.30625 = 0.25u
u = 2.30625/0.25
u = 9.225 m/s ------------ the speed at the bottom of the pot
Using
v² = u² + 2gs to calculate the height above the window
Where v = final velocity = 0
u = 9.225
g = 9.8
S = height above the window
So, we have
0² = 9.225² - 2 * 9.8 * s
0 = 85.100625 - 19.6s
-85.100625 = -19.6s
S = -85.100625/19.6
S = 4.342
If 4.342m is the height above the window and the window is 2m high
Then 4.342 - 2 is the distance above the window
4.342 - 2 = 2.342m
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);
}
honestly i think ur answer would be D because it keeps you from flying out of the window
