| 2000 | ||
3π m2, a wysokość walca jest równa 2 metry. Ma wyjść Obj = | π l. | |
| 3 |
Pc=2πrH + 4πr2
3π=2πr*2 + 4πr2
3=4r+4r2
| 3 | |
=r+r2
| |
| 4 |
| 3 | ||
0=r2+r− | ||
| 4 |
| −1−2 | ||
x1= | =−1,5 odzucamy
| |
| 2 |
| −1+2 | ||
x2= | =0,5
| |
| 2 |
| 4 | 4 | 500 | 500 | 1500 | ||||||
V= | πr3+πr2H= | π*125+25π*20= | π+500π= | π+ | π=
| |||||
| 3 | 3 | 3 | 3 | 3 |
| 2000 | |
π dm3 (1dm3=1 litr ) | |
| 3 |