PICT0290 (2)

■10

kwadrat numer 10.

= ^(2.4 f- 2.6)-300 = 375łw* t’_mt» = ^(2.6-2.3 + 0.1)- 363 - 363w’ ^_łl, = |(0j).3-0m*

V_mn = i(0J +1.6)-131 - 5 6#m‘

- i (1.6 + 3.1 * 0.8)-356 = 392m’ ^*-5(0Ut)-19 = 5m¹

»'«* —j(l.O + 0.6)-100 = 40m⁵ ^v*,n=j(0.6)-37=7 «»

- |(0.6 + 2.5 + 1.7)-397 - 381m^> -^(l.7 + 0.7)-269«16»m’ n...-|(0-7)-44-l(W K_wx, = “(0.7 ♦ 1.6-*- U) 381 - 274m^ł

Suaa objętości- wykopów 1 nasypów.

2^ V„ - 7960 + 375 + 363 + 56 + 392 + 5 = 9151 m ’

£ r„ = 8560 + 40 + 7 + 381 + 161 + 10 + 274 - 9433 m ⁴