V=96 litrow=96 dm³
V= Pp razy H
Pp- pole podstawy H wysokosc do ktorej nalano wode
H=30cm=3dm
Pp=V:H= 96:3=32 dm²
Pp=ab gdzie a i b to boki prostokata w podstawie
ab=32
2a+2b=24
a+b=12
a=12-b
(12-b)b=32
-b²+12b-32=0
Δ= 12²- 4 razy (-1) razy (-32)= 144-128=16
√Δ=4
b1= (-12+4):-2=-8:-2=4 dm
b2= (-12-4):-2= -16/-2=8 dm
a1=12-b1=12-4=8 dm
a2=12-b2=12-8=4 dm
wymiary podstawy to 4dm i 8 dm
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V=96 litrow=96 dm³
V= Pp razy H
Pp- pole podstawy H wysokosc do ktorej nalano wode
H=30cm=3dm
Pp=V:H= 96:3=32 dm²
Pp=ab gdzie a i b to boki prostokata w podstawie
ab=32
2a+2b=24
ab=32
a+b=12
ab=32
a=12-b
(12-b)b=32
-b²+12b-32=0
Δ= 12²- 4 razy (-1) razy (-32)= 144-128=16
√Δ=4
b1= (-12+4):-2=-8:-2=4 dm
b2= (-12-4):-2= -16/-2=8 dm
a1=12-b1=12-4=8 dm
a2=12-b2=12-8=4 dm
wymiary podstawy to 4dm i 8 dm