Zad. 1
[tex]1 - 2(x+3)\\2 - 2(x+6)\\3 - 2(x+9)\\4 - 2(x+12)\\\\2(x+3)+2(x+6)+2(x+9)+2(x+12)=2^2*3^3\\2x+6+2x+12+2x+18+2x+24=4*27\\8x+60=108 /-60\\8x=48 /:8\\x=6\\\\1. 2(6+3)=2*9=18\\2. 2(6+6)=2*12=24\\3. 2(6+9)=2*15=30\\4. 2(6+12)=2*18=36[/tex]
Odp. C
Zad. 7
[tex](-4+3*(-2))-(-5)=(-4-6)+5=-10+5=-5[/tex]
Odp. C
Zad. 8
[tex]15 - 21m\\3 - x\\\frac{3*21m}{15}=\frac{63m}{15}=4.2m[/tex]
Na uszycie trzech spodnic wystarczy 5m materialu. (B)
[tex]15 - 21m\\x - 33m\\x=\frac{15*33m}{21m}=\frac{495}{21}=23,57[/tex]
Z 33m materialu mozna uszyc co najwyzej 23 spodnice (C)
Zad. 9
Kwadrans: 15 minut
10km = 10000m
Predkosc Andrzeja: [tex]\frac{650m}{60s}=10\frac56\frac{m}s=10\frac56*\frac{\frac1{1000}km}{\frac1{3600}h}=\frac{65}6*\frac1{1000}*\frac{3600}1\frac{km}h=\frac{65}6*\frac1{10}*36\frac{km}h=65*\frac1{10}*6\frac{km}h=39\frac{km}h[/tex]
[tex]650m - 60s\\1m - x\\x=\frac{1m*60s}{650m}=\frac{60s}{650}=0,092s[/tex]
[tex]1min - 650m\\15min - x\\x=\frac{15min*650m}{1min}=15*650m=9750m[/tex]
9750m < 10km
Czy Andrzej przejedzie w kwadrans 10km?
Nie, poniewaz jedzie ze srednia predkoscia mniejsza niz 40km/h
Odp. B1
Zad. 10
|A| = 2+1=3
|O| = 2+1+3 = 6
[tex]P(A)=\frac36=\frac12[/tex]
Odp. D
