2. Obliczylam dla 50, 5 i 0.5%, ale tylko dla 5% sa liczby calkowite, wiec zakladam ze taka jest tresc zdania.
Dla 50% wiecej:
[tex]x+x+\frac12x=41\\\frac52x=41 /*2\\5x=82 /:5\\x=16.4[/tex]
Dla 0.5% wiecej:
[tex]x+x+\frac{5}{1000}x=41\\\frac{2005}{1000}x=41 /*1000\\2005x=41000\\x=20.4488[/tex]
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[tex]x+x+\frac1{20}x=41\\\frac{41}{20}x=41 /*20\\41x=820 /:41\\x=20\\x+\frac1{20}=20+20*\frac1{20}=20+1=21\\\text{Marek rozwiazal 20 zadan a Sylwia 21}[/tex]
3.
[tex]5x-3=4x-4\\5x-4x=-4+3\\x=-1\\[/tex]
4.
[tex]a) \\x-26=(2x-18)-(8+x)\\x-26=2x-18-8-x\\x-2x+x=-18-8+26\\0=0\\x\in (-\infty; \infty)\\\\b) \\7x-8x+7+2x=41-x\\7x-8x+2x+x=41-7\\2x=34 /:2\\x=17\\\\c) \\2x+23=16(2+0.125x)\\2x+23=32+2x\\23\neq 32\\\\d) \text{brak przykladu}[/tex]
5.
[tex]t=m-10\\t-6+m-6=26\\t+m-12=26\\t+m=38\\m-10+m=38\\2m=38+10\\2m=48/:2\\m=24\\t=m-10=24-10=14\\\\\text{Tomek ma 14 lat a Marek 24 lata}[/tex]
6.
[tex]Cp_1=10\%\\m_r_1=2kg\\\frac1{10}=\frac{m_s}{2kg}\\m_s=\frac{2kg}{10}=0.2kg\\\\m_s_2=0.2kg+x\\m_r_2=2kg+x\\Cp_2=20\%\\\\\frac{2}{10}=\frac{0.2kg+x}{2kg+x}\\2(2kg+x)=10(0.2kg+x)\\4kg+2x=2kg+10x\\4kg-2kg=10x-2x\\8x=2kg /:8\\x=\frac28kg=0.25kg[/tex]
Odp. Nalezy dosypac 0.25kg cukru
7.
[tex]J=mR\\mR=J /:R\\m=\frac{J}R[/tex]
8.
[tex]V_1=40\frac{km}h\\V_2=40\frac{km}h+20\frac{km}h=60\frac{km}h\\t_2=t_1-2h\\\\s=V_1*t=V_2*t_2\\40t=60(t-2)\\40t=60t-120\\120=60t-40t\\120=20t /:20\\6=t\\\\s=40\frac{km}h*6h=240km[/tex]
9.
[tex]t=5min=\frac5{60}h=\frac1{12}h\\t_2=t+3min=8min=\frac8{60}h=\frac2{15}h\\vt=s=v_2t_2\\\\\frac1{12}v=\frac2{15}(v-6)\\\frac1{12}v=\frac2{15}v-\frac{12}{15}\\\frac{15}{180}v-\frac{24}{180}x=-\frac{12}{15}\\-\frac9{180}v=-\frac{12}{15}\\\frac1{20}v=\frac{12}{15} /*20\\v=\frac{240}{15}=16\frac{km}h\\s=16\frac{km}h*\frac1{12}h=\frac{16}{12}km=1,(3)km[/tex]
