Proszę o jak najszybsze rozwiązanie.... Question from @patrykstudent

November 2018 1 11 Report

Proszę o jak najszybsze rozwiązanie

heheheh12 zad. 1
a) $\left \{ {{2x+2y=14} \atop {3x-y=1}} \right.$

$\left \{ {{2x+2y=14} \atop {6x-2y=2}} \right.$

$2x+6x+2y-2y=14+2$

$8x=16$

$x=2$

$\left \{ {{x=2} \atop {3*2-y=1}} \right.$

$\left \{ {{x=2} \atop {6-1=y}} \right.$

$\left \{ {{x=2} \atop {y=5}} \right.$

b) $\left \{ {{ \frac{1}{4}x+ \frac{1}{6} y=2 } \atop { \frac{1}{2}x- \frac{1}{3} y=0 }} \right.$

$\left \{ {{ \frac{1}{2}x+ \frac{1}{3} y=4 } \atop { \frac{1}{2}x- \frac{1}{3} y=0 }} \right.$

$\frac{1}{2}x+\frac{1}{2}x+ \frac{1}{3} y- \frac{1}{3}=4+0$

$x=4$

$\left \{ {{x=4 } \atop { \frac{1}{2}*4- \frac{1}{3} y=0 }} \right.$

$\left \{ {{x=4 } \atop { - \frac{1}{3} y=-2 }} \right.$

$\left \{ {{x=4 } \atop {y=6 }} \right.$

c) $\left \{ {{2(x+y)-(x-y)=4} \atop {3(x-2y)+4(x+y)=5}} \right.$

$\left \{ {{2x+2y-x+y=4} \atop {3x-6y+4x+4y=5}} \right.$

$\left \{ {{x+3y=4} \atop {7x-2y=5}} \right.$

$\left \{ {{-7x-21y=-28} \atop {7x-2y=5}} \right.$

$-7x+7x-21y-2y=5-28$

$-23y=-23$

$y=1$

$\left \{ {{y=1} \atop {x+3*1=4}} \right.$

$\left \{ {{y=1} \atop {x=4-3}} \right.$

$\left \{ {{y=1} \atop {x=1}} \right.$

d) $\left \{ {{\ \frac{x+5}{2}+\frac{2y-1}{3}=4 } \atop { \frac{3x+1}{2} +\frac{5y+2}{4}=5}} \right.$

$\left \{ {{-3x-15-4y+2=-24} \atop {6x+2+5y+2=20}} \right.$

$\left \{ {{-3x-4y+=-11} \atop {6x+5y=16}} \right.$

$\left \{ {{-6x-8y+=-22} \atop {6x+5y=16}} \right.$

$-6x+6x-8y+5y=-22+16$

$-3y=-6$

$y=2$

$\left \{ {{y=2} \atop {6x+5*2=16}} \right.$

$\left \{ {{y=2} \atop {6x=6}} \right.$

$\left \{ {{y=2} \atop {x=1}} \right.$

zad. 2
a) $\sqrt{2} + \sqrt{8} - \sqrt{12} = \sqrt{2} + \sqrt{4*2} - \sqrt{4*3}= \sqrt{2} +2 \sqrt{2} - 2\sqrt{3} =$
$=3 \sqrt{2} +2 \sqrt{3}$

b) $3 \sqrt{75} -5 \sqrt{27} + \sqrt{3} =3 \sqrt{3*25} -5 \sqrt{9*3} + \sqrt{3} =$
$=15 \sqrt{3} -15 \sqrt{3} + \sqrt{3} = \sqrt{3}$

c) $\frac{\sqrt{50}-\sqrt{18}}{\sqrt{2}} = \frac{\sqrt{25*2}-\sqrt{9*2}}{\sqrt{2}} = \frac{5\sqrt{2}-3\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{\sqrt{2}} =2$

d) $\frac{ 3\sqrt{20}-2 \sqrt{45} }{3 \sqrt{5} } = \frac{ 3\sqrt{4*5}-2 \sqrt{9*5} }{3 \sqrt{5} }= \frac{ 6\sqrt{5}-6 \sqrt{5} }{3 \sqrt{5} }= \frac{0}{3 \sqrt{5}} =0$

zad. 3
a) $-6^2+(-5)^2-(-\frac{1}{2})^{-2}=36+25-(-2)^2=61-4=57$

b) $(1\frac{1}{7})^{-1}+(1\frac{1}{3})^{-2}=(\frac{8}{7})^{-1}+(\frac{4}{3})^{-2}=(\frac{7}{8})^{1}+(\frac{3}{4})^{2}=$

$= \frac{7}{8} + \frac{9}{16} = \frac{14}{16} + \frac{9}{16} = \frac{23}{16} =1 \frac{7}{16}$

c) $\sqrt{7\frac{1}{9}} - \sqrt{1,44} *(1\frac{2}{3})^2= \sqrt{\frac{64}{9}} -1,2*( \frac{5}{3} )^2=$

$= \frac{8}{3} - \frac{12}{10} * \frac{25}{9} =\frac{8}{3} - \frac{10}{3} =- \frac{2}{3}$

d) $(2\sqrt{5})^2- \sqrt{3} * \sqrt{3} + \sqrt[3]{8} =4*5-\sqrt{9}+2=20-3+2=19$

zad. 4
a) 1928 - MCMXXVIII
b) 1410 - MCDX
c) 996 - CMXCVI
d) 1944 - MCMXLIV

zad. 5
a) $793000000=793*10^6=7,93*10^8$

b) $0,00000073=73*10^{-8}=7,3*10^{-7}$

c) $405000000=405*10^6=4,05*10^8$

d) $0,00000125=125*10^{-8}=1,25*10^{-6}$

zad. 6
a) $1,2*10^{-7}*0,3*10^5=0,36*10^{-2}=3,6*10^{-3}$

b) $3,2*10^8*0,5*10^{-6}=1,6*10^2$

c) $2,5*10^{-6}*0,2*10^4=0,5*10^{-2}=5*10^{-3}$

d) $1,4*10^{-5}*0,6*10^3=0,84*10^{-2}=8,4*10^{-3}$

3 votes Thanks 2