Răspuns :
[tex]1) \\ a^2+b^2-6a+8b+25=0 \\ a^2-6a+b^2+8b+25=0 \\ a^2-6a+b^2+8b+9+16=0 \\ a^2-6a+9+b^2 +8b +16=0 \\ ( a^2+2*(-3)*a+(-3)^2)+(b^2 +2*4*b +4^2)=0 \\ (a-3)^2+(b+4)^2=0 \\ =\ \textgreater \ \;\;\;a-3=0\;\;\;si\;\;\;b+4=0 \\ \boxed{a=3}\;\;(radacina \;dubla) \\ \boxed{b=4} \;\;(radacina \;dubla) [/tex]
[tex]2) \\ \frac{p}{100}*24=80*p \\ p*24= 80*p*100 \\ 24p=8000p \\ 8000p-24p=0 \\ 7976p=0 \\ p= \frac{0}{7976} =\boxed{0}[/tex]
[tex]3) \\ \frac{p}{100}*10=15*p \\ p*10= 15*p*100 \\ 10p=1500p \\ 1500p-10p=0 \\ 1490p=0 \\ p= \frac{0}{1490} =\boxed{0}[/tex]
[tex]4) \\ 30 \;min = 50\% \;dintr-o \;ora.[/tex]
[tex]5) \\ V = volumul \;cubului \\ d= 5\;cm=Diagonala \;cubului \\ L = latura \;cubului \\ d = \sqrt{3L^2} \\ \sqrt{3L^2} =5 \\ 3L^2=25 \\ L^2= \frac{25}{3} \\ L= \sqrt{\frac{25}{3}} = \frac{5}{ \sqrt{3} } = \frac{5 \sqrt{3} }{3} \;cm \\ V = (\frac{5 \sqrt{3} }{3} )^3= \frac{125*3\sqrt{3}}{27} =\boxed{\frac{125\sqrt{3}}{9}\;cm^3}[/tex]
[tex]6)\\\text{Daca o piramida are toate muchiile egale, } \\ \text{atunci fetele laterale sunt triunghiuri echilaterale} \\ Apotema \;piramidei = inaltimea\;triunghiului\;fetei\;laterale. \\ Inaltimea\; triunghiului\;echilateral = L* \frac{ \sqrt{3}}{2} \;\;\;\;\; dar \;\;\;L = 4\;cm \\ \\ \textgreater \ \;\;\; a=4* \frac{ \sqrt{3}}{2}= \frac{ 4\sqrt{3}}{2}= \boxed{2\sqrt{3}\;cm}[/tex]
