Question

0.8 √121 + 1.5 √0.16, 3 - 6 √(одна девятая), (0.6 √50)^2 помогите плес

Answer 1

0,8*11-1,5*0,4=8,8+0,6=9,4
3-6*1/3=3-2=1
(0,6√50)²=0,36*50=18

Answer 2

$0,8\sqrt{121}+1,5\sqrt{0,16}=0,8*11+1,5*0,4=8,8+0,6=9,4;\\3-6\sqrt{\frac{1}{9}}=3-6*\frac{1}{3}=3-2=1;\\(0,6\sqrt{50})^2=0,6^2*(\sqrt{50})^2=0,36*50=18.$