Question

como fazer essas questões ???

Answer 1

30.
P= perímetro
l=largura
c=comprimento

p=l+l+c+c
p=2l+2c
p=$2.2 \sqrt{3} + 2. \sqrt{12}$
p=$4 \sqrt{3} +4 \sqrt{12}$
p= $4 \sqrt{3} +[tex]4.2 \sqrt{3}$
p=$12 \sqrt{3}$

$\sqrt{12} = [tex]2 \sqrt{3}$

31
a)
$\sqrt{44} =2 \sqrt11}$
$\sqrt{99} =3 \sqrt{11}$


$\sqrt{44} +2. \sqrt{99}$.
=2 \sqrt11} [/tex] +$2.3 \sqrt{11}$
=2 \sqrt11} [/tex] +$6 \sqrt{11}$
=$8 \sqrt{11}$
=8x(3,32)
=26,56

b)$\sqrt{24} =2 \sqrt{6}$
$\sqrt{54}=3 \sqrt{6}$

2x($\sqrt{24})-[tex] \sqrt{54} 2x(2 \sqrt{6}$) -3 \sqrt{6} [/tex]
$\sqrt{6}$
=2,45

c) $\sqrt{252}=6 \sqrt{7}$
$\sqrt{28} =2 \sqrt{7}$

-$\sqrt{252}+[tex] \sqrt{28} = -6 \sqrt{7}$+2 \sqrt{7} [/tex]
=$-4 \sqrt{7}$
=-4 x 2,65
=-10,6

d) $\sqrt{96} =4 \sqrt{6}$
$\sqrt{112} =4 \sqrt{7}$

$\sqrt{96}+[tex] \sqrt{11}$-$3.[tex] \sqrt{112} 4 \sqrt{6}$+$\sqrt{11}$-3.4 \sqrt{7} [/tex]
4 \sqrt{6} [/tex]+$\sqrt{11}$-12 \sqrt{7} [/tex]
4x 2,45 +3,32 - 12 x 2,65
9.8 +3,32-31,8
-18,68

espero ter ajudado :))