Odpowiedź:
I. a)
[tex] \sqrt{225} = 15[/tex]
bo
[tex] {15}^{2} = 225[/tex]
b)
[tex] \sqrt{ \frac{16}{81} } = \frac{ \sqrt{16} }{ \sqrt{81} } = \frac{4}{9} [/tex]
bo
[tex]( { \frac{4}{9} })^{2} = \frac{16}{81} [/tex]
c)
[tex] \sqrt{4 \frac{21}{25} } = \sqrt{ \frac{121}{25} } = \frac{11}{5} [/tex]
bo
[tex]( { \frac{11}{5} })^{2} = \frac{121}{25} = 4 \frac{21}{25} [/tex]
d)
[tex] \sqrt{0.0081} = 0.09[/tex]
bo
[tex]{0.09}^{2} = 0.0081[/tex]
II.
Pole kwadratu =
[tex] {a}^{2} [/tex]
a - bok kwadratu
[tex] {a}^{2} = 225 \\ \sqrt{ {a}^{2} } = \sqrt{225} \\ |a| = 15 \\ a = 15[/tex]
Obwód kwartatu = 4a, czyli
4 • 15 = 60cm - odp. C
III. a)
[tex] \sqrt{121} + \sqrt{144} + \sqrt{64} = 11 + 12 + 8 = 31[/tex]
b)
[tex] \sqrt{0.04} + \sqrt{0.09} + \sqrt{0.49} = 0.2 + 0.3 + 0.7 = 1.2[/tex]
c)
[tex] \sqrt{7 \frac{1}{9} } - \sqrt{1\frac{28}{36} } = \sqrt{ \frac{64}{9} } - \sqrt{ \frac{64}{36} } = \frac{ \sqrt{64} }{ \sqrt{9} } - \frac{ \sqrt{64} }{ \sqrt{36} } = \frac{8}{3} - \frac{8}{6} = \frac{16}{6} - \frac{8}{6} = \frac{8}{6} = \frac{4}{3} [/tex]
IV.
Prostopadłościan (12 krawędzi):
2cm • 4 + 3cm • 4 + 5cm • 4 = 8cm + 12cm + 20cm = 40cm
Ostrosłup (8 krawędzi):
2cm • 4 + 4cm • 4 = 8cm + 16cm = 24cm