Odpowiedź :
Odpowiedź:
d) Tw. Pitagorasa:
a² + 4² = ( 3 √2)²
a² + 16 = 18
a² = 18 - 16 = 2
a = √2
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e) b² = ( 3 √3)² + (2 √5)²
b² = 27 + 20 = 47
b = [tex]\sqrt{47}[/tex]
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f) c² + 3² = ([tex]\sqrt{13}[/tex])²
c² + 9 = 13
c² = 4
c = 2
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Szczegółowe wyjaśnienie:
Odpowiedź:
[tex]e) (3 \sqrt{3})^{2} + (2 \sqrt{5})^{2} = {b}^{2} \\ 27 + 20 = {b}^{2} \\ b = \sqrt{47} [/tex]
[tex]d) {4}^{2} + {a}^{2} = (3 \sqrt{2}) ^{2} \\ 16 + {a}^{2} = 18 \\ a^{2} = 2 \\ a = \sqrt{2} [/tex]
[tex]f) {3}^{2} + {c}^{2} = ({ \sqrt{13} )}^{2} \\ 9 + {c}^{2} = 13 \\ {c}^{2} = 4 \\ c = 2[/tex]