Question

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Answer 1

Răspuns:

1.B; 2.C; 3.C

Explicație pas cu pas:

1.

$2\pi \ r = 12.6 \ cm$

$r = \dfrac{12.6}{2\pi} \implies \bf r \approx 2 \ cm$

2.

$\dfrac{G}{d} = \dfrac{5}{2} \iff \dfrac{G}{2r} = \dfrac{5}{2}$

$\dfrac{G}{2 \cdot 3} = \dfrac{5}{2} \implies \bf G = 15 \ dm$

3.

raza bazei este ½×diagonala pătratului

$\mathcal{A}_{4} = \ell^{2} = 8 \implies \ell = 2 \sqrt{2}$

$d = \ell \sqrt{2} \iff 2r = \ell \sqrt{2} \implies r = \dfrac{\ell \sqrt{2}}{2}$

$r = \dfrac{2 \sqrt{2} \cdot \sqrt{2}}{2} \implies \bf r = 2 \ dm$