$f\left(4\right)=122880$

$x={\left(\frac{20000}{120}\right)}^{\frac{1}{5}}\approx 2.78$

By $4^5=1024$

$0.01$

$r=10$ gives $s^2=1000$ and therefore $s\approx 31.6$ km/h.

$s=10\sqrt{r}$
So $s$ is directly proportional to $s^{\frac{1}{2}}$.

If $r=100$, then $s=10\sqrt{100}=100$ km/h. If $r=50$, then $s=10\sqrt{50}\approx \mathrm{70,7}$ km/h. The statement is false.

$V\left(2\right)=2^3=8$ cm³.
$V\left(6\right)=6^3=216$ cm³.

The volume is $3^3=27$ times larger.

$r^3=50$, so $r=\sqrt[3]{50}\approx 3.7$ cm.

$V=r^3$

$r=\sqrt[3]{V}$

Own answer.

$A\left(3\right)=54$ cm² and $A\left(6\right)=216$ cm².

The surface area becomes $4$ times as large.

$r=\sqrt{\frac{500}{6}}\approx 9.13$ cm.

$r=\sqrt{\frac{A}{6}}$

$G=7.9r^3$

$A=6r^2$

From $G=7.9r^3$ follows $r={\left(\frac{1}{7.9}\right)}^{\frac{1}{3}}\cdot G^{\left(\frac{1}{3}\right)}\approx 0.502G^{\frac{1}{3}}$.
Therefore $A=6r^2\approx 6\cdot {0.502}^2\cdot G^{\left(\frac{2}{3}\right)}\approx 1,51G^{\frac{2}{3}}$ and $c\approx 1.51$.

Do it.