$5t^2=80$ $\to$ $t^2=16$ $\to$ $t=4$ $\to$ $v=10\cdot 4=40$ m/s

$10t=15$ $\to$ $t=\mathrm{1,5}$ $\to$ $s=5\cdot {\left(\mathrm{1,5}\right)}^2=\mathrm{11,25}$ m

Ketting $s\to t\to v$ bij vraag a;
Ketting $v\to t\to s$ bij vraag b

$10t=40$ $\to$ $t=\frac{40}{10}$ $\to$ $s=5\cdot {\left(\frac{40}{10}\right)}^2$

$v=10t$ $\to$ $t=\frac{v}{10}$ $\to$ $s=5\cdot {(\frac{v}{10})}^2$ ($=\mathrm{0,05}\cdot v^2$)

$s=5t^2$ $\to$ $t=\sqrt{\frac{1}{5}s}$ $\to$ $v=10\cdot \sqrt{\frac{1}{5}s}$
($\to$ $v=\sqrt{100}\cdot \sqrt{\frac{1}{5}s}=\sqrt{100\cdot \frac{1}{5}s}=\sqrt{20s}$)

$P=2\text{π}\cdot r$; $A=\text{π}\cdot r^2$

$2\text{π}r=6$ $\to$ $r=\frac{3}{\text{π}}$ $\to$ $A=\text{π}\cdot {\left(\frac{3}{\text{π}}\right)}^2=\frac{9}{\text{π}}\approx \mathrm{2,86}$

$P=2\text{π}\cdot r$ $\to$ $r=\frac{P}{2\text{π}}$ $\to$ $A=\text{π}\cdot {\left(\frac{P}{2\text{π}}\right)}^2=\frac{\text{π}}{4{\text{π}}^2}\cdot P^2=\frac{1}{4\text{π}}\cdot P^2\approx \mathrm{0,080}\cdot P^2$

$A=\text{π}\cdot r^2$ $\to$ $r=\sqrt{\frac{1}{\text{π}}\cdot A}=\frac{1}{\sqrt{\text{π}}}\cdot \sqrt{A}$ $\to$ $P=2\text{π}\cdot \frac{1}{\sqrt{\text{π}}}\cdot \sqrt{A}=2\sqrt{\text{π}}\cdot A^{\frac{1}{2}}\approx \mathrm{3,545}\cdot A^{\frac{1}{2}}$

$G={\left(\frac{11}{\mathrm{9,1}}\right)}^6\approx 3$ gram

$G=5\cdot {10000}^{\frac{4}{3}}\approx \mathrm{1077217,345...}$ $\to$ $T=\mathrm{9,1}\cdot {\left(\mathrm{1077217,345...}\right)}^{\frac{1}{6}}\approx 92$ dagen

$T=\mathrm{9,1}\cdot G^{\frac{1}{6}}=\mathrm{9,1}\cdot {\left(5\cdot E^{\frac{4}{3}}\right)}^{\frac{1}{6}}=\mathrm{9,1}\cdot 5^{\frac{1}{6}}\cdot E^{\frac{4}{3}\cdot \frac{1}{6}}\approx \mathrm{11,90}\cdot E^{\frac{2}{9}}$

$x=3$ $\to$ $y=3^2=9$ $\to$ $z=9+7=16$ $\to$ $u=\sqrt{16}=4$

$y=x^2$ $\to$ $z=x^2+7$ $\to$ $u=\sqrt{x^2+7}$

$y\ge 0$; $z\ge 7$; $u\ge \sqrt{7}$

$u=\frac{5}{y+2}=\frac{5}{3t^2+2}$

-

$y=3t^2$, dus $y\ge 0$ $\to$ $z=\frac{1}{y+2}$, dus $0<z\le \frac{1}{2}$ $\to$ $u=5z$, dus $0<u\le 2\frac{1}{2}$;
Bereik $u$: $\langle 0\mathrm{,2}\frac{1}{2}]$

$c=a^2-1$; bereik: $c\ge \mathrm{‐}1$

$c=a$; bereik: alle waarden

$c=\sqrt{45-4a}$; bereik: $c\ge 0$ (domein: $a\le 11\frac{1}{4}$)

$c=9{\sin }^2(a)-1$; bereik: $\mathrm{‐}1\le c\le 8$

$c=\sin (3a+1)$; bereik: $\mathrm{‐}1\le c\le 1$

$c=3\sin (a)+1$; bereik: $\mathrm{‐}2\le c\le 4$

$x=\frac{300}{51}$

Gelijkvormige driehoeken geeft: $\frac{x}{2}=\frac{300}{a+2}$ (beide zijden keer 2)$\to$ $x=\frac{600}{a+2}$

$72$ km/u = $20$ m/s, dus de afstand tot de ruit neemt elke seconde met $20$ meter toe: $a=50+20t$

$x=\frac{600}{(50+20t)+2}=\frac{600}{20t+52}$

$0<x\le \frac{600}{52}$, ofwel $0<x\le \frac{150}{13}$

Eerste formule: $G=\frac{680\cdot D^4}{L^2}=680\cdot D^4\cdot L^{\mathrm{‐}2}$;
Tweede formule: $L=\frac{G}{D^2}$;
Combineren: $L=\frac{680\cdot D^4\cdot L^{\mathrm{‐}2}}{D^2}=680\cdot D^2\cdot L^{\mathrm{‐}2}$ $\to$ $L^3=680\cdot D^2$
$\to$ $L={\left(680\cdot D^2\right)}^{\frac{1}{3}}=\sqrt[3]{680}\cdot D^{\frac{2}{3}}\approx \mathrm{8,79}\cdot D^{\frac{2}{3}}$

$G\cdot L^2=680\cdot {\left(D^2\right)}^2=680\cdot {\left(G\cdot L^{\mathrm{‐}1}\right)}^2=680\cdot G^2\cdot L^{\mathrm{‐}2}$ $\to$ $L^4=680\cdot G$
$\to$ $L={\left(680\cdot G\right)}^{\frac{1}{4}}=\sqrt[4]{680}\cdot G^{\frac{1}{4}}\approx \mathrm{5,11}\cdot G^{\frac{1}{4}}$

$G=\sqrt[3]{680}\cdot D^{\frac{8}{3}}\approx \mathrm{8,79}\cdot D^{\frac{8}{3}}$

$L\approx 228$ cm, dus $23$ dm;
$D\approx 132$ cm, dus $13$ dm