27.10 Extra opgaven

Hoofdstukken > Wortels

$1,4 \cdot 4,2 = 5,88$	$2 \cdot 1,4 = 2,8$	$\frac{1}{2} \cdot 4,2 = 2,1$
$\frac{4,2}{1,4} = 3$	$\frac{1}{4,2} \approx 0,238$	$\frac{10}{1,4} \approx 7,1428$

$6 + 4 + 3 \sqrt{2} + 2 \sqrt{2} = 10 + 5 \sqrt{2}$
$\frac{1}{2} \sqrt{2} + \frac{1}{3} \sqrt{3} + \frac{1}{2} \sqrt{6} + \frac{2}{3} \sqrt{6} + \frac{1}{3} \sqrt{3} + \frac{4}{3} \sqrt{3} = \frac{1}{2} \sqrt{2} + 1 \frac{2}{3} \sqrt{3} + 1 \frac{1}{6} \sqrt{6}$
$\sqrt{2} + \frac{1}{2} \sqrt{2} - 2 \frac{1}{2} \sqrt{2} = ‐ \sqrt{2}$

$2 + 5 = 7$ cm , dus $70$ mm
$\sqrt{2^{2} + 1^{2}} + \sqrt{2^{2} + 4^{2}} = \sqrt{5} + \sqrt{20}$ cm, dus $67$ mm
$\sqrt{2^{2} + 2^{2}} + \sqrt{1^{2} + 4^{2}} = 2 \sqrt{2} + \sqrt{17}$ cm, dus $70$ mm
$\sqrt{3^{2} + 2^{2}} + 4 = \sqrt{13} + 4$ cm, dus $76$ mm

$\sqrt{\frac{1}{100}} = \frac{1}{10}$	$\sqrt{\frac{144}{100}} = \frac{12}{10} = 1,2$
$2^{4} = 16$	$8$
$\sqrt{4} = 2$	$\sqrt{\frac{225}{4}} = \frac{15}{2} = 7 \frac{1}{2}$
$\sqrt{4} = 2$	$\frac{1}{\sqrt{25}} = \frac{1}{5}$

Piet meet: $\sqrt{3^{2} + {0,2}^{2}} = \sqrt{9,04} \approx 3,0067$ meter. Het scheelt $6,7$ mm.

Linkerkolom:
$5 \sqrt{2} + 5 \sqrt{5} + 3 \sqrt{5} + 2 \sqrt{2} = 8 \sqrt{5} + 7 \sqrt{2}$
$\frac{1}{2} \sqrt{2} + \frac{1}{2} + \frac{1}{4} \sqrt{2} + \frac{1}{4} = \frac{3}{4} + \frac{3}{4} \sqrt{2}$

Rechterkolom:
$5 \sqrt{2} \cdot 5 \sqrt{5} \cdot 3 \sqrt{5} \cdot 2 \sqrt{2} = 5 \cdot 2 \cdot 5 \cdot 5 \cdot 3 \cdot 2 = 10 \cdot 10 \cdot 15 = 1500$
$\sqrt{\frac{2}{8}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

zijde vierkant is $\sqrt{10}$
diagonaal is $\sqrt{2} \cdot \sqrt{10} = \sqrt{20} = \sqrt{4} \cdot \sqrt{5} = 2 \sqrt{5}$

Lengte badvloer is $\sqrt{25^{2} + 1^{2}} = \sqrt{626} \approx 25,02$ m, dus $2$ cm langer.

Linkerkolom:
${(5 x)}^{2} = 50$
Manier 1:
$5 x = \sqrt{50} = 5 \sqrt{2} of 5 x = ‐ \sqrt{50} = ‐ 5 \sqrt{2}$
$x = \sqrt{2} of x = ‐ \sqrt{2}$
Manier 2:
$25 x^{2} = 50$
$x^{2} = 2$
$x = \sqrt{2} of x = ‐ \sqrt{2}$

${(\sqrt{2} x)}^{2} = 12$
Manier 1:
$\sqrt{2} x = \sqrt{12} of \sqrt{2} x = ‐ \sqrt{12}$
$x = \sqrt{6} of x = ‐ \sqrt{6}$
Manier 2:
$2 x^{2} = 12$
$x^{2} = 6$
$x = \sqrt{6} of x = ‐ \sqrt{6}$

Rechterkolom:
${(5 x)}^{2} = 20$
Manier 1:
$5 x = \sqrt{20} = 2 \sqrt{5} of 5 x = ‐ \sqrt{20} = ‐ 2 \sqrt{5}$
$x = \frac{2}{5} \sqrt{5} of x = ‐ \frac{2}{5} \sqrt{5}$
Manier 2:
$25 x^{2} = 20$
$x^{2} = \frac{20}{25}$
$x = \frac{2}{5} \sqrt{5} of x = ‐ \frac{2}{5} \sqrt{5}$

${(\sqrt{2} x)}^{2} = 10$
Manier 1:
$\sqrt{2} x = \sqrt{10} of \sqrt{2} x = ‐ \sqrt{10}$
$x = \sqrt{5} of x = ‐ \sqrt{5}$
Manier 2:
$2 x^{2} = 10$
$x^{2} = 5$
$x = \sqrt{5} of x = ‐ \sqrt{5}$

Twee zijden van $\sqrt{2^{2} + 4^{2}} = \sqrt{20} = 2 \sqrt{5}$ en een zijde van $\sqrt{2^{2} + 2^{2}} = \sqrt{8} = 2 \sqrt{2}$ .

Hoogte driehoek is $\sqrt{{(2 \sqrt{5})}^{2} - {\sqrt{2}}^{2}} = \sqrt{20 - 2} = \sqrt{18}$ .
Oppervlakte driehoek is $\sqrt{2} \cdot \sqrt{18} = \sqrt{36} = 6$ .

$4 \cdot 5 = 20$	$6 \cdot 5 = 20$	$25 \sqrt{6}$
$2 \sqrt{3}$	$6$	$5 \sqrt{3}$
$2 \sqrt{6} + \sqrt{6} = 3 \sqrt{6}$	$2 \sqrt{6} + 3 \sqrt{6} = 5 \sqrt{6}$	$\sqrt{1000} = 10 \sqrt{10}$
$1 \frac{1}{2} + 2 \frac{1}{2} = 4$	$\sqrt{4} = 2$	$\frac{\sqrt{33}}{3} = \frac{1}{3} \sqrt{33}$
$2 \sqrt{2}$	$\frac{1}{2} \sqrt{6}$	$\frac{3}{5} \sqrt{5}$

De diagonaal is de straal, dus heeft lengte $3$ ;
Pythagoras: $1^{2} + {hoogte}^{2} = 3^{2}$ $\to$ ${hoogte}^{2} = 8$ $\to$ $hoogte = \sqrt{8} = 2 \sqrt{2}$ (negatieve waarde voldoet niet)

$2 \cdot (1 + 2 \sqrt{2}) = 2 + 4 \sqrt{2} \approx 7,7$ cm, dus $77$ mm.

$1 \cdot 2 \sqrt{2} = 2 \sqrt{2} \approx 2,83$ cm², dus $283$ mm².

$\sqrt{2} \cdot \sqrt{6} = \sqrt{12} = 2 \sqrt{3}$

$\sqrt{{\sqrt{2}}^{2} + {\sqrt{6}}^{2}} = \sqrt{8} = 2 \sqrt{2}$

De verhouding is: $\sqrt{2} : \sqrt{6} : 2 \sqrt{2}$ . Als je door $\sqrt{2}$ deelt, krijg je: $1 : \sqrt{3} : 2$ .
$60 °$ , want $∠ A C D = 30 °$ , want driehoek $A C D$ is een $30 ‐ 60 ‐ 90$ -driehoek, blijkt uit de verhouding van de zijden van die driehoek.

De schuine zijde van de blauwe driehoek is $\frac{1}{2} \cdot 18,01 = 9,005$ meter
$x^{2} + 9^{2} = {9,005}^{2}$ geeft: $x \approx 0,30$ m, dat is $30$ cm.

$\frac{2 \sqrt{3}}{\sqrt{2}} = \frac{2 \sqrt{3} \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{2 \sqrt{6}}{2} = \sqrt{6}$

$\sqrt{6} \cdot \sqrt{6} = 6$

Zie plaatje.
$C S = D S = \frac{6}{\sqrt{2}} = 3 \sqrt{2}$
$A S = B S = \sqrt{3} \cdot 3 \sqrt{2} = 3 \sqrt{6}$
$A D = B C = 2 \cdot 3 \sqrt{2} = 6 \sqrt{2}$
$A B = 3 \sqrt{6} \cdot \sqrt{2} = 6 \sqrt{3}$

Zijde gelijkzijdige driehoek is $\sqrt{6}$ .
Lengte hoogtelijn gelijkzijdige driehoek is $\sqrt{{\sqrt{6}}^{2} - {(\frac{1}{2} \sqrt{6})}^{2}} = \sqrt{4 \frac{1}{2}} = \sqrt{\frac{18}{4}} = \frac{3 \sqrt{2}}{2} = 1 \frac{1}{2} \sqrt{2}$ .
Oppervlakte zeshoek is $6 \cdot \frac{1}{2} \sqrt{6} \cdot 1 \frac{1}{2} \sqrt{2} = 9 \sqrt{3}$ .

$p = 10 \cdot 6^{3} = 2160$ kW/u ; $p = 10 \cdot 10^{3} = 10.000$ kW/u

$2^{3} = 8$ keer

$1000 = 10 \cdot w^{3}$
$w^{3} = 100$
$w = \sqrt[3]{100} \approx 4,6$ m/s

$w = \sqrt[3]{\frac{p}{10}} = \frac{1}{10} \sqrt[3]{100 p}$

$\sqrt[3]{1000} = 10$	$\sqrt[3]{\frac{1}{8}} = \frac{1}{2}$
$\sqrt[3]{\frac{64}{1000}} = \frac{4}{10} = \frac{2}{5}$	$7$

Per speler $\frac{60.000}{\sqrt{16}} = 15.000$ gulden, totaal: $16 \cdot 15.000 = 240.000$ gulden.

$\frac{60.000}{\sqrt{10}} \cdot 10 \approx 189.737$ gulden

$b = \frac{60.000}{\sqrt{n}} \cdot n = 60.000 \sqrt{n}$

Zie plaatje.

$i = \frac{1}{3} r^{3}$

$30$ , $\sqrt[3]{900}$

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

$\sqrt{80} = \sqrt{16} \cdot \sqrt{5} = 4 \sqrt{5}$