1

a

Dan $x^{3} = \frac{1}{3} \cdot y^{2} \Leftrightarrow x = {(\frac{1}{3})}^{\frac{1}{3}} \cdot y^{\frac{2}{3}} \approx 0,69 y^{\frac{2}{3}}$

b

$x^{2} + 3 = 3 y^{3} \Leftrightarrow x^{2} = 3 y^{3} - 3 \Leftrightarrow x = \sqrt[]{3 y^{3} - 3}$ , je moet wel de wortel kunnen trekken, dus $y \geq 1$

c

$\sqrt[3]{x + 2} = y^{2} \Leftrightarrow x + 2 = y^{6} \Leftrightarrow x = y^{6} - 2$

2

a

Als $t = 2$ , dan $y = 12$ en $z = \frac{1}{14}$ , dus $u = \frac{5}{14}$

b

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

c

$0 < u \leq 2 \frac{1}{2}$

3

a

$x \to \frac{1}{x} \to {(\frac{1}{x})}^{2} \to {(\frac{1}{x})}^{2} + 2$

b

$y > 2$

c

$F_{3} \to F_{2} \to F_{1} : y = {(2 + \frac{1}{x})}^{2}$ ; $y \geq 0$
$F_{2} \to F_{1} \to F_{3} : y = \frac{1}{{(x + 2)}^{2}}$ ; $y \geq 0$
$F_{2} \to F_{3} \to F_{1} : y = {(\frac{1}{x + 2})}^{2}$ ; $y \geq 0$
$F_{1} \to F_{2} \to F_{3} : y = \frac{1}{x^{2} + 2}$ ; $0 < y \leq \frac{1}{2}$
$F_{1} \to F_{3} \to F_{2} : y = 2 + \frac{1}{x^{2}}$ ; $y > 2$

d

$F_{3} \to F_{1} \to F_{2}$ is hetzelfde als $F_{1} \to F_{3} \to F_{2}$ en
$F_{2} \to F_{1} \to F_{3}$ is hetzelfde als $F_{2} \to F_{3} \to F_{1}$ .

4

a

Dan $\sqrt{t} = 1 \frac{1}{2}$ , dus $t = 2 \frac{1}{4}$

b

$\sqrt{T - 12} = \frac{1}{6,9} \cdot D \approx 0,1449 \dots \cdot D$ , dus $T - 12 \approx {(0,1449 \dots \cdot D)}^{2} \approx 0,021 \dots \cdot D^{2}$ , dus $T \approx 12 + 0,02 \cdot D^{2}$

c

$H = \frac{6,7 \cdot k^{1,35}}{R}$ , dus $k^{1,35} = \frac{1}{6,7} H R = 0,149 \dots H R$ , dus $k \approx {0,149}^{\frac{1}{1,35}} {(H R)}^{\frac{1}{1,35}} \approx 0,244 {(H R)}^{0,74}$

d

In $S = 1000 \cdot R^{- 3}$ voor $R = {(100 + x^{2})}^{\frac{1}{2}}$ invullen geeft: $S = 1000 \cdot {(100 + x^{2})}^{\frac{1}{2} \cdot ‐ 3}$ , dus $S = 1000 \cdot {(100 + x^{2})}^{‐ 1,5}$

5

a

${(‐ 2 x)}^{3} \cdot {(2 x)}^{0,5} = ‐ 8 x^{3} \cdot 2^{0,5} \cdot x^{0,5} = ‐ 11,31 \cdot x^{3,50}$

b

$\frac{\sqrt[3]{5 x^{2}}}{\sqrt{6 x}} = 5^{\frac{1}{3}} \cdot x^{\frac{2}{3}} \cdot 6^{‐ \frac{1}{2}} \cdot x^{‐ \frac{1}{2}} = 0,70 \cdot x^{0,17}$

c

$\frac{{(‐ x)}^{4} \cdot {(‐ x)}^{3}}{‐ x^{2}} = \frac{x^{4} \cdot ‐ x^{3}}{‐ x^{2}} = x^{5,00}$