Antwort 4.2:1

Aus Online Mathematik Brückenkurs 1

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(Ny sida: {| width="100%" cellspacing="10px" |a) |width="50%" | <math>x=13\cdot\tan {27 ^\circ} \approx 6{,}62</math> |b) |width="50%" | <math>x=25\cdot\cos {32 ^\circ} \approx 21{,}2</math> |- |c) ...)
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{| width="100%" cellspacing="10px"
{| width="100%" cellspacing="10px"
|a)
|a)
-
|width="50%" | <math>x=13\cdot\tan {27 ^\circ} \approx 6{,}62</math>
+
|width="50%" | <math>x=13\cdot\tan {27 ^\circ} \approx 6\textrm{.}62</math>
|b)
|b)
-
|width="50%" | <math>x=25\cdot\cos {32 ^\circ} \approx 21{,}2</math>
+
|width="50%" | <math>x=25\cdot\cos {32 ^\circ} \approx 21\textrm{.}2</math>
|-
|-
|c)
|c)
-
|width="50%" | <math>x=\displaystyle\frac{14}{\tan {40 ^\circ}} \approx 16{,}7</math>
+
|width="50%" | <math>x=\displaystyle\frac{14}{\tan {40 ^\circ}} \approx 16\textrm{.}7</math>
|d)
|d)
-
|width="50%" | <math>x=\displaystyle\frac{16}{\cos {20 ^\circ}} \approx 17{,}0</math>
+
|width="50%" | <math>x=\displaystyle\frac{16}{\cos {20 ^\circ}} \approx 17\textrm{.}0</math>
|-
|-
|e)
|e)
-
|width="50%" | <math>x=\displaystyle\frac{11}{\sin {35 ^\circ}} \approx 19{,}2</math>
+
|width="50%" | <math>x=\displaystyle\frac{11}{\sin {35 ^\circ}} \approx 19\textrm{.}2</math>
|f)
|f)
-
|width="50%" | <math>x=\displaystyle\frac{19}{\tan {50 ^\circ}} \approx 15{,}9</math>
+
|width="50%" | <math>x=\displaystyle\frac{19}{\tan {50 ^\circ}} \approx 15\textrm{.}9</math>
|}
|}

Version vom 10:56, 8. Sep. 2008

a) \displaystyle x=13\cdot\tan {27 ^\circ} \approx 6\textrm{.}62 b) \displaystyle x=25\cdot\cos {32 ^\circ} \approx 21\textrm{.}2
c) \displaystyle x=\displaystyle\frac{14}{\tan {40 ^\circ}} \approx 16\textrm{.}7 d) \displaystyle x=\displaystyle\frac{16}{\cos {20 ^\circ}} \approx 17\textrm{.}0
e) \displaystyle x=\displaystyle\frac{11}{\sin {35 ^\circ}} \approx 19\textrm{.}2 f) \displaystyle x=\displaystyle\frac{19}{\tan {50 ^\circ}} \approx 15\textrm{.}9