Lösung 4.1:4b

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
If we use the distance formula
If we use the distance formula
-
{{Displayed math||<math>d=\sqrt{(x-a)^2+(y-b)^2}</math>}}
+
{{Abgesetzte Formel||<math>d=\sqrt{(x-a)^2+(y-b)^2}</math>}}
to determine the distance between the points <math>(x,y) = (-2,5)</math> and <math>(a,b) = (3,-1)</math>, we get
to determine the distance between the points <math>(x,y) = (-2,5)</math> and <math>(a,b) = (3,-1)</math>, we get
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
d &= \sqrt{(-2-3)^2+(5-(-1))^2}\\[5pt]
d &= \sqrt{(-2-3)^2+(5-(-1))^2}\\[5pt]
&= \sqrt{(-5)^2+6^2}\\[5pt]
&= \sqrt{(-5)^2+6^2}\\[5pt]

Version vom 08:48, 22. Okt. 2008

If we use the distance formula

\displaystyle d=\sqrt{(x-a)^2+(y-b)^2}

to determine the distance between the points \displaystyle (x,y) = (-2,5) and \displaystyle (a,b) = (3,-1), we get

\displaystyle \begin{align}

d &= \sqrt{(-2-3)^2+(5-(-1))^2}\\[5pt] &= \sqrt{(-5)^2+6^2}\\[5pt] &= \sqrt{25+36}\\[5pt] &= \sqrt{61}\,\textrm{.} \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.1:4b“