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Vectors
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Re: Vectors
by Eigenvalue » Wed Aug 26, 2026 9:33 pm
A. [tex]\sqrt{(3)²+(4)²}[/tex]=5
B.[tex]\sqrt{(-4)²+(2)²}[/tex]=2[tex]\sqrt{5}[/tex]
C.[tex]\sqrt{(-6)²+(-6)²}[/tex]=6[tex]\sqrt{2}[/tex]
D.[tex]\sqrt{ (\frac{3}{4})²+( \frac{-1}{2})² }[/tex]=[tex]\sqrt{ \frac{13}{16} }[/tex]=[tex]\frac{ \sqrt{13} }{4}[/tex]
B.[tex]\sqrt{(-4)²+(2)²}[/tex]=2[tex]\sqrt{5}[/tex]
C.[tex]\sqrt{(-6)²+(-6)²}[/tex]=6[tex]\sqrt{2}[/tex]
D.[tex]\sqrt{ (\frac{3}{4})²+( \frac{-1}{2})² }[/tex]=[tex]\sqrt{ \frac{13}{16} }[/tex]=[tex]\frac{ \sqrt{13} }{4}[/tex]
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Re: Vectors
by Eigenvalue » Wed Aug 26, 2026 9:34 pm
Additional sketch:
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Re: Vectors
by nycmath » Thu Aug 27, 2026 4:46 am
Eigenvalue wrote:A. [tex]\sqrt{(3)²+(4)²}[/tex]=5
B.[tex]\sqrt{(-4)²+(2)²}[/tex]=2[tex]\sqrt{5}[/tex]
C.[tex]\sqrt{(-6)²+(-6)²}[/tex]=6[tex]\sqrt{2}[/tex]
D.[tex]\sqrt{ (\frac{3}{4})²+( \frac{-1}{2})² }[/tex]=[tex]\sqrt{ \frac{13}{16} }[/tex]=[tex]\frac{ \sqrt{13} }{4}[/tex]
Correct. A nice geometric interpretation of vectors.
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