Wich statement is true about a line and a point?

Mathematics

2 answers:

barxatty [35]4 years ago

6 0

<span>A point is a location and a line has many points located on it.</span>

Bogdan [553]4 years ago

3 0

A line are two rays in opposite directions and go on forever, and a point is a location on any type of line segment or shape ( see pictures for examples)

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$\sin \left(\cos^{-1} \left(\frac{5}{6}} \right) \right) \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)-\cos\left(\cos^{-1} \left(\frac{5}{6} \right) \right) \sin \left( \tan^{-1} \left(\frac{1}{2} \right) \right)$

Using the Pythagorean identities $\sin^{2} x+\cos^{2} x=1$ and $\tan^{2} x+1=\sec^{2} x$ ,

$1) \sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\cos^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\frac{25}{36}=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{11}{36}\sin \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{\sqrt{11}}{6}$

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$\cos^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\frac{4}{5}+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{5}\\\left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$

This means we can write the original expression as:

$\left(\frac{\sqrt{11}}{6} \right) \left(\frac{2\sqrt{5}}{5} \right)-\left(\frac{5}{6} \right) \left(\frac{\sqrt{5}}{5} \right)\\=\frac{2\sqrt{11}\sqrt{5}}{30}-\frac{5\sqrt{5}}{30}\\=\boxed{\frac{\sqrt{5}(2\sqrt{11}-5)}{30}}$