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3 years ago

12

In a lab, a substance was heated by 16°C over a period of 4 hours at a constant rate. What was the change in temperature each ho

ur?

2 answers:

Arada [10]3 years ago

5 0

Answer:

4° C per hour

Step-by-step explanation:

16/4 will give us the temperature change every hour

ollegr [7]3 years ago

5 0

The temperature change was 4 C per hour

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Phantasy [73]

Answer:

Step-by-step explanation:

$2\sqrt{3} sin^{2} \alpha -cos\alpha =0\\2\sqrt{3} (1-cos ^2 \alpha )-cos \alpha =0\\2\sqrt{3} -2\sqrt{3} cos^2 \alpha -cos \alpha =0\\2\sqrt{3} cos^2 \alpha +cos \alpha -2\sqrt{3} =0\\cos \alpha =\frac{-1 \pm\sqrt{1^2-4*2\sqrt{3}*(-2\sqrt{3}) } }{2*2\sqrt{3} } \\=\frac{-1 \pm\sqrt{1+48} }{4\sqrt{3} } \\=\frac{-1\pm7}{4\sqrt{3} } \\either~cos \alpha =\frac{6}{4\sqrt{3} }=\frac{\sqrt{3} }{2} \\=cos \frac{\pi }{6} ,cos(2\pi -\frac{\pi }{6} )\\=cos \frac{\pi}{6} ,cos \frac{11\pi }{6}$

$\alpha =2 n\pi+ \frac{\pi }{6} ,2n\pi +\frac{11\pi }{6} (general~solution)$

$or~cos\alpha =-\frac{7}{4\sqrt{3} } \\ \alpha =cos^{-1}( \frac{-7}{4\sqrt{3} } )$

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3 years ago

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Alex73 [517]

Answer:

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the given triangle is scalene triangles because the every angle measures different.

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