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Cloud [144]
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?
Mathematics
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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2root3 sin^<a href="/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="ac9eec">[email&#160;protected]</a> - <a href
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} } )

5 0
3 years ago
Help me and pls show ur steps
Alex73 [517]

Answer:

Scalene triangle

Step-by-step explanation:

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

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3 years ago
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Select D $660 is the answer
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Is 25/1000 grater than or less than 0.205
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Equal to, I think. Hope this helps.... If not sorry.
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Fund the separate equation and angles between them
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