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schepotkina [342]
3 years ago
11

One cold night the temperature was 0 degrees at 8 p.m. The temperature dropped at a constant rate from 8 p.m. to midnight. At mi

dnight the temperature was −16.8 degrees.
What was the temperature at 9 p.m.?

Enter your answer in the box.
Mathematics
2 answers:
arsen [322]3 years ago
7 0

The correct answer is 4.2. This is because  -16.8 / 4 is -4.2 degrees.

  • Hoped this helped! :)

^-^

bearhunter [10]3 years ago
3 0
-4.2 degrees. What you would do is from 8-12 (midnight) is 4 hours. So it dropped -16.8 degrees in those 4 hours so -16.8 / 4 is -4.2 degrees. So it dropped -4.2 degrees every hour but it was also -4.2 degrees at 9:00 p.m.
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Answer:

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Step-by-step explanation:

The expression is

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antiseptic1488 [7]

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Step-by-step explanation:

4 0
3 years ago
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Serggg [28]

Answer:

\boxed{\sf \huge \boxed{ \sf ?} = 6}

Step-by-step explanation:

\sf Solve  \: for \: \boxed{ \sf ?} :  \\  \sf \implies \frac{\boxed{ \sf ?}}{5}  \times  \frac{10}{8}  = 1.5 \\  \\  \sf  \implies \frac{\boxed{ \sf ?} \times 10}{5 \times 8}  = 1.5 \\  \\  \sf \frac{10}{5}  =  \frac{ \cancel{5} \times 2}{ \cancel{5}}  = 2 :  \\  \sf \implies  \frac{ \boxed{ \sf 2} \times \boxed{ \sf ?} }{8}  = 1.5 \\  \\  \sf \frac{2}{8}  =  \frac{ \cancel{2}}{ \cancel{2}  \times 4}  =  \frac{1}{4}  :  \\  \sf \implies \frac{\boxed{ \sf ?}}{ \boxed{ \sf 4}}  = 1.5 \\  \\  \sf Multiply  \: both  \: sides \:  of \:  \frac{\boxed{ \sf ?}}{4}  = 1.5 \: by \: 4 :  \\  \sf \implies \frac{4 \times \boxed{ \sf ?}}{4}  = 4 \times 1.5 \\  \\  \sf \frac{4 \times \boxed{ \sf ?}}{4} =  \frac{ \cancel{4}}{ \cancel{4}}  \times \boxed{ \sf ?} = \boxed{ \sf ?} :  \\  \sf  \implies \boxed{\boxed{ \sf ?}} = 4 \times 1.5 \\  \\  \sf 4 \times 1.5 = 6 :  \\  \sf \implies \boxed{ \sf ?} = 6

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3 years ago
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