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pentagon [3]
3 years ago
9

Jay traveled from Miami to Jacksonville, a distance of 320 miles. He left Miami at 8:00 AM, stopped for lunch for 30 minutes, an

d stopped for two 15 minute breaks at rest areas. He arrived in Jacksonville at 2:15 PM. What was his average rate of speed during his travel time?
Mathematics
1 answer:
fredd [130]3 years ago
5 0
Distance Rate times Time
320= R 5.25
R= 60.9
R= 61 mph
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Step-by-step explanation:

<u><em>Explanation</em></u>

Given expression

           =     \frac{(4g^{3} h^{2}k^{4} )^{3}  }{8g^{3}h^{2}  } - (h^{5} k^{3} )^{5}

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