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Arlecino [84]
3 years ago
8

6. Let's say the countdown for a space shuttle launch has begun. At "T minus 27 hours" (that is, 27 hours before launch), a prob

lem occurs. If the technicians have not fixed the problem by T minus 8 hours, the launch will have to be scratched. How much time do the technicians have to correct the problem?
Mathematics
1 answer:
nataly862011 [7]3 years ago
5 0

Answer: the technicians have 19 hours to correct the problem.

Step-by-step explanation:

The countdown started at "T-minus 27 hours" that means that the clock starts ticking backwards and if an hour passes away the lime left for the launch will be 26 hours and the limit the technicians have is T minus 8 hours so the problem can be solved with a simple substraction 27 hours minus 8 hours equals 19 hours.

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otez555 [7]
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3 years ago
Given Sine (t) = StartFraction 7 Over 25 EndFraction for StartFraction pi Over 2 EndFraction less-than t less-than pi, what is C
lys-0071 [83]

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Step-by-step explanation:

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4 years ago
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Let A(t)A(t) be the area of a circle with radius r(t)r(t), at time tt in min. Suppose the radius is changing at the rate of drdt
torisob [31]

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Step-by-step explanation:

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3 years ago
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