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Anna11 [10]
3 years ago
10

An airplane left Miami, FL. At the same time another plane left Santiago, Chile. The two planes flew toward each other at rates

of 625 mph and 575 mph. If Miami and Santiago are 4200 miles apart, how long will it take until the planes pass each other?
Mathematics
1 answer:
Evgesh-ka [11]3 years ago
7 0
To solve the above question we use the concept of relative speed.
The rates of the two planes are 625 mph and 575 mph. 
Relative speed will be:
[speed of plane A]+[speed of plane B]
=625+575
=1,200 mph
Distance=4200 miles
Thus the time taken for them to meet will be:
Time=distance/speed
=4200/1200
=3.5 hours
We therefore conclude that the planes met after 3.5 hours

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Let u=x^2-4 and v=4x-5. By the product rule,

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By the power rule, we have (u^5)'=5u^4 and (v^4)'=4v^3, but u,v are functions of x, so we also need to apply the chain rule:

\dfrac{\mathrm d(u^5)}{\mathrm dx}=5u^4\dfrac{\mathrm du}{\mathrm dx}

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\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=10xu^4v^4+16u^5v^3

Replace u,v to get everything in terms of x:

\dfrac{\mathrm d((x^2-4)^5(4x-5)^4)}{\mathrm dx}=10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3

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