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Zigmanuir [339]
2 years ago
11

In the United States a notable hang glider flight was a 303 mile, 8 1/2 hour flight from new york to Kansas. What was the averag

e rate during that time?
Mathematics
1 answer:
Vsevolod [243]2 years ago
4 0
There are several information's already given in the question. Based on the information's provided, the answer can be easily deduced.

Distance covered by the hang glider = 303 miles
Time taken by the hang glider for covering the distance = 8 1/2 hour
                                                                                        = 17/2 hours
Then
Average rate of flight = Distance/ Time
                                  = 303/(17/2) miles per hour
                                  = (303 * 2)/17 miles/hour
                                  = 606/17 miles/hour
                                  = 35.65 miles/hour
I hope the procedure is clear enough for you to understand.<span />
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1 year ago
Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 5y = sin(x)
vichka [17]

Answers:

a = -6/37

b = -1/37

============================================================

Explanation:

Let's start things off by computing the derivatives we'll need

y = a\sin(x) + b\cos(x)\\\\y' = a\cos(x) - b\sin(x)\\\\y'' = -a\sin(x) - b\cos(x)\\\\

Apply substitution to get

y'' + y' - 5y = \sin(x)\\\\\left(-a\sin(x) - b\cos(x)\right) + \left(a\cos(x) - b\sin(x)\right) - 5\left(a\sin(x) + b\cos(x)\right) = \sin(x)\\\\-a\sin(x) - b\cos(x) + a\cos(x) - b\sin(x) - 5a\sin(x) - 5b\cos(x) = \sin(x)\\\\\left(-a\sin(x) - b\sin(x) - 5a\sin(x)\right)  + \left(- b\cos(x) + a\cos(x) - 5b\cos(x)\right) = \sin(x)\\\\\left(-a - b - 5a\right)\sin(x)  + \left(- b + a - 5b\right)\cos(x) = \sin(x)\\\\\left(-6a - b\right)\sin(x)  + \left(a - 6b\right)\cos(x) = \sin(x)\\\\

I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.

The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero

a-6b = 0

a = 6b

At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)

-6a  -b = 1

-6(6b) - b = 1 .... plug in a = 6b

-36b - b = 1

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b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

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