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2 years ago

The table shows the distance Randy drove on one day of her vacation

1 answer:

4 0

Given the table showing the distance Randy drove on one day of her vacation as follows:

$\begin{tabular}
{|c|c|c|c|c|c|}
Time (h)&1&2&3&4&5\\[1ex]
Distance (mi)&55&110&165&220&275
\end{tabular}$

The rate at which she travels is given by

$rate= \frac{distance}{time} \\ \\ = \frac{55}{1} =55mi/h$

If Randy has driven for one more hour at the same rate, the number of hours she must have droven is 6 hrs and the total distance is given by

distance = 55 x 6 = 330 miles.

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vichka [17]

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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

-37b = 1

b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

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2 years ago