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olga2289 [7]
3 years ago
5

Two trains leave a town at the same time heading in opposite directions. One train is traveling 12 mph faster than the other. Af

ter two hours, they are 232 miles apart. What is the average speed of each train?
Mathematics
2 answers:
vichka [17]3 years ago
5 0

Answer:

Step-by-step explanation:

I'm going to paint you a picture in words of what this looks like on paper.  We have a train leaving from a point on your paper heading straight west.  We have another train leaving from the same point on your paper heading straight east.  This is the "opposite directions" that your problem gives you.  

Now let's make a table:

                 distance       =        rate      *       time

Train 1

Train 2

We will fill in this table from the info in the problem then refer back to our drawing.  It says that one train is traveling 12 mph faster than the other train.  We don't know how fast "the other train" is going, so let's call that rate r.  If the first train is travelin 12 mph faster, that rate is r + 12.  Let's put that into the table

               distance        =        rate        *        time

Train 1                                        r

Train 2                                    (r + 12)

Then it says "after 2 hours", so the time for both trains is 2 hours:

           

               distance        =        rate        *        time

Train 1                                        r           *          2

Train 2                                  (r + 12)       *          2

Since distance = rate * time, the distance (or length of the arrow pointing straight west) for Train 1 is 2r.  The distance (or length of the arrow pointing straight east) for Train 2 is 2(r + 12) which is 2r + 24.  The distance between them (which is also the length of the whole entire arrow) is 232.  Thus:

2r + 2r + 24 = 232 and

4r = 208 so

r = 52

This means that Train 1 is traveling 52 mph and Train 2 is traveling 12 miles per hour faster than that at 64 mph

Ahat [919]3 years ago
4 0

Answer: its A    ^3^❤

Step-by-step explanation:

that guy above me probably got yall confused

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We are looking for the length of a curve, also known as the arc length. Before we get to the formula for arc length, it would help if we re-wrote the equation in y = form.

We are given: 36 y^{2} =( x^{2} -4)^3
We divide by 36 and take the root of both sides to obtain: y = \sqrt{ \frac{( x^{2} -4)^3}{36} }

Note that the square root can be written as an exponent of 1/2 and so we can further simplify the above to obtain: y =  \frac{( x^{2} -4)^{3/2}}{6} }=( \frac{1}{6} )(x^{2} -4)^{3/2}}

Let's leave that for the moment and look at the formula for arc length. The formula is L= \int\limits^c_d {ds} where ds is defined differently for equations in rectangular form (which is what we have), polar form or parametric form.

Rectangular form is an equation using x and y where one variable is defined in terms of the other. We have y in terms of x. For this, we define ds as follows: ds= \sqrt{1+( \frac{dy}{dx})^2 } dx

As a note for a function x in terms of y simply switch each dx in the above to dy and vice versa.

As you can see from the formula we need to find dy/dx and square it. Let's do that now.

We can use the chain rule: bring down the 3/2, keep the parenthesis, raise it to the 3/2 - 1 and then take the derivative of what's inside (here x^2-4). More formally, we can let u=x^{2} -4 and then consider the derivative of u^{3/2}du. Either way, we obtain,

\frac{dy}{dx}=( \frac{1}{6})( x^{2} -4)^{1/2}(2x)=( \frac{x}{2})( x^{2} -4)^{1/2}

Looking at the formula for ds you see that dy/dx is squared so let's square the dy/dx we just found.
( \frac{dy}{dx}^2)=( \frac{x^2}{4})( x^{2} -4)= \frac{x^4-4 x^{2} }{4}

This means that in our case:
ds= \sqrt{1+\frac{x^4-4 x^{2} }{4}} dx
ds= \sqrt{\frac{4}{4}+\frac{x^4-4 x^{2} }{4}} dx
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ds= \sqrt{\frac{( x^{2} -2)^2 }{4}} dx
ds=  \frac{x^2-2}{2}dx =( \frac{1}{2} x^{2} -1)dx

Recall, the formula for arc length: L= \int\limits^c_d {ds}
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L= \int\limits^9_5 { \frac{1}{2} x^{2} -1 } \, dx = (\frac{1}{2}) ( \frac{x^3}{3}) -x evaluated from 9 to 5 (I cannot seem to get the notation here but usually it is a straight line with the 9 up top and the 5 on the bottom -- just like the integral with the 9 and 5 but a straight line instead). This means we plug 9 into the expression and from that subtract what we get when we plug 5 into the expression.

That is, [(\frac{1}{2}) ( \frac{9^3}{3}) -9]-([(\frac{1}{2}) ( \frac{5^3}{3}) -5]=( \frac{9^3}{6}-9)-( \frac{5^3}{6}-5})=\frac{290}{3}


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