Applications And Problem Solving

Applications And Problem Solving-55
\[\begin25t\ 20\left( \right)\end\] At this point a quick sketch of the situation is probably in order so we can see just what is going on.In the sketch we will assume that the two cars have traveled long enough so that they are 300 miles apart. That means that we can use the Pythagorean Theorem to say, \[ = \] This is a quadratic equation, but it is going to need some fairly heavy simplification before we can solve it so let’s do that.

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Let \(t\) be the amount of time it takes the first machine (Machine A) to stuff a batch of envelopes by itself.

That means that it will take the second machine (Machine B) \(t 1\) hours to stuff a batch of envelopes by itself.

Also, as we will see, we will need to get decimal answer to these and so as a general rule here we will round all answers to 4 decimal places.

So, we’ll let \(x\) be the length of the field and so we know that \(x 3\) will be the width of the field.

Now, we also know that area of a rectangle is length times width and so we know that, \[x\left( \right) = 75\] Now, this is a quadratic equation so let’s first write it in standard form.

\[\begin 3x & = 75\ 3x - 75 & = 0\end\] Using the quadratic formula gives, \[x = \frac\] Now, at this point, we’ve got to deal with the fact that there are two solutions here and we only want a single answer.

Now, from a physical standpoint we can see that we should expect to NOT get complex solutions to these problems. Two hours later the second car starts driving east at 20 mph.

Upon solving the quadratic equation we should get either two real distinct solutions or a double root. How long after the first car starts traveling does it take for the two cars to be 300 miles apart?

The word equation for this problem is then, \[\begin\left( \begin\ \end \right) \left( \begin\ \end \right) & = 1\ \ \left( \begin\ \end \right)\left( \begin\ \end \right) \left( \begin\ \end \right)\left( \begin\ \end \right) & = 1\end\] We know the time spent working together (2 hours) so we need to work rates of each machine. \[1 = \left( \right) \times \left( t \right)\hspace \Rightarrow \hspace = \frac\] \[1 = \left( \right) \times \left( \right)\hspace \Rightarrow \hspace = \frac\] Note that it’s okay that the work rates contain \(t\). Plugging into the word equation gives, \[\begin\left( \right)\left( 2 \right) \left( \right)\left( 2 \right) & = 1\ \frac \frac & = 1\end\] So, to solve we’ll first need to clear denominators and get the equation in standard form.

\[\begin\left( \right)\left( t \right)\left( \right) & = \left( 1 \right)\left( t \right)\left( \right)\ 2\left( \right) 2t & = t\ 4t 2 &= t\ 0 & = - 3t - 2\end\] Using the quadratic formula gives, \[t = \frac\] Converting to decimals gives, \[t = \frac = 3.5616\hspace\hspace\hspace\hspacet = \frac = - 0.5616\] Again, the negative doesn’t make any sense and so Machine A will work for 3.5616 hours to stuff a batch of envelopes by itself.


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