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College Algebra's Blog

May 02, 2008

We get alot of students searching for homework help at our site, that are having trouble with algebra. Whether the textbook is confusing or they just need some extra help, we have a bunch of resources to help students get the homework help they need:

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August 09, 2007

Dear Professor Cram:

One automobile starts out from a town at 8am and travels at an average speed rate of 35mph. Three hours later a second automobile starts out to overtake the first. If the second automobile travels at an average rate of 55mph how long before it overtakes the first?

Holly F., Madison Area Technical College

Thank you for your interest in College-Cram, Holly, and thanks for your question.

Your question is a variant of the classic "A train leaves Chicago going..." question you see on tests like the SAT, but don't let that alarm you. These vary with direction and whether we are solving for how long it takes in time or distance. In this case, you have two vehicles travelling the same direction, with one leaving first and a second having to go faster in order to catch up and we need to know the time it takes.

The basic formulas relate time and velocity (speed as a rate, such as miles per hour) with distance travelled:

    Distance = velocity x time

(For example, at 60 mph for two hours you go 120 miles.)

Now, back to your problem. In order to solve these two simultaneous equations we have to see when the distance travelled by both are the same, so we will need a formula representation for each one, and then set them equal to each other and solve for time. We know the second vehicle left three hours after the first, so we know the time of the second one is three hours less than the time the first one travels.

For the first vehicle, Distance = t hours x 35 mph. For the second vehicle, Distance = t-3 hours x 55 mph. At the point the second overtakes the first, the distance is the same, so let's make them equal:

    t x 35 = (t-3) x 55

We'll want to solve for t-3 (how long the second vehicle took to catch up) in hours.

  • 35t = 55t - 165
  • 165 = 55t - 35t
  • 165 = 20t
  • 8.25 = t
  • 5.25 = t-3

It took 5.25 hours for the second car to catch up.

As a check for your values, see if both cars really travelled the same distance. Car 1 went 8.25 hours at 35 mph, or 288.75 miles. Car 2 went 5.25 hours at 55 mph or 288.75 miles. It worked!

I hope this helps. Let us know if you need anything else.

Good Studying,

Professor Cram

Keywords: Algebra, Algebra Problem, equation, math, time, Two Cars Algebra Problem, velocity, word problem

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Dear Professor Cram:

First I must tell you, that I am so happy to have found your site, it has given me hope that I can learn what I need to advance in my college studies. Here is my question: how do you find the square root of a radicand?

D. B., University of Louisiana

Thank you for your feedback, we love to hear from people we are helping.

To answer your question, there are several ways of finding a square root. The easiest is to use a calculator -- just enter the radicand, press the square root key, and *bam* there's your answer. Barring a calculator, there is a long, drawn-out method for calculating square roots that looks a bit like long division and hasn't been taught in schools in a long, long time.

For a quick and dirty approximate answer, the Babylonian Method is your best bet. Basically, you guess at the answer (let's call your guess 'A'.) Divide your radicand by 'A' and average that with 'A', with that answer being your new 'A'. Repeat until you get the same answer two times in a row.

For example, say you're looking for the square root of 69. We'll guess 8. Do the averaging (I'm only going to 3 decimal places):

  • 69/8 = 8.625
  • A = (8 + 8.625)/2, or 8.313

Now repeat, with 8.313 as the new guess:

  • 69/8.313 = 8.300
  • A = (8.313 + 8.300)/2, or 8.307

And again, with 8.307 as the new guess:

  • 69/8.307 = 8.306
  • A = (8.307 + 8.306)/2, or 8.307

We got 8.307; if you use your calculator, you'll see that the square root of 69 is in fact 8.3066238629180748525842627449075. Not too far off, those Babylonians.

Good Studying,

Professor Cram

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Dear Professor Cram:

Find the slope of the line described by the equation: y + x = -3(14 - x/3)

KG, US Navy

KG, thank you for this algebra question. This is a pretty straight-forward question, so here's how we tackle it:

The Slope-Intercept form of the equation of a line is y=mx+b, where m is the slope and b is the y-intercept. If we can put your equation into this format, then we should be able to find m.

  • Y + x = -3(14 - x/3)
  • Y + x = (-3)(14) - (-3)(x/3)
  • Y + x = -42 - (-3x)
  • Y + x = -42 + 3x
  • Y = 3x -x -42
  • Y = 2x - 42

Thus, the slope is 2 and the y-intercept is -42.

Good studying,

Professor Cram

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Dear Professor Cram:

I am taking a distance learning college algebra course, so I am left on my own to some extent to learn the material. I'm just beginning to work problems like: (5x²/6y²)³

The book I'm using doesn't give me an example of this type of problem. Where do I start?

Marilyn D., Lancaster, CA

Thank you for using College-Cram.com and for your Algebra question.

The best way to expand this rational expression is to recall that raising anything to the power of three means to multiply it by itself three times. Thus, your expression becomes:

  • (5x²/6y²)(5x²/6y²)(5x²/6y²)

Since these are fractions, let's reorganize it so we have the product of the numerators (numbers on top) divided by the product of the denominators (numbers on the bottom):

  • ((5x²)(5x²)(5x²)) / ((6y²)(6y²)(6y²))

To multiply these out, multiply the coefficients and add the exponents:

  • 125x6/216y6

This result is equal to your original rational expression, only expanded.

I hope this helps. Let us know if you need anything else.

Good studying,

Professor Cram

Keywords: algebra, college algebra, Expanding a Rational Expression, exponents, polynomial, Rational Expression

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Dear Professor Cram:

600 children attended summer camp, 58% girls, 42% boys (overall). 25% attended dance camp. If dance camp was exclusively attended by girls, how many girls attended dance camp? The choices are a)150, b)95, c)102, d)118, and e)123. The answer is supposed to be A, but I can't get it. I took 25% of 600 and got 123, then took 58% of 123 but it's not an option. Please help! There must be some underlying factor I'm missing...

Kaye B, Riverdale

Thanks for your question, Kaye. This is a classic word-problem trick -- giving too much information. Let's review the facts first:

  • 1. 600 children attended summer camp
  • 2. 58% of summer camp attendees are girls
  • 3. 42% of summer camp attendees are boys
  • 4. 25% of summer camp attendees went to dance camp
  • 5. All dance camp attendees are girls

The question asks how many dance camp attendees are girls. Well, fact #5 says that all the attendees were girls, so we just need to know how many attendees went to dance camp.

Facts #2 and #3 don't tell us that, so they are just extra information intended to throw us off the answer.

Fact #1 says there were 600 children, and fact #4 says 25% of them went to dance camp. Therefore, 600 x 25% or 150 children went to dance camp, and they are all girls, so the correct answer is A.

Word problems can be tricky. The key to solving them is to first remember that they may have more information than you need. By avoiding the TMI trick (too much info), you'll have a much better shot at finding out the correct answer.

Hoping this isn't Camp Crystal Lake,

Professor Cram

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