Charles Y.

Former college instructor and computer programmer who loves math

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Question:

$$b = 2.35 + 0.25x$$ $$c = 1.75 + 0.40x$$ In the equations above, b and c represent the price per pound, in dollars, of beef and chicken, respectively, x weeks after July 1 during last summer. What was the price per pound of beef when it was equal to the price per pound of chicken?

Charles Y.

Answer:

This was a practice question I saw at some point, so I'll work through it here. Let's parse the question itself - What was the price per pound of beef when it was equal to the price per pound of chicken? We want to know the value of $$b$$ when it was equal to $$c$$. How can we know the value of $$b$$? We have a formula for it, but it depends on $$x$$, so it might be really helpful to know the value of $$x$$ that makes $$b = c$$. Let's start by setting $$b = c$$, then. $$b = c$$ ...now, substitute the formulas for b and c in there, and we have an equation where we can solve for x... $$2.35 + 0.25x = 1.75 + 0.40x$$ ...subtract 0.25x from both sides... $$2.35 = 1.75 + 0.15x$$ ...subtract 1.75 from both sides... $$0.60 = 0.15x$$ ...divide both sides by 0.15... $$4 = x$$ So when x = 4, it turns out that b = c. What's the value of b when x = 4? Just plug it back in! $$b = 2.35 + 0.25x$$ ...evaluate for x = 4... $$b = 2.35 + 0.25(4)$$ $$b = 2.35 + 1$$ $$b = 3.35$$

Pre-Calculus

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Question:

How many different ways are there to arrange the letters in the word "ALASKA?"

Charles Y.

Answer:

This is one of my favorite topics to talk about, and it has a pretty decent number of real-world applications. Let's start at the very beginning - a very good place to start. When you read, you begin with A, B, C. Now, how many ways can we arrange the letters ABC into different orders? We can list them all pretty easily: ABC, ACB, BAC, BCA, CAB, CBA That makes six. You can verify that there aren't any more. Take a few seconds to convince yourself. Now, let's go a bit further. I've seen a lot of people say that the longest word in the English language that does not repeat any letters is "UNCOPYRIGHTABLE," which has 15 letters. How many ways are there to arrange those letters into different orders? You can start listing possibilities, but we'd be here for many thousands of years before you finished (I'm not kidding - I'll verify this later). There has to be a better way, and it turns out, there is. Let's go back to ABC. We've verified that there are six ways to arrange those letters in different orders. How can we get there without just listing all the possibilities? Well, take a look at that list. How many letters can we start with? Three, right? For each starting letter, how many different second letters are there? For example, take a look at the sequences starting with "A." How many different letters can be in the second position? Two - B and C! After that, there's only one letter left. So let's review that for a second. There are three possible starting letters. FOR EACH of those possible starting letters, we have two possible second letters (specifically, the two letters that weren't used as the starting letter - for example, if we start with A, then we can choose B or C as the second letter). FOR EACH of those second letters, there is one letter left, and it has to be third. We have three ways to pick a first letter, and for each of those, we have two second letters, and for each of those, we have one third letter, so we have $$ 3 * 2* 1 = 6 $$ total ways. If you wish, you can take a few minutes to verify that there are $$ 4 * 3 * 2 * 1 = 24 $$ ways to arrange the letters of "MATH" into different orders. Similarly, there are only $$ 2 * 1 = 2 $$ ways to arrange two different letters (AB and BA, for example). So if we extend the pattern, it turns out that there are $$ 15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 $$ ways to arrange the letters of "UNCOPYRIGHTABLE." That is a total of $$ 1,307,674,368,000 $$ ways! Now, I said I'd verify that we'd be here awhile if you tried to list all of the ways. If it took you five seconds to write down each possibility, and you did this all day and all night, never stopping to eat or sleep or do anything else, you'd be here for $$ 1,307,674,368,000 * 5 = 6,538,371,840,000 $$ seconds. There are 86,400 seconds in a day, so you'd be here for $$ 6,538,371,840,000/86,400 = 75,675,600 $$ days, which, at 365 days per year, would keep you here for $$ 75,675,600/365 = 207,330 $$ years (rounding down, of course). OK, so it turns out, there's a term for the product of all positive integers from 1 to N, inclusive. It's called a factorial, and it's written as $$N!$$ So we define $$ N! = 1 * 2 * 3 * ... * (N-1) * N $$, for all positive integers N (e.g. $$ 4! = 1 * 2 * 3*4 $$). By the way, it turns out that $$ 0! $$ is also defined. What do you think it equals? If you said 0, that's a good guess, but it turns out, it's actually defined to equal 1. If you think about it, it sort of makes sense - how many ways are there to arrange 0 letters? Isn't it 1? An empty "word" still counts! OK, so we've established the definition of a factorial, and we can talk about the original question. How many ways are there to arrange the letters in the word "ALASKA?" Well, if the letters were all different, we've established that it would just be $$ 6! = 720 $$. However, note that I've thrown a twist into the problem because "ALASKA" has 3 A's in it, while the letters in words like "ABC," "MATH," and "UNCOPYRIGHTABLE" are all different! Why is this important? Well, let's look at a simpler example. Let's take the word "EEL." It has three letters, but two of them are the same, and we can't tell them apart. It turns out, there are only three (and not six, like with "ABC") ways to arrange the letters: EEL, ELE, and LEE. If you could tell the E's apart (let's call them $$E_1$$ and $$E_2$$), you'd similarly have six ways to arrange the letters: $$E_1E_2L$$ $$E_2E_1L$$ $$E_1LE_2$$ $$E_2LE_1$$ $$LE_1E_2$$ $$LE_2E_1$$ In this case, we have the familiar 3! = 6 ways. However, note that each arrangement is duplicated by switching the orders of the two E's (e.g. $$E_1E_2L$$ is the same as $$E_2E_1L$$, as we're not going to be able to tell them apart). In other words, each arrangement is listed twice. Why twice? Because there are 2! = 2 ways to arrange the E's! So we take our familiar 3! = 6, and divide it by 2! = 2 to get 6/2 = 3 arrangements. Let's go back to the word ALASKA, now. There are 6 letters, so if we could tell them all apart, we could arrange them in 6! = 720 ways. However, there are 3 A's, which can be arranged in 3! = 6 ways, so if we listed all 720 ways to arrange the letters in ALASKA, each one would actually get listed 6 times, once for each arrangement of the 3 A's, so we have to divide the 6! by 3! to get the actual number of distinguishable arrangements. Thus, the total number of ways to arrange the letters in the word "ALASKA" is $$\frac{6!}{3!} = \frac{720}{6} = 120$$. We can actually extend this for multiple repeated letters. For example, if we had the word "BANANA," which has 6 total letters, but two N's and three A's, how do you think we could calculate the number of arrangements? Right! There are 6! total arrangements, 3! ways to order the A's, and 2! ways to order the N's, so we divide 6! by both 3! and 2!. $$ 6!/(3!*2!) = 720/(6*2) = 720/12 = 60 $$ If you want to take a few minutes to convince yourself that there are actually 60 listings by actually writing it out, feel free. I've listed them alphabetically below, though: AAABNN AAANBN AAANNB AABANN AABNAN AABNNA AANABN AANANB AANBAN AANBNA AANNAB AANNBA ABAANN ABANAN ABANNA ABNAAN ABNANA ABNNAA ANAABN ANAANB ANABAN ANABNA ANANAB ANANBA ANBAAN ANBANA ANBNAA ANNAAB ANNABA ANNBAA BAAANN BAANAN BAANNA BANAAN BANANA BANNAA BNAAAN BNAANA BNANAA BNNAAA NAAABN NAAANB NAABAN NAABNA NAANAB NAANBA NABAAN NABANA NABNAA NANAAB NANABA NANBAA NBAAAN NBAANA NBANAA NBNAAA NNAAAB NNAABA NNABAA NNBAAA

Algebra

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Question:

John has two large bins of different kinds of ground coffee. Albert's Discount Coffee is valued at $5 per pound. Bob's Special Coffee is valued at $8 per pound. John would like to create a 10-pound bag of coffee that is a mixture of the two coffees, valued at $7.40 per pound. How much of each coffee should he put into the bag?

Charles Y.

Answer:

Let's suppose that we want to put A pounds of Albert's and B pounds of Bob's into this mixed bag. Since Albert's costs $5 per pound, what is the value of A pounds of Albert's coffee? It's just $$ 5A $$, right? What's the value of B pounds of Bob's, at $8 per pound? Right - it's $$ 8B $$. Now, we want 10 total pounds of coffee valued at $7.40 per pound. What's the cost of this bag? Multiply 10 by 7.40 and see that the value of the entire bag is $74. OK, so we have the total value of the bag, and we have expressions for the value of Albert's and Bob's coffee. What can we say about these values? How can we relate A, B, and the value of the bag? Right! We have: $$ 5A + 8B = 74 $$ OK, so let's leave that there for a second. The total weight of the bag is 10 pounds. Why is that special? Right - we have: $$ A + B = 10 $$ ...and... $$ B = 10 - A $$ With that, we can substitute for B in that previous equation: $$ 5A + 8B = 74 $$ $$ 5A + 8(10 - A) = 74 $$ Then, we use the distributive property... $$ 5A + 80 - 8A = 74 $$ ...add like terms... $$ -3A + 80 = 74 $$ ...subtract 80 from both sides... $$ -3A = -6 $$ ...and then divide both sides by -3. $$ A = 2 $$ ...and since we had $$ A + B = 10 $$, we now have $$ B = 8 $$. So we have 2 pounds of Albert's and 8 pounds of Bob's coffee!

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