Homeschool Math Blog

Math Mammoth Thanksgiving sale is already in full swing... You'll get 20% off of all the downloads and CDs sold at Kagi store. To take advantage of this offer, enter the coupon code THANK2010 at checkout. Enter the coupon code on the shopping cart page (not the first order page). Coupon valid till November 29, 2010, only at Kagi store. Not applicable to printed copies. Order pages: Blue series , Golden and Green Series ; Light Blue complete curriculum products ; Packages (CDs or downloads) ; Make It Real Learning workbooks ; Versiones en español .

I just found this via Math Teachers at Play ...   Carolin's Notation for Solving Equations Carolin is a student from Germany. I just wanted to note that that is exactly how I was taught (in Finland) to note what is done to each side of the equation, and I really like the notation. I don't know if it's used in all Europe... Basically, you note in "the right side margin" what you're going to do to both sides of the equation in your next step. The "margin" is made by writing a vertical line to the far right of your actual equation solving process. 6x - 5 = 2x | -2x 4x - 5 = 0 | +5 4x = 5 | ÷4 x = 5/4 I just wanted to pass this on in case some of you who are teaching how to solve equations find it useful with students.

Math Mammoth Statistics & Probability is now available -- and it's the last book I'm intending for the Math Mammoth Blue Series . So the Blue Series is now complete! Price: $5.30 (download), $10.30 (printed) This book starts with the easiest topics, such as reading different graphs . These lessons are meant for 5th grade. Some are useful even in earlier grades. Then we study histograms and various kinds of line graphs , including how to draw them. The next section deals with mean, median, and mode —the three measures of central tendency—how to use them, and how these concepts relate to line and bar graphs. To create circle graphs , the student needs to calculate percentages, and it is assumed here that the student has already mastered how to calculate those. The lesson on data analysis presents various types of graphs for students to read, and reviews some other percent-related topics. Then students learn how to make stem-and-leaf plots . Stem-and-leaf plots

Here are two problems for you to solve... OR to learn from me when I solve them. Both problems are for algebra 1, and use a system of 2 linear equations. By the way, the comments have some wonderful ideas for solving these mentally, without using algebra. So please read them too! Problem: John bought red pens for $4 apiece and blue pens for $2.80 apiece. If John purchased a total of 24 pens for $84, how many red pens did he purchase? Solution: This is a typical problem that will have two variables and two equations. Let r be the amount of red pens he buys, and b be the amount of blue pens he buys. We get our first equation from this sentence: "He bought a total of 24 pens." So, r + b = 24. We get the second equation from the fact that his total purchases were worth $84, and red pens cost $4, and blue pens cost $2.80 4r + 2.8b = 84 Now, just solve this system of two equations using your preferred method. r + b = 24 4r + 2.8b = 84 I will multiply the top e

Here's another one of those job / workers word problems (inverse or direct variation). Try and see if you can solve it using the "table" method instead of equations: A certain job can be done by 18 clerks in 26 days.  How many clerks are needed to perform the job in 12 days? Again, we can set up a table and reason this out. Initially set it up like this: jobs | clerks | days -------------------------- 1 | 18 | 26 -------------------------- 1 | | -------------------------- 1 | ? | 12 Then think of the "days" column. We want to "go" from 26 to 12. You could use a proportion here... or first figure out how many clerks are needed to do this job in 2 days, and then from that go to 12 days. If 18 clerks do it in 26 days, then how many clerks would do it in 2 days... which is 1/13 the amount of time.... so we need 13 times as many clerks. 13 x 18 = 234 clerks are needed. jobs | clerks | days -------------------------

Simplify the ratio 186:403. The answer is 6:13. How do we get it? To simplify ratios (or fractions), we need to find COMMON FACTORS of the two numbers. So, one way to do it is to first find the GCF (Greatest Common Factor) of 186 and 403. Then divide 186 and 403 by it. Alternatively, just find ANY factor of 186 and 403, and divide both by it, to simplify the ratio somewhat, and to get started. Then repeat the process. Okay, 403 is not divisible by 2, 3, 4, 5, 6. This I know by divisibility tests. Maybe it's divisible by 7... need to try (calculator). No, it isn't. Maybe by 11? No. Maybe by 13? YES. My calculator helps. 403 = 13 x 31. I happen to know both of these are primes, so therefore 403 doesn't have any other factors. Then 186.... is it divisible by 13 or 31? By 13, no. By 31, YES!  186 = 31 x 6 So since 186 = 6 x 31 and 403 = 13 x 31, then the ratio 186:403 simplifies to 6:13. Clearly that's as far as we can get, as it's simplified to the lowest

I just stumbled onto a pretty neat resource for all math teachers (parents too). It is Glencoe's Online Study Tools . What they have is online, multiple-choice tests, quizzes, and standardized test practices, for ALL middle and high school topics. They are randomly generated so you can get as many different quizzes as you like. Just choose any state you wish (it doesn't matter), then a book or course (such as algebra or geometry), and then you'll see the list of topics. In other words, you don't have to own the book in order to do the quizzes and tests.

Just yesterday I read this interesting article, so I thought I'd pass it on. It's a list of 9 most mathematically interesting buildings in the world ... each one has something different that is interesting about them - from pi and symmetry to asymmetric hyperbolic paraboloids ! It includes the Great Pyramid, The Eden project, Sagrada Familia, the Gherkin, and some interesting-looking buildings I had never heard of.

Today I want to feature a great video from a fellow blogger and Youtuber, Dave Marain. He solves this question, which at first can sound intimidating: If 10 workers can build 3 houses in 60 days, how many workers are needed to build 5 houses in 40 days? It sounds like you'd need algebra, proportions, inverse or direct variation, etc. And true, you could use those. But his method completely avoids all that and is based on setting up a simple TABLE, AND using common sense! | HOUSES | WORKERS | DAYS | ---------------------------------- | 3 | 10 | 60 | ---------------------------------- | 1 | 10 | | ---------------------------------- | 2 | 10 | | ---------------------------------- | 5 | | | ---------------------------------- See the video below: And here's a link to his other Youtube videos: www.youtube.com/user/MathNotationsVids .