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When it comes to online math programs, the choices seem endless. However, the service, the offered courses, and the quality of those programs vary a lot, and extra care is needed when choosing the right program. Here are some things to consider:
Are you looking for a full-year program or a summer program? Some programs are self-paced and can be completed in any timeframe, but others need to be done at a certain pace. These are usually programs that include sessions with an instructor, which is another criterion to consider. One program that is self-paced but still offers instructor support is SCOUT by the University of California. What is the quality and rigor of the course? One of the most rigorous programs I have worked with is the Art of Problem Solving (AOPS). Their courses are not for everyone. On the other side of the spectrum, you will find BYU courses, that are more or less rubber stamp courses with material that is lacking in quality. One provider of good-quality moderate-level courses is APEX. For those students who are a bit more ambitious, they offer honors courses, too. Is the math program meant to be enrichment, support to an ongoing class at school, or as a replacement/advancement course? In the last case, you want to check whether the courses are approved by colleges or your school before you enroll. Finally, the cost of such programs varies widely. Some programs are free, and others match the costs of college courses. The prices do not always correlate with the quality. E.g. Khan Academy offers excellent free courses (no certification, though). Once you have found the program that is right for you, you can start having fun learning!
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Imagine what we could do if we could reprogram our cells: We could correct what goes wrong in different diseases or counteract inherited weaknesses. Such a reprogramming tool was identified a few years ago and won the involved researchers Jennifer Doudna and Emmanuelle Charpentier a Nobel prize in 2020. The name of the underlying molecular machinery is CRISPR, and it is used to cut DNA at specific sites, to then reconnect the DNA, but to make that piece of DNA unusable in the process. Since its discovery, the method has been developed further to be used as a molecular therapeutic, and the first such therapeutic has just been approved in England.
One of the major strengths of this new technology is however also a major concern: The DNA is permanently changed in the process. So, what if something goes wrong? What if the wrong piece of DNA gets cut? What if there are unintended long-term effects that cannot be undone? When both DNA copies of a gene are cut and gone, they are gone for good. In a new article published in the journal Cell (Tieu et al., 2024), researchers have modified the original CRISPR approach: Instead of cutting DNA, they cut the RNA that serves as a messenger and template when proteins are made. The DNA remains unchanged, and as a result, all of the changes made are reversible. This is a much safer way to reduce or eliminate the production of specific unwanted proteins. In this work, the method was used to downregulate RNA in so-called T-cells of the immune system that are involved in the fight against tumor cells. T-cells that normally would show an exhaustion response to their permanent fight did much better when certain RNA molecules were cut and removed and the proteins were no longer made by the cells. As a result, tumors in mice treated this way shrunk more than in control mice. The researchers showed that with the used technique they can downregulate up to 10 genes at a time and finetune protein production by switching the CRISPR tool on and off with an antibiotic. This mean, that they can optimize the tumor response by the T-cells. If this method proves feasible as medication, it is another step on the way towards a targeted and optimized cancer treatment. A lot of times math problems look a lot more difficult than they are. Take complex fractions: \(\frac{\frac{2}{3}}{\frac{4}{5}}\) At first glance, this problem looks very intimidating, and there isn’t a student who doesn’t panic a bit seeing those for the first time. At a second glance, the complex fraction can be simplified with methods that 5th graders learn: You find an equivalent fraction that is simpler. To find equivalent fractions we need to multiply both the denominator (i.e. the bottom) and the numerator (i.e. the top) by the same number. E.g. \(\frac{1}{2}\) is equivalent to \(\frac{2}{4}\) and \(\frac{3}{6}\). The trick is to multiply by the right number: To simplify complex fractions, we multiply the numerator and denominator by the denominator's reciprocal (i.e. the flipped fraction). \(\frac{\frac{2}{3}\cdot \frac{5}{4}}{\frac{4}{5}\cdot \frac{5}{4}}\) When a fraction is multiplied by its reciprocal, the product is 1 (try it out). And there we go: the denominator is now 1 and can be ignored. \(\frac{\frac{2}{3}\cdot \frac{5}{4}}{\frac{4}{5}\cdot \frac{5}{4}}=\frac{\frac{2}{3}\cdot \frac{5}{4}}{1}=\frac{2}{3}\cdot \frac{5}{4}\) All that is left is the numerator which is a product of two fractions. In the last step, we multiply across and find \(\frac{2}{3}\cdot \frac{5}{4}=\frac{10}{12}=\frac{5}{6}\) In short: To simplify complex fractions, find an equivalent fraction with a denominator of 1. And don't let the intimidating looks of a problem deter you. Is improving your study habits one of your goals for the new year? Or would you like to help your students or your kids to improve their study habits? Here are some ideas:
Thank you to all the dedicated, hardworking teachers who ensure their students learn and thrive. You have one of the most important jobs in the world, and you are changing lives because of what you do. Also, thank you to all the parents who provide the support and environment for their kids so they can aim for their goals. I hope you have a happy, fulfilling, and successful year 2024.
Now is the time to start thinking about the AP exams that will come up in about two months. You don’t want to cram all the preparations into the last two weeks before the exam, but give yourself time to review the concepts and get used to the kind of questions typically asked. There are a lot of ways to prepare besides the practice that the college board provides. Here are a few resources you can use to get ready. My experience with them is in math and science, but you can check them out for other AP courses as well.
Let's talk about three of the most important numbers in math. They are so important that they got their own symbol. First, there is \(\pi = 3.1415…\). \(\pi\) is the area of a circle with radius 1. It is also half the circumference of that circle. \(\pi\) is probably the first irrational number students learn about in school besides square roots: its decimal digits never end and do not repeat. Another irrational number that is important enough to get its own symbol is the natural base \(e = 2.7182...\). \(e\) is used as the base of exponential functions that describe continuously compounded interest. An exponential function with base \(e\) is the most fundamental exponential function, because it is identical to its derivative. And finally there is \(i\) which is the symbol for \(\sqrt{-1}\) and therefore essential for the set of complex numbers. So here you have three numbers that play roles in completely different areas of math, two of which never end and one of them is not even real. You might think that these numbers have nothing to do with each other. But here comes the magic: The relationship between these numbers is \(e^{i\pi}=-1\) Isn't math beautiful! Which of the following equations looks easier to you? \(\frac{3}{4}x + \frac{1}{2} = -\frac{2}{3}x + \frac{5}{6}\) or \(9x + 6 = -8x + 10\) If you, like me, think that the second equation looks easier, you can simply use the fraction busting method to convert the first into the second equation: Multiply all terms by the Least Common Denominator (LCD) of all the fractions, and the fractions miraculously disappear. In this example the LCD is 12. \((12)(\frac{3}{4})x + (12)(\frac{1}{2}) = (12)(-\frac{2}{3})x + (12)(\frac{5}{6})\) \((\frac{\enclose{horizontalstrike}{12}}{1})(\frac{3}{\enclose{horizontalstrike}{4}})x + (\frac{\enclose{horizontalstrike}{12}}{1})(\frac{1}{\enclose{horizontalstrike}{2}}) = (\frac{\enclose{horizontalstrike}{12}}{1})(-\frac{2}{\enclose{horizontalstrike}{3}})x + (\frac{\enclose{horizontalstrike}{12}}{1})(\frac{5}{\enclose{horizontalstrike}{6}})\) \((\frac{3}{1})(\frac{3}{1})x + (\frac{6}{1})(\frac{1}{1}) = (\frac{4}{1})(-\frac{2}{1})x + (\frac{2}{1})(\frac{5}{1})\) \(9x + 6 = -8x + 10\) Why does this work? The LCD is a multiple of all the denominators. Therefore, all products of the LCD and the fractions can be cross-simplified, and all denominators can be reduced to 1. You don’t have fractions, but decimals? No problem, the method works here, too (decimal busting): \(1.45x - 3 = 0.9x + 2.92\) You multiply all terms by the same power of 10 so that all decimals disappear. \((100)(1.45)x - (100)(3) = (100)(0.9)x + (100)(2.92)\) \(145x - 300 = 90x + 292\) I know that my students love this method, but let me know how it worked for you □ I love using smart ways to do math because it can make math so much easier. Unfortunately, many times those little tricks and methods are considered less important than the big concepts and therefore get less attention in math class than they deserve. Let’s look at how much of a difference a smart simplified sequence of steps can make to the difficulty of a problem. As an example, we will multiply radicals. Here is the problem: \(\sqrt{84} \cdot \sqrt{175}\) The brute force approach is to multiply the radicands. To find the simplest form of the resulting radical, we then need to factor the radicand and find all squares that can be pulled out of the square root. \(\sqrt{14700}\) \(\sqrt{7 \cdot 2 \cdot 2 \cdot 3 \cdot 5 \cdot 5 \cdot 7}\) \(7 \cdot 2 \cdot 5 \cdot \sqrt{3}\) \(70 \sqrt{3}\) Now let’s do the same problem but differently. We don’t start by multiplying but by finding the factors of both radicands right away. Then we write the product of those factors as one radicand and pull out the squares. \(\sqrt{7 \cdot 2 \cdot 2 \cdot 3} \cdot \sqrt{5 \cdot 5 \cdot 7}\) \(\sqrt{7 \cdot 2 \cdot 2 \cdot 3 \cdot 5 \cdot 5 \cdot 7}\) \(7 \cdot 2 \cdot 5 \cdot \sqrt{3}\) \(70 \sqrt{3}\) The difference lies in the first step. With the second method, the product of 84 and 175, 14700, never appears. We skipped the multiplication completely and went to factoring right away instead. This does not only save us the work of multiplying large numbers but factoring 84 and 175 is also so much easier than factoring 14700! Conclusion: Always try to find a smart way to make the math easier, and you can save a lot of time and avoid mistakes, too. Three-fourths of all students at Union High School live within 2 miles of the school. Of all the students that live within 2 miles of the school, one third takes the bus. Four-fifths of the students living further away from the school take the bus. Which fraction of all students at Union High School take the bus? |
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