数学Content Review I

我正在计划一场足球比赛之后的聚会。现在我知道球员将真的非常饿。因此，我发现我每三个球员都需要大约两个馅饼。但是这里有36名玩家来了，所以我需要多少个比萨饼？好吧，在这一集中，我们将讨论可以帮助我们解决这个问题的主张。我们还将讨论您将在该法案中看到的其他顶级代数概念。
让我们开始谈论您需要查看该法案数学部分的概念。现在，佩姆达斯（Pemdas），“请原谅我亲爱的莎莉姨妈”，就像您的老师可能告诉您的那样。操作顺序，这不是直接在该法案上进行测试，对于代数前的代数不会有PEMDAS问题，但是当您做问题时，这将在许多问题中出现。您将需要确保正确执行操作顺序。因此，PEMDA代表括号，您将首先进行，指数，乘法，除法，加法和减法。您将要按照此顺序进行操作。好的，让我告诉您一些非常重要的事情，我看到学生做了很多事情。他们知道Pemdas，并且对此感到非常好，他们已经在您的代数之前看到了它，但是他们没有正确使用您的计算器来帮助他们进行计算。这是我的一个学生中的一个例子。因此，您将她交给了她，她已经完成了整个复杂的代数问题。 This was is what was left to calculate. Okay 12 plus 29 over 132. Right 29 divided by 132. So you know PEMDAS, you know okay 'division' where do you go before addition so you would do this first and then you would add 12. But when you put it in your calculator, you have to make sure to tell your calculator to do that. What she did was input her numbers just from left to right into her calculator. And she kind of thought her calculator would just know what to do. If you put it into your calculator, your calculator will think, 'Oh it's 12 plus 29, whatever that is and all of that divided by 132,' which is a completely a different number than 12 plus parentheses on your calculator 29 over 132, signifying to your calculator that's a separate expression, a separate fraction or division that needs to happen first. So really watch this you guys, watch your calculators and think about PEMDAS. Alright let's look at the first pre algebra concept that you need you need to know that's actually is going to be straight forwardly tested on the ACT.
好的，学生经常问我，你知道我看到了一个整个言语的问题，我如何将其更改为代数，以便我可以解决它。您知道您会看到什么是50％或29％的12％。因此，他们在那里有很多单词，而不是代数。只要您记住这些规则，实际上并不困难。当您看到问题中的“什么”时，将是您的X，当您看到的是，这将是您在代数方程式中的平等标志。这是“始终表示乘法”的“始终表示乘法”的单词“在任何百分比问题中”的关键，这始终是需要乘以的两个​​部分之间。因此，如果您曾经思考过，我需要在这里繁殖什么，“'''of to之间。现在，这听起来可能仍然令人困惑，所以让我们尝试一个真正的问题。如果x的50％为150，则x的2％是多少？因此，您看到我们在这里实际上有两个百分比的问题。让我们逐步把它放在一片中，首先我们需要找到'50％的x'。 For we have to find what 'X' is if '50 percent of X' is '150' and next we've to find out what's two percent of 'X', the number that we find first. Alright let's hoop right in and then you'll see how I use that knowledge about 'what', 'is' and 'of' that I talked about earlier. If 50 percent of X is 150, how is it algebra? Well remember 'of' is always between the two things we need to multiply together. So 50 percent of X is 150. 0.5 and remember 50 percent in algebraic terms is 0.5. And if you need a review there is a little sheet about that in your bonus material. So 50 is 0.50 times X right because you've got your 'of', is equals 150. Great and now we can just solve it. It's just a simple algebra problem, you know divide both sides by 0.5 you can cross it off, divide here by 0.50 and we know that then X is equal to 300. Okay so let me circle that. But we're not done right? We next need to know, what's two percent of X? What's two percent of 300? So let's do this again. Okay what is two percent of 300? So 'what' that's our X is, equal, two percent of X. Again 'of' between the two things that need to be multiplied so 0.2 times X I'm sorry times 300 because we already know, so times 300 the number we just found. Okay so use your calculator. 0.2 times 300 is equal to six. So now we know that our answer is B, six. And again if you need more practice there is a lot of practice problems in your bonus material for percents.
我们继续吧。因此，我们讨论了一些基本的易案剂，还将对涉及percents的一些东西进行测试，即增加百分比和减少。该法案喜欢测试，与百分比有关的事情增加了或减少。我们将在一秒钟内谈论它的外观。公式的公式是第一个金额和第二个金额之间的变化量除以您开始使用的数量。让我向你展示我的意思。“如果沙龙在3月在Twitter上有65位关注者，四月份有95位关注者，她的追随者有多少人增加？”因此，您会看到这是一种不同的百分比问题，在我们发现某物的百分比之前，我们想要“您的百分比在第一个数量和第二次数量之间减少或增加了”。好的，请记住，请记住我们拥有的公式，我们的变化数量超过了他开始的数量。好的，当我们有65岁的时候，我们最终获得了95，而变化的数量是多少？ Well 30 right, from 65 to 95 perfect. Very straight forward, so 30 is the amount of change and what did we start with? Well we started with 65, 30 over 65. And then when you use your calculator and you'll get something about 0.46. And remember how switch percents from algebra into percent terms, 0.46 is about 46 percent so B will be the right answer. Okay let's keep going.
让我们转到我最喜欢的代数主题之一，可怕的插件问题。这些就是您会看到的。您可能会从希望参加的练习测试中记住他们。他们看起来很恐怖，您会看到一些非常非常可怕的方程式，您会想：“哦，天哪，我不知道我必须知道这一点，有关音量的信息，关于计算温度的事情。”您认为：“哦，天哪，我无法解决这个问题。”实际上你可以。这将是一个简单的直接插件和问题，您将给定的数量插入可怕的方程式中。让我们看看我的意思。半径为R的体积V的公式为V等于四分之二的PIE R Cubed。而且您可能已经在感到胸痛，“哦，天哪，这是什么。” Right? If the radius of the pitch ball is 12 inches, what's the volume to the nearest cubic inch? Never be intimidated about these problems. You'll see three to four of them and they're really, really not bad because look we've got all the parts we need to easily solve his problem. We know that the radius is 12 and so we've got that the volume and they tell you the volume B when you have your radius R, so we would just plug-in our radius into R, no big deal. Let's go ahead and solve it. So we have four thirds times pie times 12 cube, easy okay. And so you see it's not intimidating after all. Now it's just a straight forward algebra problem. Remember PEMDAS, we're going to do exponents first. So we've four thirds times pie times 12 cubed, Let's figure out what that is. So I got my calculator here and 12 cubed is 1728 and hopefully you're doing this along with me at home. So we've got four thirds times pie times 1728. And now that we've got everything is multiplied across we can just multiplied across. By the way because the answer choices don't have pie in them, that's a sign to us that we actually have to calculate pie. We can just call it three point one four, okay. So I'll just multiply across. I'm just going to use my calculator and we'll see what we end up with. So we have four thirds or one point three, three about times three point one four, you can do this on a calculator, times 1728 okay. If you're doing this along with me hopefully you ended up with about 7234, let me write that down, 7234 and now if you're trying to worry oh my gosh nothing looks identical take a closer look at the problem it does tell you what's the volume to the nearest cubic inch so we just want the closest, the closest answer choice which it looks like it would be E, 7238 great. So that's it for scary looking plugging problems never be intimidated by these just look and see what do you have to plug in.
好的，让我们看一下我看过的ACT测试的比例，其中比例将显示三到四个甚至四次，所以我希望您对比例感到非常好。什么比例？好吧，比例是一个方程式，表明两个比率相等。这有点像一个分数，然后是分数的简化形式，您知道它们相当于彼此，例如，四分之三的分数与一半正确'，因为如果您减少了四分之三的八分之一，您最终会得到一半，关于比例的很酷的事情是您可以跨越倍数，它们应该彼此相等。因此，二次是八次，八次，您可以使用它来解决ACT问题。
让我们去看披萨问题，然后我们将查看一个真实的ACT问题测试比例，好吧，如果您记得我参加我的聚会的每三名足球运动员需要两个披萨，那么三十六六十六个披萨需要多少个比萨饼足球运动员注意这确实可以帮助您在现实生活中。因此，每三个足球运动员都可以每三个披萨，所以我们可以将其写成一小部分，或者确实是一个比例，在这里确实是同一件事，所以每两个比萨饼每两个比萨饼都在每个两个比萨饼上，三个或两个右边的三个或两个要养活三个人，我们知道我们需要一个同等的分数，以便我们要喂多少人，对我们需要多少比萨饼来喂36个人。好的，如果您需要每三个足球运动员每三个足球运动员需要两个披萨我们可以有两个等效的分数。现在就像我说的那样，我们可以越过乘数实际解决x，所以我们有两次36等于三x，让我写下来，两个次36等于三x，所以两个x oke as s o twos 36，72 so 72等于三x我们'只是将双方划分为三个右侧，所以X将等于24好。因此，如果我需要三个足球运动员需要两个披萨，我将要看一下，我将需要24个比萨饼来喂36名足球运动员，但现在我知道了。
让我们看一下真实的行为问题测试比例，朱莉和迈克在每五秒钟旋转30度的轮播上。他们将在一分钟内旋转多少度？好的，所以我们知道我们将在这里拥有等效的比率或等效的分数，我们每五秒钟有30度，因此每五秒钟30度，这将等效于等于多数度，但是它们在一分钟内旋转了一分钟这只是简化的分数。因此，他们将在一分钟的井中旋转多少度X度，尽管您有60秒的时间持续了一分钟，因为当您比例时，等效零件必须在同一单元中。好的，现在我们可以继续进行交叉乘数，所以我们有30次60倍五x，所以30倍60、1800等于五x，因此，如果您将双方除以五个以隔离您的变量x，则将等于360.好的，我们看看播出的旋律是否在整个分钟内每五秒钟旋转30度，它们旋转整个360度。C是这里的正确答案选择。
Okay let's move on to another concept taking a look at average problems these will always show up on the ACT and usually you'll see them two or three or sometimes even four times. The formula hopefully you're familiar with already but let's just go over it, sum of terms divided by the number of terms meaning adding up all the different parts you have that you want to take the average of and divided by how many different things there are. We're going to look at two example average problems because there are two types of average problems that show up, a really basic average problem and then a slightly more complicated one.
首先，让我们看一下更容易的人，汤姆在六个高中生物学测试中获得了89、72、54、50、80和69的得分。大约汤姆班上的平均水平是多少？希望您看起来真的很熟悉，所以我们有很多不同的术语，我们知道公式是您将所有术语加起来并除以您找到平均水平的数量，让我们这样做。因此，我们有89加72加54加50 GOSH这是这个家伙参加的很多测试，再加上80加69，我做对了，是的。因此，我们将所有内容汇总为我们有多少，所以让我们划分六个，编写六个不同的测试，这些测试将等于x，好的。因此，让我们继续拿起计算器，因为这要添加很多，所以我们有89加72加54加50，希望您与我一起做，再加上80，另外69个好，好的，等于414414超过六个将是我们的平均水平，所以让我们这样做414除以六，这个人的平均水平为69，很棒，所以答案选择B。
Now let's take a look at a slightly more difficult average problem that you'll see, be on the lookout for this you will definitely see this problem or this problem type and you can totally nail it if you know what to do. Tom scored an 89, a 72, 54, 50, 80 and 69 same as before and six high school biology tests tomorrow he's going to take a seventh biology test if Tom needs an average of 65 to pass the class what's the lowest possible score he can get on the test and still pass the class. Okay so let's think about this for a second, we've got the scores for six things right, but we're missing a score we need the score for the seventh, this is like I said there is something you're going to see the ACT loves to test average problems where what you're missing is actually something at the top of your equation remember what our equation is everything added up divided by how many you have that gives you your average right. Here we know what we want the average to be we want him to get a 65 we also know how many tests there are, there are seven let's write that down so we've got our typical formula everything added up is going to go here divided by how many terms okay I mean there are going to be seven, seven total tests and we know his average needs to be a 65 so you see how interesting right, we actually have two of the parts but we're going to miss, we're missing part of the top. But we've got the rest of the test we know he got an 89 plus a 72 plus that 54 plus that what else 50 plus the 80 plus the 69, okay guys plus X.
Alright so just watch up for this trail over here, this is our whole top so you see how we're missing that seventh right but we know everything else added plus that seventh test whatever it is I'm calling X divided by seven the amount we have it can give you 65. A lot of students look at this and they think, "Oh my gosh what do I do", don't worry about this all you're going to do is treat it like you would a typical algebra problem remember if you have something going on in the bottom you have division and you have an equal sign with something on the other side you can multiply the seven to the other side you remember this if you think about algebra if I had multiplied both sides by the same thing which is allowed, multiply both sides by seven I can cross of my sevens here and this becomes 65 times seven.
Okay so now this whole chunk here is going to be equal to 65 times seven let me just figure out what 65 times seven which will make things a little bit clearer here. So 65 times seven is 455, okay so this whole chunk over here is going to equal 455 I don't want to rewrite this whole chunk because remember on the last problem we added this all up and they're equal to 414 right we did that for the last problem. So now we know 414 plus X is equal to 455 okay so you see what we did we just added up all the test scores for leaving our X our missing score, but it just turned into a really straight forward algebra problem all of this you move this to the other side so 414 plus X is 455 now we can just subtract to figure out what is, X is going to be 455 minus 414 that is 414 okay, let's use our calculators for that minus 414 and wow a 41. So lucky Tom just a pass if that's all he wants to do all he needs is a 41 on this test great and that looks like answer choice A.
现在我想这样做，以便你们看到我如何设置这个方程式，但请记住我们的节奏策略和我们的后门策略，我们谈论的是在您做这些问题时要记住的，有时还有一种较短的方法来做到这一点，您仍然可以需要基本了解如何设置方程，但请记住，您也可以在此类问题中从答案选择中向后工作。因此，我想直接做到这一点，以便您可以看到它的外观，但是您可以做的另一件事是，一旦您意识到一切都加起来了，这是七个给您的65，您可以从答案的选择，说还好吗x 41，这是您知道的全部，您知道41在这里除以七个，给了您65，等等，依此类推，从答案中挑选。
因此，有时候或实际上经常需要记住的事情是一件很酷的事情，有几种不同的方法来解决一个问题，这是一种非常有力的感觉，可以拥有一堆工具，您可以将其拔出并在问题上使用。太好了，这就是您将反复看到该行为的顶级代数问题，我希望您对此感到非常好在60个问题中，数学14将是代数。如果您需要更多的练习，请确保前往奖金材料，然后再练习，我认为您会做得很好。