تصحيح الامتحان الجهوي 2025 في الرياضيات | الثالثة إعدادي | جهة الدار البيضاء سطات (الحل الكامل)

Hinzugefügt: 2026-06-01

[00:00:00]Peace, mercy, and blessings of God be upon you, followers of the Success Platform for School Support.

[00:00:05]Welcome to this new video, which is aimed at third-year middle school students who are about to take the unified regional exam to obtain their middle school certificate. In this exam, I will be providing the answer key for the mathematics exam, 2025 session, Casablanca-Settat Regional Academy. Regarding regional exams, they are almost identical across all regions, meaning the same questions and the same framework. Therefore, if a student understands an exam from a specific region, they can easily answer any exam from any region. Having already worked on social studies exam models, we will now work on mathematics models. We will begin with the first question. The first question asks us to solve this equation: 9x - 4 = 3x - Two, here's an easy one.

[00:01:05]When I have an equation like this, what do I need to do? I need those unknowns, i.e., the x-values.

[00:01:11]I'll add them to one side, and the numbers or common numbers to the other side. So here I have xx and 3x. If I take 3x, then 3x has a plus sign. So when I take it to the other side, it becomes minus. So I have 9x - 3x, and this is minus 4. I'll take this to the other side, and it will become plus 4. So the equation will be 9x - 3x = -2 + 4. 2x - 3x = 6x - 2 + 4 = 4. So now we'll find what xx equals. It equals 2/6. What is 2/6? If we simplify, it's 1/3. So what is the solution? x equals 1/3. So always when This question asks me what to do. I need to add the numbers containing x in one place and add the regular numbers in another. Agreed? Now, for question two, I'm asked to verify that x raised to the power of two, or x squared minus three, x plus two, equals x minus one multiplied by x minus two. So, when it tells you to verify or show me an equation, it means you have to start from here to show this, or from there to show this.

[00:02:36]Agreed? So, we'll start from x minus one multiplied by x minus two. So, here we'll develop. If I have x times x, it's x raised to the power of two. I have x times minus two, it's minus two x. And I have minus one. If we multiply it by x, it gives minus x. And minus one times minus two, it's plus.

[00:03:01]So, I have x raised to the power of two minus two x minus x plus two So, we'll calculate if x to the power of 2 is -2x - x, which is -3x + 2. Therefore, x to the power of 2 - 3x + 2 is what I have here, and they asked me to find it. So, it's verified that x - 1 multiplied by x - 2 equals x to the power of 2 - 3x + 2. So, we'll move on to question number 2. We're still on question 2, and we're looking for the equation. So, this equation, x to the power of 2 - 3x + 2, they asked me to solve it. So, here I need to pay attention. The first method I have is the delta method, which we haven't studied yet. And the second method is that we 'll solve this equation starting from here, agreed? So, we have x to the power of 2 - 3x + 2. What does that equal? ​​It equals x - 1 multiplied by x - 2. So, we'll substitute this with x - 1 x - 2 equals zero. So, to solve this equation, we need to make x - 0 equal to zero and x - 2 equal to zero. Therefore, we have x - 1 equal to zero and x - 2 equal to zero. Now, let's find the solutions. Here, x - 1 equals zero.

[00:04:34]This - 1, we'll bring to the other side, x equals 1. And here, x - a equals 0. When we bring to the other side, x - a equals 0.

[00:04:44]So, what are the solutions to the equation? Are x equal to 1 or x equal to a? The solutions to the equation are: x equals 1 or x equals a.

[00:04:56]Now, let's move on to question number three: solve the inequalities. So, just as we solve the inequalities (i.e., just as we solve the equation), we solve these inequalities.

[00:05:04]This inequality is the same. So, we do the same operation: we add the numbers containing x on one side and the numbers without x on the other side. So, here we have 6x + 1 greater than 4x - 3 What do we do?

[00:05:19]We bring that 4x to the other side, and it becomes plus fourx. So, when we move the first one, it becomes minus fourx. Therefore, sixx minus fourx is greater than or equal to minus three. And this one I have here, when it comes to this place, will become minus one because there is a plus here. So, 6x minus fourx is twox, and minus three minus one is minus four. So, what does x equal? ​​It equals minus four divided by two, and minus four divided by two is minus two. So, x is greater than minus two. So, what are the solutions? They are all the numbers that are strictly greater than minus two. So, when I want to write the solution, I will say that the solutions are the numbers that are two, one after the other. Now, for question number four, add a system. So, give me a system here that I need. We have two ways to solve a system. The first way is to find an x function, like this: for example, if x + ≠ 40, I'll make x = 40 - ≠ 40 and substitute it into the second equation. Or, I can multiply.

[00:06:40]I perform the multiplication operation. For example, notice here you have two x's and here you have x's. So, if I want to remove this x's, what do I do? I multiply it by -2x's. So, if I multiply, I multiply it by -2's, and it becomes -2x's -2's, which equals -80. I add the sum to remove this -2x's with these two x's and find the x's, and then substitute. But this isn't the way we'll work.

[00:07:05]We'll work the easier way: we'll find one function of the other. So, notice this first one: x + ≠ 40. What does it become? It becomes x = 40 minus this plus we brought it here. So now I have x, which is 40 minus. And we'll come to these two equations, and wherever we find x, we'll replace it with 40 minus. So what will I have? Here's 2. This 2 you have here multiplied by x and x, we said it's 40 minus. So now I have 2 times 40 minus plus 5, which equals 155.

[00:07:45]Let's develop here: 2 times 40 is 80. 2 times minus is minus 2 plus 5, which I have here equals 155. So you'll still have 80 minus 2 + 5, which is plus 3, which equals 155.

[00:08:03]Where did we get 75 from? So minus 2 plus 5 is 3. And this 80 you have here, what did we do to it? We brought it here, so you get 155 minus 80, which equals 75. So what does it equal? ​​It equals 75 divided by 3, which is 25 net.

[00:08:27]Here, what did we find? We found that it equals 25. Now that we've found it, what am I required to do? I'm required to find the solution, or rather, find what x equals. So, we'll come to these questions you have here. In them, x equals 40 minus 25, and we found it equals 25. So, what does x equal now? It equals 40 minus 25, which is 15 net. So, x equals 15, and x equals 25. So, this is the solution to the system. After that, we'll come to question number five.

[00:09:10]Always, when they give me a problem here, it has a relationship with that system they gave me before. They always have a connection. So, let's go to question five and try to understand the problem. Amina, posi 40, pi, combo, d, pi, d, 20, d, g, and pi The 50 dirham note, etc. Anyway, let's try to explain the problem to you. Amina has 40 bills. How many bills does she have?

[00:09:37]40 bills, right? Some of them are 20 dirham notes and some are 50 dirham notes. 40 bills are a mix of 20 dirham and 50 dirham notes, but we don't know how many 20 dirham notes she has and how many 50 dirham notes she has. So, the total value of those 40 bills is 1550 dirhams. We calculated it and found it to be 1550 dirhams. Agreed?

[00:10:02]The question is, if we have 20 dirham notes, we want to find the number of 50 dirham notes and the number of 20 dirham notes. Let me repeat the problem so we understand it better. We have Amina with 40 bills. The total value of those bills is 1550 dirhams. The banknotes they have are only 20 and 50 dirham notes. They don't have, for example, 100 or 200 dirham notes.

[00:10:29]The question is, we want to know how many 20-dirham notes she has and how many 50-dirham notes she has. Okay? So, what we'll do is: we'll say X is the number of those 20-dirham banknotes and R is the count of those 50-dirham banknotes. Okay? Because I want to know how many there are in each one. Clear?

[00:10:55]Next, what we'll do is use a system. We know that the number of 20-dirham banknotes plus the number of 50-dirham banknotes equals a total of 40 banknotes. So, we'll count the banknotes in her hand. She has 40 banknotes. And here's the second calculation: we know she has 20-dinar notes and 50-dinar notes, but we don't know exactly how many she has. We don't know exactly how many she has of 20-dinar notes and how many she has of 50-dinar notes. Those 20- dinar notes are what we called x. So x = 20 + 50 = y. That is, multiplied by the number of 50 dirhams, it equals 1550. Agreed?

[00:11:41]I think the problem is clear. This 20x + 50y = 1550.

[00:11:49]If we divide it by 10, we divide 20x by 10, 50y by 10, and 1550 by 10. We get 2x + 5 = 155. Notice this calculation: 20x + 50 = 1550.

[00:12:06]We'll substitute this because I divided everything by 10. 10.

[00:12:09]That's 2x + 5 = 155.

[00:12:14]So, the question or the system, what's in it? It has x + 5 = 40 and 2x + 5 = 155. What does this system remind us of? It reminds us of the system we solved in question number four, and we found its solutions. So, without going into detail, I'll just say that in question 4, we found x = 15 and r = 25. What is x? What does x mean? It's the number of 20 dirham notes and r = the number of 50 dirham notes. So, when they ask me how many 20 dirham notes she has, I'll tell them she has 15 20 dirham notes and 25 50 dirham notes. So, that's the answer: 15 notes of 20 dirhams. 20 and 25, 50 dirham banknotes. So we'll have finished the first exercise. All regional exams in mathematics have this exercise presented in this way, almost the same way. So I need to do the exercise well so I don't lose five points.

[00:13:16]Now let's move on to exercise two. Exercise two: If it has -1, then it has 1, and so has 5.

[00:13:32]Question number one: What did they ask me to do?

[00:13:35]Place the points, so I need to draw a grid and represent the points on it. Okay, so here it is, I have a grid, this is it. So I need to come and represent the points. The first point has -1, so I'll go to -1 in physics, here it is, -1, right? And at the 1, here it is, and I'll come here to find my point, which is -1. For the second point, it has 1. So I'll come to the first number, which is the x. From x, look for it, here it is, 1, and here I have 2, so Donk Kenji, here at two, one with two, I'll make a point like this, a point like this, so I'll get to this point whose cords are one. The same thing for the C, it's a mess.

[00:14:29]Donk, I'll go to the X, I'll look for five, this is it, and I'll go to the L and look for four, this is it, and I'll get to this point where they meet here.

[00:14:39]This is the C whose cords are five and four. I go to question two, he told me to calculate the cords of point M, the millimeter of the segment.

[00:14:49]He told me to calculate the millimeter of the segment B and PC. So, I have C, which is a segment, I want to calculate the cords of M, which are its millimeter. I need to know the relationship of the millimeter. So, when I want to calculate the millimeter of a segment, what do I do?

[00:15:05]XP + XC / 2, AR + C / 2 If they gave me, for example, one sigma, what would I do? I'd write x_b. For example, if they gave you here, you would write x_a + x_b / 2, or x_a + x_b / 2. Agreed? So this is the rule I need to memorize. I'll go and substitute. How much do I have for x_b? I'll go to the point and see how much x₁ has. Here it is, I wrote it as 1C. I'll go to C_xC, sorry, x_C. How much does it have?

[00:15:45]Here it is, 5. Here it is, 1 + 5 / 2. The same thing. I'll go to the point and take its value, which is 2. Here it is, I wrote it. And your value for C is 4. Here it is, I wrote it. So 2 + 4 / 2. If we calculate, 1 + 5 is 6 divided by 2C, which is 3. 2 + 4 is 6 divided by 2C, which is 3E. So we have found the values ​​for the 10_b.

[00:16:07]Now I'm required to do the decimals.

[00:16:09]When you're given a vector AB and told to calculate its values, what do you do? For example, if I have AB, I go to point B and write x, b - x, a, r, b - a. Now, if they gave me a CD and told me to calculate its values, what would I do?

[00:16:28]I would write x, d - x, c, r, d - r, c.

[00:16:33]Now, let's substitute. So, I have x.

[00:16:38]What do I do at point B?

[00:16:41]Its x is 1, and the x of a is -1. So, it's 1 - -1. 1 - -1 is 1 + 1, which is 2. I'll go like this: br, b, b, b, a is 1, 2 - 1 is 1. So, the values ​​of AB are 2 and 1. Understood? These are them. Now, let's move on to the next question. The next question is question number four. What did the questioner tell me? The equation for a is a = 1x + 3. That's why he told you to show me that the equations The equation AB is A = 1/2x + 3/2. He gave it to us and told us to figure it out.

[00:17:29]First, when we want to figure it out, we need to calculate A. So, A is this coefficient. We need to calculate this 1/2. We need to find it. What do we know? We know that this coefficient is the direct coefficient. What do we do? For example, if we want to work on A, we do π - π/x, π - x, π. And if they gave you, for example, CD, you would do dy - cy/x, dy - x, cy. So, we need to memorize this relationship.

[00:18:02]Now, we'll substitute how much π we have. Here's its arc, which is 2. And how much A we have, which is 1. So, 2 - 1. And below, how much x we ​​have, which is 1. How much x we ​​have, which is -1. So below, you'll have 1 - -1.

[00:18:23]What does that equal? ​​2 - One is 1, one minus one is two. So we found A, we found A which is one over two. Here we are, we found this one over two. So what's left? I need to keep showing that over two. I need to find it.

[00:18:39]We know that equations are written in the form A = x + 1. We found A is one over two, so it equals one over two x + 1. We still have the problem with this B. We need to look for it. What do we know? We know that these points belong to that line, right? So we'll take either point A or the point I want and substitute it into this equation to find it. We'll do an example. For example, I'll take B. I'll say that B(1/2) belongs to that line. So we'll substitute here.

[00:19:15]How much do I have? Notice how much I have for B( 2). So we'll come to Let's write 2 = 1/2x.

[00:19:22]What do you have x for 1? I replaced this x with 1, so now we have 2 = 1/2 * 1 + b. Now we're going to look for b, so this is... Notice this equation: 1/2, you'll get 2, so what do you get? 2 - 1/2 = b. We'll find a common denominator: 2c * 2c is 4 - 1 is 3/2. So what is b equal to? It's 3/2. So we found a is 1/2, and now we find it's 3/2. Therefore, it equals 1/2x + 3/2.

[00:20:12]This is what they asked me to show here. Understood? Let's move on to the question: Construct the parallel points.

[00:20:20]To show that these parallel points (i and c) are collinear points, they must belong.

[00:20:26]We know that a and b belong to the same line, AB. So, I need to show that c belongs to them. Therefore, c must belong to that line AB. So, what do I do? I'll take these equations I found and take the values ​​for c and substitute them here to see if they work. So, what do I have for c in the given information? Here it is, 5 and 4. And the equation I found is y_r = 1/2x + 3/2. Now I'll substitute and see if it works. What do I have? It's 4. Here it is, 4, right? It equals 1/2x + 3/2. So, what do I need to show? I need to show that y_r, meaning 4, equals this. I'll substitute 1/2x and 1/x. How much x do I have? It's 5. Now I have 1/ If we agree on 5, then plus 3 over 1, so this is what we mean. This is 1/2 x + 3/2. When I substitute x with 5, when I calculate, it should give me 4, it should give me 1/2, it should give me this 4 I have here. So I have 5 times 1/2, which is 5/2 plus 3/2. The denominator is 1, which is 2. 5 plus 3 is 8/2. So 8/2 is how much? It's 4. So it came out correct. If, for example, I got a different number, like 7 or something, I would say that that point doesn't belong to that line, so it's not linear. Do you understand the idea? Okay, if they give me, for example, this point and give me its denominators, for example 2 or 5, and tell me to go and check that it's linear with those points, what am I going to do? I'm going to substitute In the equation, we'll do 1/2 x, which is 2. We'll do 1/2, then 2 + 3/2, and we'll see if we get 5. If we get 5, then it's true. If we don't get 5, then it's not true. Understood? So we said that this point is to prove the equation is true.

[00:22:40]The question after that is, what am I required to do? I have to prove that 5. Or they gave me a equation called d. Here's its equation: it equals -2x + 4. In question A, they told me to prove that d, this d they gave, its equation is -2x + 4, and that the equation we were given at the beginning is equal to 1/2x + 3/2. They showed that the two equations are perpendicular to each other. You have to remember a rule when you want to prove that two equations What I do is take their coefficients and their direct coefficients from the first equation and multiply them by the coefficient of the second equation to see if I get -1.

[00:23:32]I need to show that a_f and a_p equal -1. If we take this first equation of ab, what do we have? a is 1/2, which is the one attached to x. And this second equation we have here, what do we have? a is -2.

[00:23:46]So we do 1/2 times -2 and see what we get. So the a of ab is 1/2 and the a of d is -2. So 1/2 times -2 is -2. 2/2 is -1. So I got -1. And if I get -1, then I say that these two equations are how they are. So we show them that the question after it tells you to draw them.

[00:24:17]It tells you to draw them. The first thing I have is that you pass through them. From zero to four, and then from two zeros [music] and we'll go and draw it.

[00:24:34]Let's get to the question, or let's get to the third exercise. The third exercise is statistics. The table I have here, what does it represent? It represents the baggage allowance of 25 days. So we have 25 days. We went and found out how old each one is. Let's go to this table. We found that 10 of the players are... sorry, sorry, we found that the number of players who are 10 years old is 8, those who are 11 are 6, those who are 12 are 1, those who are 13 are 1, those who are 14 are 4, those who are 15 are 3, those who are 16 are 1. The first question took me to the mode of the game, so I have to calculate the mode. What is the mode?

[00:25:22]Look here, the effective Maximal, you go to the effective number and see what your maximum is.

[00:25:28]Who is the highest effective? So, my highest effective is... and you go to the character and see what number is opposite that effective number you have. So, what character is opposite that effective number? 8 is 10. So, that's the mode. The mode is that number or that character that is most suitable for the highest effective number. Okay, the question that led me to the median is... so, I want to calculate the median of this series. The first thing I need to do is find out the total number. It's 25.

[00:26:01]Where did we get the total number? I go to this number, here it is: 25. So, how does this number work? When you have it and you want to calculate it, you divide by one, add one, and divide by... So, when we add one, we divide by... and get 13.

[00:26:25]Here you need to... Write them down on one piece of dirty paper.

[00:26:29]How many eights do you have?

[00:26:32]How many one? How many are 10 years old? How many eights? How many are 11 years old? I'll explain it to you properly. We'll come here. We said how many are... How many are 10 years old? Eight. So how many eights do I have? 10 times. One, two, three, four, five, six, seven, eight in order.

[00:26:59]Now, what follows eight? It follows six. How many are six? 11. One. So one, two, three, four, five, six, seven, eight, in order. Now, what follows eight? It follows six. How many are six? 11. One. So, one, two, three, four, five, six, seven, eight, nine, ten, eleven. So you understand it well. And I need to continue, but I won't continue. I'll come here. How many did I find the mean? The mean I found is 13. So, I start counting from here, from here, from the eight: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 12 and 13 came here. So, what is the number opposite 13? It's 6. And that six we found is... What is the corresponding number? It's 11. So, this is the average. So, I got 11. So, my median is 11.

[00:28:16]Question three asked you to calculate the mean of statistical analysis. So, I need to calculate the mean. When I want to calculate the mean, it's easy. What do I do? I go to the sum of x, i, n, i, and t. What is this x, i, i, i? Look, 8. You do 8 times 10, plus 6 times 11, plus 2 times 12, plus 1 times 13, 4 times 13, 3 times 15, 1 times 16, divided by the total. How much do I have? How much do I have? The total is this: 25. This is the sum of those variables. The 25 millimeters I have, we'll calculate it and get 300 divided by 25, which equals 12. Understood?

[00:29:09]So, we'll continue and move on to exercise number four. If exercise four is the function F(x), then the F of two is -1. You gave me the function definition. The function F, so what do I do? We know that the F of two is -1, and since we have a function, the function is written as F(x) = x. So, what am I required to do? I'm required to look for A.

[00:29:43]Understood? So, I have 2x = -1.

[00:29:46]How did we get 2x = -1?

[00:29:50]Because this F(x) = x, and I have the F of two = -1. So, automatically, this -1 you have here equals 2x. What you have here is F(x). So, what does A equal? ​​It equals -1 divided by 2. Net concept. Let's go to the question.

[00:30:15]He told you, "Ekrig F Do X On Funxion Do X." Here, he asked you to find me other than... The formula for F, so it tells you to find only A, then it tells you to write F(x) as a function of x. We know that F(x) equals ax. What did we find? A is -1/2. So, we'll substitute it here, and we'll have F(x) equal to -1/2x.

[00:30:37]The question is to construct delta as the graph of the function F, so I want to draw delta. Delta is the representation of F. We have F equal to -1/2x. The graph is the linear function, since it's a linear function, so it passes through the logarithm from 0. So, the code for the first point is Z. So, √x = 0, F(x) = 0. We already know this.

[00:31:10]I need to find another point, so √x = 2. I'll take x = 2. And now we'll substitute here. So, how much do I have? F of two is -1, which is what they gave me at the beginning. If you want to take another point, you have the right to take it and substitute it in the function and see what you get. So, if we substitute two here, wherever we find it, look with me here, you have F of x equals -1/2 x. If you substitute F of x here, meaning wherever you find x, you substitute it with 2. So, if you substitute x here with 2, you will have -1/2 times 2. So, it's -2/2, which is -1. So, we found the second point, which is 2 -1. So, we found two points. Since it's a function, it passes through Z and passes through 2 -1. So, here we come to the graphic. We said here, it passes through Z to Z and passes through 2. It's two minus one. Here you have minus one. Here you have minus one. So you'll go to two minus one, draw a point, and that line will pass through that two minus one and through Z. So this is the graph aspect.

[00:32:21]We'll come here. He told you to take the graph of 3. That drawing we drew showed the G. We went to the drawing, and the image of the image of 3 is minus. Here he told you what number has an image of two.

[00:32:42]Either I work it or I go and work it from that function I have. Okay, the G function I have, but I work it graphically. We work it from the graph. So G is equal to 2. So what is the number whose image equals two? It's one x prime G is x. So when I want to work G is x, I have the lines. I'll take two points that pass through from Let's calculate the coefficient. What is coefficient 0 minus four? It's the value of this first point, which we'll call, for example, DC. It's DC minus A of C divided by X, which is -X of C. So it's 0. 0 minus 4 divided by 2 minus 0 is -A divided by C, which is -A. We found it. So now that we've found it, what do I need to do? I need to find the value of A.

[00:33:40]You have x, right?

[00:33:46]Plus A equals X plus A.

[00:33:55]What did we find? A is -2 x plus A. Now we'll take a point, that point we 'll go to find B. We'll take a bar. That bar has X and X. So, for example, let's take this bar 0. What do we do? We'll substitute B and X with 0 to find A. So we'll find A equals 4. If we substitute A with 4, what does A equal? It equals, it equals, minus two, right? Minus six times x, and x is 0. So, minus x is zero plus this. That's it. I still have what it equals: 4. It equals minus a, x plus. Let's go to x, five. So, x, five, what's required of me? I'm always given one here.

[00:34:55]Here, they gave me k, so they gave me x, which is k. Okay, remember this: it's a square.

[00:35:28]This is t. When he said calculate x times x, he said calculate how much it equals. Understood? Right? We know something. We know that the volume, meaning the size of that pygm, equals 1/3 times y. What is the area of ​​the base times what, which is the height? This is a base we know. So, this is the area of ​​the base. And what is the base you have? He told you the base is a square.

[00:35:36]And what is the area of ​​a square? It's side times side. So, this is... We substituted it with pi, which is two, since its area is squared (side by side). So, we have a value equal to 1/3 times abi + voa to the power of A.

[00:35:56]What is the power of A? It's the height. So, the power of A we want to calculate is the height. So, we have a volume equal to 1/3 times the base, which we called ab, voa (meaning abi and square), and the height.

[00:36:13]We want to find that height.

[00:36:14]How much volume do I have? They gave me 216 cm. I substituted it, and it equals 1/3 times the height. So, how much do I have? This is how much side I have. It tells you that abi equals six. So, 6 to the power of two times the power of A, which is what I'm looking for. So, what does 216 equal? ​​If you calculate this, 6 to the power of two is 36 divided by 1/3, which is 36 divided by 3, which is 12. So, you have 216 equals 12. So, what does this equal? ​​It equals 216 divided by 12. If we calculate it, we get 18. Understood? And we'll have found the ext., which is 18. Now, let's move on to the next question. I'm required to calculate the x-axis. Here it is. Calculate the difference: x-axis. Here it is: x-axis. Okay, x-axis. Here it is. Calculate this difference.

[00:37:15]Focus carefully on how we'll calculate it. We have a x-axis. Wait, let's come here. We'll start from the base. If we want to calculate the x- axis, follow along.

[00:37:39]What does it equal? ​​It equals x-axis. This is x-axis plus x-axis plus x-axis. So, we're working on this triangle. Here's why. What is it? This is the x-axis of the triangle. This is the x-axis of the triangle. This is the x-axis of the triangle.

[00:38:03]When I have a right triangle... The angle is x. I have that hypotenuse, which equals the sum of the two sides. So I have s⁻², because this is a right angle here at A. This is a right angle at A. So you will have that s⁻², which equals s⁻² plus s⁻².

[00:38:26]I have s⁻², which I calculated in the previous question; it is 18. It is 18⁻². The problem I have is that I need to find x. Where do I get it?

[00:38:39]Since I have this square, notice with me, I don't work on this, I work on this x, because this is a square. So here I have a right angle. So ac x equals s, which equals ab.

[00:39:00]Here is x. Here I work on this. Here I have a right angle. So this ax you have here equals this. The sum of these sides equals this plus this, and this is six. And this is six.

[00:39:14]Agreed? And I will find what it equals. So we found it to be six. The square root of two is 2^(2^(18^2)), and the vertices of 2^(2^(18^2)) are 324. The vertices of 2^(6^2)) are 72, totaling 396. So, when I want to calculate the square root of 2^(2), I take the vertices of 396, which is 6, and the vertices of 11, and we will have found the square root of 2^(2).

[00:39:45]After that, he asked me to calculate the ratio report, since he said ratio, like he said that factor of reduction. So, I have a position of 3. What does that equal? ​​Because when we came here, what did he tell you? Pay attention to what he told you. He told you to give him the volume of this pygmide.

[00:40:05]This small pygmide is how much? It is 8 cm³, and the large one is how much? It is 216. So, when he tells you here to give him the ratio report of reduction, you take that small volume. Dividing by that large number gives you a value called the logarithm of the deduction. So, it equals 1/27 because if you divide 8 by 216, you get 1/27. If you want to find it like this, it's 1/3. The calculation we'll do is try to remove these two values, the ratio 3. So, we'll do that ratio like this, the ratio 3 will give us 1/3. So, this is what's called the logarithm of the deduction. So, what do we have? It's 1/3. Now, it tells you to calculate the difference (S).

[00:40:58]You have this whole difference called S_A, and this S_O here is like the reduction of this S_A. And you found that the reduction factor for the deduction is 1/3. So, what do you do?

[00:41:13]Notice that we have S_A and S_O, meaning this large S_O equals So, that coefficient is multiplied by what? What do I have? That coefficient is 1/3.

[00:41:28]What do I have, the power of the letter A? It's 18. So, what do I have, the power of the letter S? It's 18 times 1/3, which is 6. So, the power of the letter S equals 6 cm. So, we've finished this exercise. Now, let's move on to exercise number 6.

[00:41:43]We have a pigeonhole (p < 0.g., ABCD) in pigeonholes. We have a pigeonhole whose center is π, and π is the density which is converted to the density of the letter C. So, I need to draw that pigeonhole. It's a pigeonhole like what? Its name is ABCD.

[00:42:06]Its center is this.

[00:42:08]Here, the center will be... The first question is: Consume point π, my image is const. The second question is: Image of point π, image of the density. And we'll show that point π is the millimeter of π. So, we'll go little by little.

[00:42:29]Notice with me in this figure ABC Here it is, here it is, ABCD. This is the first shape we started with. He told you to draw the point, or draw it. This is the image of this, or the image of B with splicing. And what is the splicing of T? It is D and C. So from this, or this is the image of B with this splicing from D to C. So from D to C, we go from left to right. So the image of it too, I have to go from D to C.

[00:43:07]And this distance that I will have here, I have to have here so that we will have drawn. And the question after it told you to draw the image of F with splicing.

[00:43:18]Here, and we want to draw its image, which we call F. It is the image of splicing.

[00:43:23]When we say splicing, it means from D to C. So from D to C, I go to the right. So I come to B and I have to go to the right, but You'll tell me where to stop it, here or there, etc. I see this distance, DC. The distance DC is the distance from O to F, which will come to me approximately here, and we will have drawn our image here.

[00:43:51]Question number: between me and this, what is the value of O and C? So what do I have? I have B and O equals DC, and O and F equals DC. We extracted them from this shape, so we have.

[00:44:12]Notice with me, B and O equals DC, and O and F equals DC. DC repeats here and repeats here, so B and O equals O and F. So I have B and F in PaGalilograms. So O and F equals B and O. We extracted this from here because this is PaGalilograms, and we have O, which is the value of AB. What does CD give me?

[00:44:36]It gives me that O equals O and D, and we have D and C and F, which is PaGalilograms. So O and D. So this O and D equals C and C. This equals √F. So, what do I need to deduce from this? Notice something I need to deduce from here: it equals √F and √B, and it equals √D. So, we will deduce that √F equals √D, and we know that √D equals √F. So, √F equals √F. So, consequently, we found that √F equals √F. So, √F is the millimeter of that segment √F.

[00:45:30]This point that is being repeated here is the millimeter of that segment √S. You had a vector, for example, a_b, right? It equals, for example, √D in a pagalilogram. What is the point that is being repeated to us? So, what will we deduce?

[00:45:52]We will deduce that it is the millimeter of a. That's it. That's it. So, we will be