This video provides a complete solution to the 2025 unified regional mathematics exam for the third year of preparatory school (الثالثة إعدادي) in the Marrakech-Safi region, Model 2. The exam covers fundamental algebraic concepts including solving linear equations and inequalities, applying algebraic identities such as (a-b)(a+b) = a² - b², solving systems of equations using elimination and substitution methods, and working with linear functions including evaluating functions at specific points and finding intersection points. The statistics section covers calculating mode, median, mean, and percentages from frequency distributions. The geometry section includes vector calculations, proving parallelograms, finding line equations and their intersections, and calculating the volume of pyramids using the formula V = (base area × height) / 3. The video demonstrates step-by-step problem-solving methods for each exercise type, providing detailed explanations of the mathematical reasoning and calculation processes.
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تصحيح الامتحان الجهوي 2025 في الرياضيات | الثالثة إعدادي | جهة مراكش آسفي |النموذج 2 (الحل الكامل)Added:
Peace, mercy, and blessings of God be upon you, followers of the Success Platform for School Support. Welcome to this new video in which we will be providing the answer key for the unified regional exam for the preparatory school certificate.
This video is intended for students in the third year of preparatory school, whether they are enrolled students or independent candidates preparing for the regional exam.
We have already provided the answer key for the social studies regional exam, using two models, and we have also provided an answer key for mathematics for the Casablanca-Settat region.
Now we will be working on another mathematics exam for the Marrakech-Safi region for the June 2025 session. As for the regional exams, they are generally very similar. If you look at the exam for the Casablanca-Settat region or the Rabat-Salé- Kénitra region, it is very similar to this exam for Marrakech. Sorry, the marking is almost the same, the questions are almost the same, etc. So, let's try to understand how we work on the math exam. Let's start with the first exercise. So, what's required of me in the first exercise? The first question will add the equations. So, I'm required to solve this equation I have. Agreed? The equation I have is 3x - 1/2 = x + 3/2. So, when I have an equation, what do I have to do? Those that contain x, I have to add them next to each other, and the numbers or ordinary numbers that don't contain x, I add them next to each other. But when I move one side to the other side, I have to pay attention to the sign. Agreed? For example, now I have 3x. 3x = -1/2 = x + 3/2, right? So, when I move this x to the other side, it had a plus sign, so what will it have now? It will have a minus sign, so it will become 3x - Here, I have 3/2c, which I left as is.
But here, I have -1/2, which I need to bring here. When I bring it here, what will it become? It will become +1/2. And we continue the calculation. So, 3x - x is 2x, and over c + 1/c is 4/a, and 4/2 is 2. So, what will I have? I will have 2x, which equals 2. So, what does x equal? It equals 2 divided by 2, which is 1. That's it. So, we've finished the first question. All exams give us the same question; they only change the equation. So, the method is the same. Let's move on to the next question. They told me to figure out what to do. They gave me an expression: tx - 4 equals 3x - t multiplied by 3x + t.
They told me to show that this expression equals this expression. Always, when you give me this type of question, I go and start with this second expression that you have here. This is the one where you can put in it Developing is easier than factorization, so we always tried to start from this to show that it equals this first expression, agreed? So we have 3x - 2 multiplied by 3, which is 3x + 2. So we'll develop them.
Notice with me, I have this: 3x - a multiplied by 3x + a.
When you reminded us, you reminded us that it equals what? It equals a - a multiplied by a + a. This is the third identity we memorized, the third important identity. Let's apply it.
Here I have 3x, which is a, and two is a. So notice with me, here I have a - a and here I have a + a. So when you have a - a multiplied by a + pi, what does it equal? It equals a to the power of 2, which is - pi, which is a. So we said that a is this 3x, so we will have 3x. Okar minus bi, okar minus bi, how much do I have? It's two minus two raised to the power of two. So, three x or two is how much three? This exponent here, this two, is for three and for x. So, to the power of two is x₆ or to the power of two is x₆. So, what I get is nine x₆ squared or x to the power of two minus two raised to the power of two, which is four. So, what I got is x to the power of two minus four, and that's what I wanted to show. Here, we agree at the beginning, so we explained this expression and made sure it's correct, as they told me. Let's go to the question.
What did they ask me for? They told you to find the solutions.
What are the solutions they gave me?
x okar minus four. They told me to find the solutions for these questions. I always help myself with the previous question. Notice with me now, when I want to solve this, it will be difficult for me. So, what should I do?
First, what did I show? Notice with me here, I showed that this T(x) * 2 - 4 is 3x - 2 multiplied by 3x + 2. So instead of solving the equation T(x) * 2 - 4, which equals 0, what should we solve? We'll solve the equation of 3x - 2 multiplied by 3x + 2, which equals 0. That's the one that will be easy. So, this is it. So, when I have 3x - a multiplied by 3x + a, which equals 0, we learned that a * equals 0 means that a equals 0 or equals 0. This gives me 3x - a equals 0 or 3x + a equals 0. So, 3x - a equals 0 or 3x + 2 equals 0. And we'll solve each equation separately. So, 3x here is the first one. This is 3x - 2 equals 0. What did we say?
We said that There are x's in them, I need to keep them in one place, and the ones without x's, I need to add them in another place. Now, here I have -2 -2.
When it comes here, what will it become? It will become +2. So, I'll still have 3x, which equals 2. So, what does x equal? It equals 2/3. Agreed? Let's go to the second case. I have 3x + 2, which equals 0. So, this +2 had a plus sign. When we move it here to this side, what will it become? It will become -2. So, I'll have 3x = -2. So, what does x equal? It equals -2/3. That's it. So, what are the solutions we found? 2/3 or -2/3? So, the solutions are -2/3, 2/3, and 2/3. Agreed?
Now, let's go to question number three. What did question number three tell me? It told you to add a swift link.
What is the difference between these links? The inequality and the equation have the same answering method; we answer them the same way.
We find the solutions to the equations of the questions, and we find the solutions to these. So, what do I have? 5x - 1 is greater than or equal to x + x. As we said, I always add the xs in the same place and the ones without x in a separate place. So, here I have 5x. I'll bring this x to it.
When this x comes here, what will it become? It will become -x. So here it becomes 5x - x, which equals 7. I leave it in its place. But this -1 I have here, when I bring it here, what will it become? It will become +1 because I change its sign. So it will become 7 + 1. So, 5x - x is greater than or equal to 7 ≤ 1. So, what is 5x - x? It's 4x, and 7 + 1 is 8. So, what is x?
This is 8 divided by 4, and 8 divided by 4, which is 2. So we found that x is greater than 2, agreed. What are the solutions? What will we write in the solutions? We will write the solutions as the solutions of all numbers. All numbers greater than -2 will take all numbers greater than -2. We say the solutions are 3 and 4. The sentences are complete.
Question number 4. Question number 4 is this, and they asked me to solve this system. I need to solve this system.
We learned that when we want to solve a system, we have two methods. The first method is either to go, for example, and find x. Here, what is it equal to? It's a function. For example, you will find x, which is 33 - 2 divided by 3, and then substitute it into the second equation. This is the first method. The second method is that I try to multiply the above and below. In the same way, in a simple equation, let's give you an example. For instance, if you multiply 3x by 5, you get 15x. And if you multiply 5x by negative 3, you get negative 15x. So, if you add that 15x to negative 15x, you get zero. So, these x's are gone, we'll calculate something else, and then we'll solve. Pay attention here, how am I going to do it? Wait a minute, excuse me, the first question didn't ask me to figure out the system. They'll ask me about the system in question B. In question A, they asked me to verify that this couple is the solution to that system. Now, they gave me this system and they gave me a couple and told me to show that it's the solution to this one. They told me to show that it's the solution to this one. So, what am I going to do? We'll go and substitute in that first equation. Notice with me this couple you have here: t 3 t So, we'll substitute x and 3, and then we'll see what happens.
If I have 3 times x, what did we say x is? So it will become 3 times t + t x.
What is 3? That is, plus t x 3 x 9 is 27.
Plus 2 x 3 x 6, we got 33. Agreed?
Now the first equation is good. I need to double- check the second equation. I'll do the same thing. I'll substitute t and t into this second equation. Here I have 5 times x, so 5 x 9 + 4 x 3 = 45 + 12 = 57. But here I have 61, and here I got 57. So this is not a solution. So, when I find that equation doesn't work in the first equation and the second equation, I say Yes, it's not a simple solution, so what's the answer? We have 57, not 61. So, we say that the point you gave us, 3, is not a simple solution for the system.
Agreed?
Now, let's get to the question. They asked me to do the solution for that system. Notice what I have here. Look, follow along carefully: 3x + a equals 33, 5x + 4x = 61. I need to remove them so that only one remains. I just need to find the easiest one. Agreed? So, the first thing I'll do is remove the two. Follow along carefully. Here you have x and here you have 4.
Agreed? So, 2 and becomes 4. When I multiply by 2, I need to multiply this by 2, and this... this... I need to multiply by... Okay, so I have 3x times 2, which is 6x times 2 times 2, which is 4, and 33 times 2, which is 66. Agreed? Now what do I do? In the second equation, I'll take this second equation you have: 5x + 4a, which equals 61. I'll subtract it from these first equations. And these first equations are... these are the ones I have now, right? So what will I have? I'll have 6x + 4a, which equals 66. I'll subtract 5x + 4a from it, which equals 61. How do I do this? We subtract x with x, x with y, and numbers with numbers. 6x + y, here it is. I'll subtract 5x + 4a from it, which equals... and this 66, I'll subtract 61 from it. So what is this? This is 6x + 4a.
This is 6x + 4a. -5x - 4, I write it for you as 6x + 4, then -5x. Why is it minus? Because the parentheses are preceded by a minus sign. So, if we change the sign, -4a, -4a, -4, something with 4, i.e., 6x - 5x is x. And here, 66 - 61 is 5. So, how much does x equal? It's 5. So, I found that x equals 5. Here, I found that x is this. Next, what do I need to do? I need to find the value of x. How do I find x?
I'll choose one of these equations I have, either this one or this other one, and I'll substitute x with this 5 to find it. So, I chose to work with the first one: 3x + 2a equals 33. I'll substitute x, and what is x? It's 5. So, I'll have 3 times 5 + 2 x = 33 times 5, which is 15. So, this will become 15. Plus 2 equals 33, and this 15 we'll bring to the other side next to 33, so you end up with 2 equals 33 minus 15. So here we have 15 plus 2 equals 33. We'll leave this here, and this 15 will come here and become negative because it's a plus and has to become negative, so it will become 2. What does 33 minus 15 equal? And what is 33 minus 15? It's 18. So what is 18 divided by two, which is nine? So it's nine. So we found x is five and we found arc is nine. So what is the solution to this system I have? It's five. Agreed?
Now let's move on to the next question. This question tells me there's a problem. Let's try to read it to understand it, and always pay attention. This problem that you have here is related to this. The system in the question before it, as always in all exams, so they told me we have a project, a play, that suggests two entrance fees: one for adults and one for children.
They told us that for adults, there's one fee for teenagers and one for children.
For example, we have a family with two adults and three children. That family paid 330 dirhams. For the family with four adults and five children, the five children paid 610 dirhams.
Take me to the theater, as it's easy to understand.
We have a family who went to a theater and wanted to So, this family, A, has two members ( Lizadelt) and three members (Lizonfou), and they paid 330 dirhams. The other family, B, has four members (Lizadelt) and five members (Lizonfou), and they paid 610 dirhams. The C family went to pay, but we don't know how much they owe. So, we need to help them. When the C family wants to pay, the question that will come to their mind is: how much do the members pay, and how much do the members pay? The problem we have is that we don't know how much the members pay and how much the members pay. So, in this case, what should I do? I need to calculate how much the members pay and how much the members pay so that I can then find out how much the C family owes.
I think the problem is clear. So, the first thing we start with is the price of the ticket. So, x is the price of the ticket for the child. And yk is the price of the ticket for one child. That's the first thing, agreed on. Now we need to create a system. Where will we get that system from? We'll get it from those two families because those two families are the ones who will help me. Notice with me: for family A, there are two children, so there are two xs, two children, two xs. And how many children do they have? Three. So, 3 x plus 3 a, how much did they pay?
330 dirhams. So, this is the first family, there are two children and three children, they paid 330 dirhams. So, we will write this as an equation: 2 x plus 3 a equals 330 Let's look at the family. The family has four " laysadilt" (low-valued units) and five "laysunfun" (low-valued units). The "laysadilt" are represented by the letter "x" (x). So, 4 x + 5 laysunfun. The "laysunfun" is their ticket. So, plus 5 x, what does that equal? It equals 610 dirhams. Understood?
610 dirhams. So, this system will help me find x. What are x and y? It's the price of the "laysadilt" ticket and the price of the " laysunfun" ticket. So, this system will help me find x. 2 x + 3 equals 330, and 4 x + 5 equals 610. Notice here, this system... does n't it resemble a system we studied? Here it is.
Look, it resembles this system I have here. So, what's the difference? Here I have 33, here I have 330, here I have 61, and here what? I have 610.
Understood? So, what am I doing? Okay, so I need to solve this system.
First, I'll use the solutions that came up here. What's the difference?
What do I need to do with those solutions? I multiply them by 10. Why? Because 33 became 330 and 61 became 910. So, to get from 33 to 330 or from 61 to 610, what did we do? We multiplied by 10. So, those solutions you got, multiply them by 10 to find the ticket price. If you don't know, you'll solve the system normally. So, here's the method for the system, and so on.
In the end, we'll find that earlier in the system we came up with, we found that is equal to 5 and x is equal to 9. That's what I told you. Now it's 50 and x is 90, like those solutions. We multiplied them by 10 because here it was 33 and became 330, and here it was 61 and became 610.
If you don't know, try to calculate the system the way we did it before. So, we found X and A. We found X is 90 dirhams and A is 50 dirhams. So, we found how much the child will pay and how much the adult will pay.
Now, what do I need to do? I need to go back and find out how much this family I have needs to pay.
This family, if it has two adult children, we represented them with X, so it's 2 x 90 plus how many extras they have (2 + 2). And we represented the extras with 2 x 50. So, what will I get? I'll get 2 x 90 plus 2 x 50, which is 180 plus 100 is 280, so how much will the family pay us? They'll pay us 280 dirhams. So, the exercise seems easy and understandable.
Let's go now to equation 2. Equation 2 is also easy, worth four points. What's required of me?
Define the function of line. What did they give me? The function of x equals negative one divided by 3 x. And one piece of information they gave me here: they told me it's a function of line. What does a function of line mean?
It means that it passes through the point Z. Okay, let's get to the first question: Calculate the function of 3. The function of x is given to you, and they told you to calculate the function of 3. It's easy. What will you do? Wherever you find x, you'll substitute three. And when they told you to calculate the function of negative x, what will you do? Wherever you find x, you'll substitute negative x and calculate. So, let's start with the function of 3.
For 3, I have the F of 3. Here it is. Wherever we find x, we substitute it with 3, which equals -1/3 times 3.
So 3 times 1 is 3. Therefore, this is -3 divided by 3, which is -1. So, what did we find the F of 3 - 1? Now, let's look at the F of -x.
Wherever we find x, we substitute it with -6. So, we get -1/3 multiplied by what x. And x is what? It's -6. -6 times -1. -6 with -1 will become +6. 6 times 1 is 6 divided by 3, which is 2. So, what do I have the F of -x? It's 2. So, we have calculated the F of -x. Now, let's go to the question after it. You took me to the next question... I gave it to you, but who is that number whose result is -2/3, like they gave us the inverse of the previous question? So how do I do this? I have f(x), what does it equal? It equals -2/3.
Where did we get it from? Here it is because they told me that f(x) equals -2/3.
They gave it to you, and we know that f(x) here in the given information is -1/3. When you have f(x) equals -3 and f( x) equals -1/3x, what does that mean?
It means that -2/3 equals -1/3x.
So I'm going to look for this x. So what do I do?
I multiply 3x by this -1. Notice this: this -1 is the same as -1 times x, which is -x. So, to make the explanation easier for you, it equals 3. We multiply it by -2, which here gives us -6 divided by 3, and -2 divided by 3 is -2.
So, -x equals -2. Therefore, what will x equal? We remove the quotient here. So, x equals 2. Therefore, x equals 2. This is the answer to the question. I think it's easy. When they tell you to calculate the result, it's easy, and when they tell you to find the number whose result is that, it's easy. Now let's go to question 2: the function √(2x) = √(2x).
Here, they gave me some information. They told me that √(5) equals 5 and √(2x) equals -4.
This is what I know about this function, and I know it's √(2x), but they didn't give me any other information. What did they ask me to do in the question? They told me to show that this function equals 3x - 10. The first thing is, when you hear this, you The function "a" needs to be written as f(x) = ax + b.
The linear function is also written as f( x) = ax, but where is a written as ax + b? What do I need to do? I need to find out what a is equal to and what b is equal to. So pay attention: here's the function a, where we write it as g(x) = ax + b. The problem I have is that I need to find a. So, the first thing is we need to find out what a is equal to. We have g(5), which is 5, and we have g(2), which is -4. So, I'll come here and substitute.
See what I did: ax + b, ax + b. What did I do? I substituted 5 for this x and 5 for this g(x). Now, I'll go and find out how to do a. Follow along carefully. This is the first one. Here we are. We've reached the first one. Now let's move on to the second. We have the x of 2 equals -4, meaning the x of 22 equals -4.
What do we do? We go to x and substitute it with 2. Here it is.
This is what we didn't do here; it's -4. So here you have 5a + 5 equals 5, and here you have 2a + 5 equals -4. So here, what can we do? We create a system and get out. So what do we do?
When we wrote the system, we wrote 5a + 5 equals 5, and we wrote the x of 5 equals -4. We didn't write them, right? Take this first one and subtract these from it. You have 5a - 1, which is 3.
We agree, right? -1 is 0. So 5 - 4 is 5 + 4, which is i. So now I have 3a equals i. So what does i equal? It equals x divided by 3, which is 3.
You found i = 3. What do you do? You put it back here.
So, g(x) = 3x. You'll need to go back to one of these equations to find it.
For example, let's take 5a + b = 5.
We'll substitute a with 3 to find it. Or, let's take this and substitute a with 3 to find it. For example, we used the second equation: 2 + b = -4. 2a = 3. So, we get 2 times 3 + b = -4.
2 times 3 is 6 + b = -4. So, what is b?
Find this 6, which becomes -6. So, it equals -4 minus x, which is -10. Therefore, b is -10 and equals 3. You'll go back and substitute into the equation you have here. So, g(x) = ax + b, which is 3. b is -10, so g(x) = 3x - 10. So, we've shown what they asked us to show here in the question.
Here we are today. 3x * g * x = 3x - 10. After that, they asked me to calculate g of 3 in the question g of 3. If you didn't answer this question, it's not a problem, you can answer it because they gave you the equation here. They gave us the equation: g * x.
Here, x is 3 * x, which is 3 * 3 - 10. 3 * 3 is i - 10, which is negative. So, what did we find? G of 3 is negative one. I think it's easy. Let's move on to the question about the representation of g. They asked me to show that this line g * 3 - 1 is the intersection of these two lines.
They told me that this line g * 3 - 1, I want to show it is the intersection of those two lines.
What do I need to do?
Try to understand the method of answering with me because it's the same questions. They get used to applying the same method, but only the numbers change. So, I'll come here. I have f of 3. Now I want to show that three minus one is the point of intersection of dy and delta so that... The point of intersection must belong to both line and line θ.
For example, this is line θ. So, this point must belong to both line θ and line θ to be the point of intersection. So, what am I going to do?
I'm going to go to the first function of f(x), which equals -1/3, and I'm going to substitute these. I know that f(x) is given to me as f(3). I'm going to verify it. Look here, f(x) equals -1/3. Go calculate f(3). If you get -1, then this point, √3, belongs to that line, agreed? So, wherever I find x, I'll substitute 3. So, I'll have -1/3 multiplied by 3, and I'll calculate it: 3 times -1, which is -1 divided by 3. So, when I substituted x with 3 and found that it equals -1, which is correct. So, since it came out correct, what should I say? I should say that this point, y, belongs to the line dy.
Here's the first piece of information: y belongs to the line dy. So, y belongs to dy. Now, we'll do the same thing for the second function, which will be y of x equals 3x - 10. So, what will you do? You'll go to x and substitute it with 3, right? And you'll see what you get: -1. So, I have 3x - x, how much is 3? So, it's 3 times 3 - 10, which is 9 - 10, which is -1. So, -1. Is this it? Yes. So, y belongs to delta. I take x and substitute it, and I see if it comes out. If it does, then it's correct. So, we find that this point, y - 3, belongs to dy and at the same time belongs to delta. So, that line segment G3 - 1 belongs to both of them, and since it belongs to both of them, it's their point of intersection, which is this. So, G3 - 1 is the intersection point of d and delta. That's how I answer. Now, let's move on to the next question, which asks you to draw that d and delta. The drawing is easy. I have the first function, f(x), which equals -1/ 3x. We know it's a line, so it passes through the first point, which is Z. I need to find another point, and what is the other point we found that belongs to that line? For that function, it's that G that we were basing 3 - 1 on. So, here's the construction of delta. Now, let's look at delta. Delta is written like this. Here's what I do when I want to draw it: I take a point from my own, as follows... For example, if I take a point, let's say I assign it x². Here, I'll substitute 2 here and see what I get. It will be 3 times 2, which equals 6 minus 10, which is minus 4. So, the first point, whose x I chose, has an x of 2, and its x is minus 4. Take any point you want, for example, take another point and assign it 5. The next 5 will be 3 times 5, which equals 15 minus 10, which is 5. So, when I took 5, what did I get? It's 5. Therefore, when I want to draw this line, this line delta must pass through 2 minus 4 and 5 minus 5, and this line dy must pass through 0 and 3 minus 1. That's how I'll draw it. Here's another example. Look at this table. For example, we'll work here on an example. Look at this. The equation is x = -1/3. This is the equation of x = 0.
We can write it as x = 0.
I know how it's structured; it's a line, so it goes from zero to zero. This is certain. The second point is, what do I do? I take a number and substitute it. For example, if I want to substitute x with 3, if I substitute x with 3 here, it will give me -1. So, when I draw, what do I do? I draw the first point, which has z. Here it is: 0. Normally, this should pass through here, so it's 0. The second point is 3 minus what? It's 3 minus 1. It's x = 3. I look for x = 3, which is 1, 2, 3. Here it is: -1. Here it is. This is -1. So, 3 minus 1, here it is. It will come here, so I draw this line. I do the same thing for the equations. The second one, I have 3x - 10. We choose two points, for example, we'll take x = 2 and then substitute the x = 2. It will give me -4.
So, I substituted x = 2 and it gives me -4.
If I take x = 5, it will equal 5. Then we'll draw it. We'll go to x and look for x = 2.
Yes, it equals -4. So, x should equal 2.
Anyway, this is just an approximation because there's a problem with the drawing. x = -4.
This is the first point. The second point is 5/5. We'll look for 5/5.
This is x/5. It should be here. This is the second point. We'll draw these lines, and when we draw these lines and these lines together, we'll find they intersect. The line at this point is 3 - 1. I explained it in the previous question.
I hope the explanation is simple and clear.
Now let's move on to Lexgesis 3 Statistics.
Lexgesis 3 is easy. Here I have a table with a lot of information, which is the number of cities visited, and the effective is the number of tourists, meaning the total number of visitors. So, the number of visitors who visited one city is 16, the number of visitors who visited two cities is 12, the number of visitors who visited three cities is 8, and the number of visitors who visited four cities is four. First, it tells me to identify the mode.
What do I need to show? I need to show the mode of this series.
When I want to show the mode, what do I do? For the positive mode, I look in the table for the highest effective. What is the highest effective in the table? It's 16. What is the appropriate number?
16 or the one opposite 16? It's one. So, what The modulus is one net, always. When you hear this word here, don't miss the point.
You go to a table and look for the maximal and see which is the corresponding value. Understood? Now we'll calculate some problems. Let's try to understand what to do when I want to calculate the total. First, I need to calculate the total number, which is 40. Where did 40 come from? The total number is the number of the maximal. The sum of the maximal is 16 + 12 + 8 + 4. This will give me 40.
First, I do the second thing. I try to calculate the maximal sum. What is the maximal sum? Here's a table. This table is a table of 4. Here's the sum: 16 + 12 + 8 + 4. I go and calculate the maximal sum.
What is the maximal sum? I get the first value. Here in the effective value, you write it here. Here it is. You had 16. Here it is. Where do we find the second value? Look what I do. We come here. Here, how much do you have? 12. You add 12 to 16. So, 12 added to 16 gives you 28.
When you want to look up this value again, what do you do? You take this value, which is eight, and add it to 28, giving you 36. When you want to look up this value, what do you do? You take this 4 and add it to this 36, and you get 40. This is how we calculate the effective cumulative value. So here, what do I have? The negative cumulative value. This is the effective cumulative value we worked on. If the effective cumulative value is 20, what is the largest effective cumulative value here in the table?
Greater than 20, why did we take it greater than 20? That's what you got here, that's the number here, that's the total number.
When you want to calculate, look, let me repeat it. When you want to calculate the median, you go to the total number, that's the total number, you divide it by two. Here's 40, we divided it by two, it gave me 20.
Then you go and calculate the summative effective, right? And you come looking for what? The summative effective greater than this 20 that you found. So who is 28? Because here we have 16 and the biggest one is 28. Now, 28, 36, and 40 are all greater than 20, but who is the first one we found greater than 20? It's this 28. Okay, what is the error opposite this 28? This 28, what is the caricature error that Its counterpart is two.
So, how much do I have? The median is two.
I think the problem is clear. So, when it tells you to calculate the median, what do you do? You calculate the total number and divide by two. Then you calculate the total effective number and see which is the first total effective number greater than that total divided by two that you found. Try to repeat this exercise to understand it well.
After that, we'll come to the mean. So, the mean is easy. What do I do? I take this number of the effective number and multiply it here by the cart, plus this by this, plus this by this, plus this. So, it's 16 times 1, plus 2 times 12, plus 3 times 8, plus 4 times 4, divided by the sum we calculated. What is it? It's 40. So, that's what we wrote. Once we calculate it, we get 1 times 16, which is 16. 2 times 12 equals 24, and so on. This will give you 80 divided by 40, which is two. So, this is the answer for the average.
After that, I need to calculate the tourist visitor representation.
3 times. Follow along with what they asked me to do. They told me to calculate the percentage of visitors who visited at least three cities. When we say "at least," we mean the minimum number of cities they visited is three. So, I have three or four here. So, the minimum number of cities they visited is three. So, they will have visited more than three: 3, 4, 6, and so on. So, how many visitors do I have who visited the cities? We wrote them down as 8. How many visitors do I have who visited the four cities? We wrote them down as 4. So, I will calculate their total: 8 plus 4 equals 12. So, the number of visitors who visited the three cities is more than 12. What will I do? I will divide it by The total number of visitors is 40. So, 12 divided by 40 multiplied by 100 equals 30. Therefore, the percentage is 30%.
Now let's move on to question number four.
What does exercise number four tell us? Sorry, not the exercise question. So, I have a calculation for the angle of inclination.
First, look, follow along here. It tells you that the angle between the two points equals 30. I have a rhombus, and this angle is 30 degrees. Look, this angle we have here equals 30 degrees. They told me that this is the intensity of the point.
This is it. This is the intensity that you are converting. We call it inclination. The first question tells you to copy the diagram you have. It tells you to copy it again. So, when you want to transfer it onto the paper, you have to transfer it while respecting the angle and say that this angle equals 30. Then he told you to draw a point, which is the image of the deduction.
What is deduction? It's what you do when we say deduction, meaning from where you go from the right side, the side of the deduction. And the distance from the deduction, we have to remember what I have here.
When I want to draw deduction, what you're transferring to, what he told you to draw? Look, follow carefully.
He told you to draw the point, which is the image of the deduction. It's the image of the deduction. I come to the deduction, here it is, I want to find its image, its image, so that I don't make a mistake in the deduction. The deduction that you're transferring to, when I want to go from the deduction, did I go on the right side or the side?
From the left, I went to the right side. So, from D to that point, I'll go to the right side. The question is, where will I stop?
What will the distance be? I'll calculate the distance from A. It's the distance from D to that point where I'll stop, and that point is O. So, O is the image of the deduction of A. I think it's clear.
Let's move on to the next question. You gave me the image of the deduction.
It tells you to determine the image of I. Here it is. What is the image of I in terms of deduction? I mean, from I. In the diagram, we see that the image of I is C. But how will we demonstrate it?
Notice that I is the midpoint of AC. This is the first piece of information we need to know. I mean, I is the midpoint of segment AC.
If I is the midpoint, then I have I equal to AC. Here it is: I am the midpoint of AC. Don't let y equal yc when you have a point that is the midpoint of a segment. So, the three points are linear, meaning they are collinear points. So, I have yc, how are they linear? And when yc is linear, it means that the vector y equals the vector yx. The first time, we started from the distance and said that the point is linear, and when it's linear, then the vector equals the vector. The fusion of the vector is yx. How did we know this? Here's the fusion of y. What does it do? It transforms y into y. So, we answered the question and said that the mass of the fusion.
This reading is written like saying the mass of the fusion is y. You don't write this in the exam. Write the sentence and say the mass of y. Let's go to question number [number missing]. It asked you to show us that y is here. This distance y equals The distance that we drew is okay, the distance. Sorry, Victor told you, show me the vector E. The second idea is that if Milio Dial E S Donk A is equal to AC. This is where we got it from question two. Here it is from here. E is equal to AC. Notice with me here I have or equals E. Here I have A equals to AC. I mean, I have A. E.
equals to AC. I mean, I have A. E.
equals to E. Lee EC, which is the answer to the question, we have Namchio's concept of kistion 4, he told me Ditghmini la miz longle c io qallak henaaya justtifi al-gibbons qallak hadd li c ha hiya c io io ura kant 3ana huna fi al-shakl qallak c io sahs li shakal ghakun muwaftin ana 3rifin bil iyad kit 30 dunk huma qallak c io kit ghakun nabi nahyu baynuha shanu qulu 3rifin 3rifin bil iyad kit li alay hadhi qaliha fi al-awdal bayna qali anna hadhi iya hiya limage iya bil tumslassion iya bil tonslassion iya hiya hadhi li likritig yak wa ana shnu 3rifin bil thikk al-bonslassion iya btawsil li alay hadhi qaliha fi al-awdal bayna qali anna hadhi iya hiya limage iya bil tonslassion iya hiya hadhi licritig yak wa ana shnu 3rifin bil bulslassion iya btawsil li alay hadhi qaliha fi al-awdal bayna bayna anna hadhi iya hiya limage iya bil tonslassion iya hiya hadhi licritig yak wa ana shnu 3rifin bil bulslassion iya btawsil li alay ya katif l sah zawiya sah sah zawiya fa th zawiya iya di ra ktub li si I or its concept, and we know that IAD equals 30. Where did we get it from? We got it from the given information. Here, IAD equals 30. So, how much will CE be? It will also equal 30. So, we will have finished exercise number four. Let's move on to exercise number five.
This is an easy exercise, worth four points. They gave us one of the points: A3 3 B0 A4 C0 D5 - and asked me to answer these questions. So, the first question is: calculate the distances of the vector BX. They told you to calculate the distances of BX and then show that this distance, or make sure, is equal to two vertices of a cent. First of all, when I want to calculate the vector of a point, I do XC - XBRC - RB. And they gave you a vector ab Yak told you to calculate the coordinates for it. What do you need to do? You'll do XP - XA RAB - ARIC.
If they gave you a CD and told you to calculate the coordinates, you'll do XD - XCAR D - C. So that's what we did for PC. We did XC - XP C - B.
Where do I get this XC? I'll go to the point C.
Here's its X, which is two. So it's two minus how many ARICs do we have? It's 0. So two minus zero. How many Cs do I have? It's 0. And how many Ds do I have? It's four. So zero minus four. So I have two minus zeros and I have Y0 minus four. So 2 minus zeros is 2, Y0 minus four is minus A. So what are the coordinates for PC? They are 2 minus four. After that, I need to show that this distance PC equals two points. When I want To calculate the distance, I need to know the relationship.
This is the relationship of what takes me, and this is the relationship of the distance. So, for PC, I make two large vertices: xc - xb, octal + c - rb, octal. It's the same as x - xb, c - rb.
What's the difference? I just add a square here and a square here, and I put them between the vertices. Then I'll substitute. How much do I have for xc? It's 2. How much do I have for xb? It's 0. So, 2 - 0 raised to the power of 2. Plus how much do I have for c? It's 0. How much do I have for ric per square to the power of 2? So, 0 - 4 raised to the power of 2. So, 2 - 0 is 2 raised to the power of 2. So, 4 raised to the power of 2 is -4 raised to the power of 16.
Because this square, these two vertices, remove the decimal points. So, -4 raised to the power of 16 is 16. So, the vertices of 4 + 16 are the vertices of 20. And what is 20? It's 4 times 5 if we calculate it with a calculator. The answer to the question, which was based on the quadratic equation AB, is 5. We'll have confirmed that the question they asked me for is: The question was about the quadratic equation AB/CD, which is a PaGalilogram. So, to prove it, I need to calculate the coordinates of AB and DC. I need to find the coordinates of the vector AB and DC. Why did I say AB/DC and not AB/CD? Because they need to have the same direction and the same direction, going from side B. So, I need to go from side C. Therefore, I'll calculate their coordinates.
What did we say? I'll do x₀ - x₀, AB - a, and DC.
I need to do x - x₀, DR - d.
That's what we did. What do I have? x is 0, and x₀ is -3. So, you'll get -3. What do I have? The value of b is 3, so 4 - 3 is 1. Where do I get the values of the points? We have them here in the given data.
We'll use the same method to calculate DC: 2 - 5 = 0 - 1.
Why? Because I have the value of C, where C is 0, minus the operation. The value of D is -1, so -1 is 1. Therefore, we found DC. Notice what we found for AB: -3 = 1. And what we found for DC: -3 = 1. So, what do we conclude?
That this AB equals the x vector, and the ab equals the DC vector. We know that if we have two sides, we have a quadrilateral and a quadrilateral. If we have two sides, for example, this is the quadrilateral. If these two sides are opposites, then this is the quadrilateral. If they have the mime vector, then they have The same vector, donk donk, I'm saying that it's a pagalilogram. I actually worked on this thing, this is its opposition, it's DC. When I calculated their vectors, I found -3, 1, -1. So they have a vector m. So what do we conclude? It's a pagalilogram.
Let's go to question number three.
What am I required to do? I need to show that the equation for the numbers PC is A = -2x + 4. So they asked me to show that the equation for PC is A = -2x + 4. The first thing we need to know is what the equation is. It's written in the form A = x + A = x + 4. So I need to calculate it, see what it comes out to, and then calculate it. It's called the direct equation. How do I calculate it?
Now we're working on PC, right? So What is the equation A is Ac - RB / xc - XP. For example, if I was given a straight line DC and I wanted to calculate its direct correspondence, what would I do? I would do C - Adi divided by x - xd. If they gave you, for example, abi and you wanted to calculate what you would do, you would do b - A, right? x divided by XP - xa. So, we should try to memorize this relationship because it's important. We'll substitute from where we get xc, yb, xc, yb, xc, and XP. When we get them, we'll get from here, from these points. How much RC do I have? I'll go to its yc, which is 0. How much b do I have? It's 4.
How much xc do I have? It's 2. How much XP do I have? It's 0. So, we substitute and calculate. It's 0 - 4 divided by 2 - 0 - 4 divided by 2 - 2. So, we found A So, we found the equation, and it equals -2x + 4, since it's written as ax + 0. Anyway, this isn't very clear. We found ax + 0, so we said it's -2. We still need to find b. What do I do? I take a point from these, either x or c, and I take their codes and substitute them into the equation. Now, what do I have? The equations are: ax + r equals -2x + pi. What do I still need to do?
Take either pi or c, whichever you want. For example, if you take x, you'll substitute pi of x to find pi. If you take c, you'll substitute c, meaning ax + c, and then look for pi.
For example, we'll take pi, which belongs to that group, so we'll take x.
How much do I have for pi? It's four. I'll substitute x with four, which equals Minus x and what I have x is 0, so minus two times 0 plus, so here four minus two times zero is zero plus, so what does x equal? It equals four. So what I have now is a equals minus two x plus 4, and that's what they asked me to show above. Okay, now let's go to question number four. What is the interrogation of two lines?
No, question, excuse me, question four. I want to show the interrogation of these lines, delta. Okay, what did they give me? They told me delta passes through a point on BC. The information they gave me was that delta passes through a point, meaning they gave me a point A that belongs to delta, and they told me it's perpendicular to BC. When we say it's perpendicular to BC, then it intersects that line BC at a point. Understood? So I have the interrogation Delta Pass Bag A 3, they gave it to me, or it's a bigbond on PC. What is the direct coefficient of x? From that equation we came up with here, it equals -a + 4. What do I have the direct coefficient of x? It's -2. Understood, right? And we know a relationship. We know that when I have two digraphs, the direct coefficient of the first multiplied by the second equals -1. That's it, because that's what we'll use. I found the direct coefficient of x to be -2. From here, it's -2.
Here's the relationship: a, b, y, b equals -1. So the first coefficient is -2. I don't have the second coefficient of delta, but since they are bigbonds, the coefficient of delta, which is called a, multiplied by the coefficient of PC, which is -2, should equal -1. And I'll go and look it up.
On the line �... Bring it here, donk, what will it be? It's 3 - 3/2. If we calculate it, 2c times 3 is 6 - 3, which is 3/2. So what does b equal? 3/2.
So now I have the equation for delta.
A equals 1/2x plus 3/2.
Let's move on to exercise number four. It tells you to show that point A is the intersection point of BC and point B is the intersection point of delta.
How do I show it?
First, when I want to show it as the intersection point, we saw in a previous exercise that when you want to show it as the intersection point, you have to show that A belongs to BC and that A belongs to delta. So, when you want to show that A belongs to BC, you substitute into the equation of BC, and when you want to show that A belongs to delta, you substitute into the equation of delta. Let's start with BC.
What is BC? A equals -2x plus 4. What do I do? I substitute what I have. I substitute x We write that x equals 1, meaning we go here to x and substitute it with 1. What do we get? It's -2 times 1, which is -2 plus 4, which is 2. So, when we substituted x with 1 here, what did we get? The result is 2. So, here, does it equal 2 or not?
Yes, this is correct. We'll do the same thing for delta. We found this equation: the delta function equals 1/2 x + 3/2. What do you do? You take x, which is 1, and substitute it here. If it gives you 2, know that this value belongs to this delta function. So, here, x, which we'll substitute with 1, becomes 1 times 1/2, which is 1/2 plus 3/2, which is 4/2, and 4/2 is 2. So, 1/a plus 3/2 is 4/a, which is 2. When we replaced x with one, we got two, right? If so, then that one belongs to delta. So here we found that what belongs to PC, and here we found that what belongs to delta. So what should I say? I'll say that those two lines intersect at a point which is one, two. So what is one, two? It's the directional line of PC and delta. So the exercise is clear. Let's go to question number five. It asks you to show me what the two lines of the equation are equal to. When they say what they mean by it, they mean the distance.
We know the relationship of distance: x = x - x, and x + x - x. And that line, here it is, it gives us what it is, one, two, and what it is, three, three. And we'll go and substitute. How much do I have x? What is one? And how much do I have x? What is three? So it's one minus three raised to the power of two. How much do I have a? What is two? How much do I have a? What is 3? So 2 minus 3 is two. So 1 minus 3 is 2, which is 4. And c is two, which is one. So 4 plus is 5. So what is a? What is the vertices of five? Let's go calculate the area of the palliogram AB CD.
That's ABCD. I need to calculate its area. What do we know? We know that its area equals the base. Right? The base means the base multiplied by the height.
Base multiplied by the height.
We know that BC is the base. What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? What is BC? The number five, so two verts of five times the two verts of five, two verts of five times two verts of five, the square root times the square root is the square root raised to the power of two, which cancels out, so five, so it's two times 5, which is 10, so how much do I have? It's 10. Let's go to the last exercise, which is exercise number six. It's an easy exercise. Without the figure, what do I have? BC 2 OF G. What is a cup? So when they tell you a cup, it means its base is a square.
Its base is a square. When you hear "cup," know that its base is a square.
That square's side is called ab, and it equals six. Okay, the first question asked you to calculate the square. Where is the square? Here it is. This square equals six. The two verts of two, we'll work on the square. This square is the square.
This is what we'll work on. Why are we working on it? Because we want to calculate this.
What did we say? We said that this is a cup, so its faces are squares. So this is a square.
This is a square and a square.
How can its angle be a right angle? If its angle is a right angle, then this is a square. Here I have a right angle. So this is a right triangle. And since it is a right triangle at the right angle at the right angle, what do I have to do? I have to apply Pythagoras. Either I will use this, or I can use D. What did I do? What does it say? A right triangle at the right angle. I didn't have a right triangle. I applied Pythagoras. What does Pythagoras tell me? If we use this, then here we are. If we use D, then this is a right triangle. Here, if I have this side, this is called the hypotenuse, then its length equals D + D Here it is, so if π equals π, then π plus π, then π, then how much do I have? π is six. Where did I get it from? They told me that π equals six. The side equals six, and since I'm working with a cup, all its sides will be equal. So they all equal 6. So π equals six,... This pyramid is drawn here with these dashed dots.
Its volume, we want to show, is 72. What do we know about the base? When we look at this pyramid, what is its base like? It's a square. So, this base is a square. What is its side? It's six. So, what is the area of this base of this pyramid? This pyramid I have has a square base. What is the area of this base? It's six times 6, so it's 36. We found it. And then, what is the height of this pyramid? This pyramid I'm working on, what you can see is its height.
This height is six because we said all sides are six. So, six. Let's repeat. We want to calculate the volume of the pyramid.
What does volume equal? It equals the height multiplied by the area of the base. The base is a square.
We calculated its area, which is side times side, and it's 6 times 36. We multiply that by the height, which is 6, so 36 multiplied by 6 gives us what it is divided by three.
Why is it divided by three? Because here I have a relationship.
What does it tell me? It tells me the area of the base times the height divided by three. Here I told you it's just the area of the base times the height. So here I got a slip of the tongue.
Sorry. When we say the volume of the pyramid, it's what it is: the area of the base times the height divided by three.
The area of the base we got here is 36 because it's side times side, and the height is 36. So 36 times 6 divided by 3 is 216 divided by 3 is 72. We confirmed that this equals 72. Now let's move on to question number three. What am I required to do? This is the equation: ` abi`, `k`, `a`, `ab`, `ki`, `agh`. What did they tell me about it? They told me ` lamid`, `of`, `f`, `g`. What did they tell me? This is like saying the smaller of this, the larger of this, the larger one. We all agree on how much. They gave me `abi`, which equals 3. They gave me `abi`, which equals three, meaning they wanted to tell me this is the reduction of this by three times. Understood? And they asked me to show that it's not, excuse me, less than three times smaller than it. They gave me a side whose `abi` equals three.
The terms, we'll stop here. Here, `abi` equals three.
They asked me to show that the coordination of the direction is `k`, which equals one over two. Understood? Right?
What do we do? In the exercise, they told us that the smaller of this is to work on `abi`, `a`, and how much do I have, 3, and how much do I have, `a`, 6, right? So, let's calculate how much I have. We said it's three, and how much I have is six. So, six divided by six is one over two. Therefore, one over two is the same.
They told me the coefficient is like the reduction coefficient. Okay, let's repeat. So, when we wanted to calculate the reduction coefficient, what did we do? We took the side they gave us, which is A, and divided it by the whole side, which is A and B. That's how much? Three divided by six is one over, which is 3 over X, which is one over two. So, one over two is what I have. Then they told you to calculate the volume in the Pyramid of this small pyramid. A is the volume of the large pyramid, which is 72.
And we found that the reduction coefficient is one over two. So, how do we find A is the volume in the Pyramid?
Here, I have it in the Pyramid. The value of ka equals 3 times the volume. We used ka because we were working with volume, and if we were working with service, we would have used ka or square. Therefore, the value of the prime equals ka raised to the power of 3 multiplied by the volume of ka. We calculated ka and found it to be 1/2 raised to the power of 3. The volume of that large number is 72. Therefore, 1/2 raised to the power of 3 is 1/8 multiplied by 72, which is 9. So, the value of the prime is 9 cubic centimeters. I hope I was able to explain the exam well. This exam is for the Marrakech region. We also worked on an exam for the Casablanca-Settat region; the same exercises and questions are repeated. If you work on these two or three exam models and master the rules, you will answer very correctly in the regional exam. It's available on the channel.
We will also be adding the corrections for the social studies exams, and we will also add the corrections for the other subjects: Arabic, Islamic education, physics, etc. Every day, God willing, we will post sample exams for the third year of middle school. So, if you are waiting for us to make another video, then... Yeah.
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