BasicMaths

For everyone who is helping me solve this problem, yes, the cross-sections are parallel to the yz-plane.

Okay, so firstly we need to find an input to substitute for n to take advantage of the information we already have. If we let n=3 and plug in, we get  

f(3)=3f(1) - 2f(2)

From that value, we can plug in the information we had into the equation, giving us

f(3)=3(-1) - 2(3) =>    f(3)= -9

Then, we repeat these steps. We now know the values of f(3), f(2), and f(1). Can you take it from there?

All the best.

BasicMaths

Hello! I'll give you a hint: use the PEMDAS method. This stands for Parentheses, Exponents, Multiplication and Division from Left to Right and Addition and Subtraction from Left to Right. Hopefully, this makes sense! - BasicMaths

Translation:

Привет! Подскажу: используйте метод PEMDAS. Это означает круглые скобки, экспоненты, умножение и деление слева направо, а также сложение и вычитание слева направо. Надеюсь, в этом есть смысл! - Основы математики

Asking my teacher....I will respond ASAP when I get the answer.

Also, I didn't know how to make this post off-topic, so yeah.

Hey Guest. I just wanted to clarify some possible misconceptions you might have about this forum. Most importantly, the main goal of this forum is not to do people's work for them but to help them with any problems or struggles they might have with certain questions. We would love to help you with problems that you need help understanding, but we cannot do all the work for you. And we definitely do not want to be responsible for your grades on your assignments. Using this forum as a free pass out of homework isn't fair to the contributors that work hard answering your problems and it is not fair to yourself as well. School isn't just about passing grades, it's about learning and grasping the concepts. By posting the problems here without attempting them yourself isn't helping you learn anything about what was taught in class. Just to end this on a positive note, we have plenty of contributors who are glad to help you any time of the day; just remember that we do not do your work for you. Hopefully, this all made sense to you.

All the best,

BasicMaths

There are many methods to solve this problem, but the method I'm using today is the one shown below. As you can see, drawing a rectangle around the perimeter of the triangle highlights 3 smaller triangles that we can use to help solve the problem. First, we find the area of the rectangle, which is 4x6=24 sq units. Next, we find the area of triangles A, B, and C, which are 3x2/2=3 sq units, 3x4/2=6 sq units, and 6x2/2=6 sq units, respectively. Now we can subtract the areas of those triangles from the area of the rectangle, 24-6-6-3=9 sq units, giving us the final answer: 9 sq units. Hopefully, this helps! - BasicMaths

"Mathematics is not about numbers, equations, computations, or algorithms: it is about understanding." - William Paul Thurston

Okay, so this is a problem involving systems of equations. We would set up this system as follows:

n+d=21 (number of coins)

5n+10d=150 (amount of money)

For the second equation, I find it easier to remove the decimal places altogether and write it in cents instead. We can then rearrange the first equation as n=21-d and substitute that into the second. The second equation then becomes 5(21-d)+10d=150. Writing the terms out and simplifying gives us 5d=45, which becomes d=9. Therefore, we have 9 dimes. Plugging that value into the first equation gives us n=12. This means we have 12 nickels. We then double-check, and it all works out. Hopefully, this helps! - BasicMaths

(note: in the system of equations above, n stands for nickels and d for dimes)

"Mathematics is not about numbers, equations, computations, or algorithms: it is about understanding." - William Paul Thurston

Huh, thanks Guest, I will check it out. :)

Nope, I checked. Or maybe I just didn't look that far back! 😂

Ahh dang it EP! You beat me to the answer! Anyways, I totally agree that you should remember the d=rt formula for problems like these. It makes it a lot easier when computing if you know how to switch out variables for the specific one you are looking for.

Yeah, but unfortunately, I couldn't access my teachers laptop; I should have asked him .-.

This thread is comedy GOLD! 😂 Anyways, I'll tell you the thing you were missing. When the string stretches, the point the ant was on moves with it. So in theory, the ant will travel a little more than an inch and will eventually overtake the string! You can even view this in Desmos for all you visual learners.......at least that's what my teacher did! 😂 But most people do actually think that way; the ratio will always stay constant; but you must think outside the box for this one!

Good job everyone!

BasicMaths

CPhill, that's a very intuitive guess!

However, there's one factor you haven't taken into account. Try to see if you can find it :)