# Double-Checking Maths Homework for a 100% Semester Grade

• MHB
• ertagon2
In summary, the conversation is discussing a homework assignment in which the person needs to get at least 6 out of 10 questions correct in order to receive 100% in the math module, which is worth 20% of their final grade. They ask for help with question 3 and question 4, and there is a brief discussion and clarification on the correct answers. The conversation also includes a discussion on how to solve question 3 using a continuous growth model.
ertagon2
So here is a thing. If I get at least 6/10 questions in this homework I will get 100% in this semester's homework which is equivalent to 20% of the maths module. So could you double check these questions? Also I wouldn't mind some help with q.3 I got a weird answer of 11,8 billion.
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View attachment 7585

Workings:
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2.) Correct.

3.) I would begin with:

$$\displaystyle \frac{1}{N}\d{N}{t}=\frac{1}{25}$$ where $$\displaystyle N(0)=N_0$$

Integrate:

$$\displaystyle \int_{N_0}^{N(t)} \frac{du}{u}=\frac{1}{25}\int_0^t \,dv$$

$$\displaystyle \ln\left(\frac{N(t)}{N_0}\right)=\frac{t}{25}$$

$$\displaystyle N(t)=N_0e^{\Large{\frac{t}{25}}}$$

With $N_0=3,\,t=35$, we find:

$$\displaystyle N(35)=3e^{\Large{\frac{7}{5}}}\approx12.17$$

5.) Correct.

MarkFL said:
2.) Correct.

3.) I would begin with:

$$\displaystyle \frac{1}{N}\d{N}{t}=\frac{1}{25}$$ where $$\displaystyle N(0)=N_0$$

Integrate:

$$\displaystyle \int_{N_0}^{N(t)} \frac{du}{u}=\frac{1}{25}\int_0^t \,dv$$

$$\displaystyle \ln\left(\frac{N(t)}{N_0}\right)=\frac{t}{25}$$

$$\displaystyle N(t)=N_0e^{\Large{\frac{t}{25}}}$$

With $N_0=3,\,t=35$, we find:

$$\displaystyle N(35)=3e^{\Large{\frac{7}{5}}}\approx12.17$$

5.) Correct.

Ah c=4 in q4
So f(2) = 2^2 + 2 + 4
= 10 ?
Thanks.

ertagon2 said:
Ah c=4 in q4
So f(2) = 2^2 + 2 + 4
= 10 ?
Thanks.

Yes, that's what I got as well. :)

This is what I did:

$$\displaystyle N(t)=Ae^{kt}$$

$$\displaystyle N(0)=Ae^{k(0)}=3$$

$$\displaystyle A=3$$

$$\displaystyle N(1)=3e^{k(1)}=3*1.04$$

$$\displaystyle 3e^{k}=3.12$$

$$\displaystyle e^{k}=\frac{3.12}{3}$$

$$\displaystyle (k)ln(e)=ln(1.04)$$

$$\displaystyle k=ln(1.04)$$

$$\displaystyle N(t)=3e^{ln(1.04)t}$$

$$\displaystyle N(35)=3e^{ln(1.04)(35)}$$

$$\displaystyle N(35)~11.84$$

Why is this wrong and what does$$\displaystyle \frac{1}{N}\frac{dN}{dt}$$ mean in the context of the question?

ertagon2 said:
This is what I did:

$$\displaystyle N(t)=Ae^{kt}$$

$$\displaystyle N(0)=Ae^{k(0)}=3$$

$$\displaystyle A=3$$

$$\displaystyle N(1)=3e^{k(1)}=3*1.04$$

$$\displaystyle 3e^{k}=3.12$$

$$\displaystyle e^{k}=\frac{3.12}{3}$$

$$\displaystyle (k)ln(e)=ln(1.04)$$

$$\displaystyle k=ln(1.04)$$

$$\displaystyle N(t)=3e^{ln(1.04)t}$$

$$\displaystyle N(35)=3e^{ln(1.04)(35)}$$

$$\displaystyle N(35)~11.84$$

Why is this wrong and what does$$\displaystyle \frac{1}{N}\frac{dN}{dt}$$ mean in the context of the question?

Your formula can be simplified to:

$$\displaystyle N(t)=N_0\left(1+\frac{1}{25}\right)^t$$

This is mathematically equivalent to an investment that is earning 4% annual interest compounded once per year. However, if the interest is continuously compounded then we have:

$$\displaystyle N(t)=N_0\lim_{n\to\infty}\left(\left(1+\frac{1}{25n}\right)^{nt}\right)=N_0e^{\large{\frac{t}{25}}}$$

Your model is discrete, whereas we are told to use a continuous growth model. :)

## What is the purpose of double-checking math homework for a 100% semester grade?

The purpose of double-checking math homework is to ensure accuracy and completeness in the assignment, ultimately leading to a higher grade for the semester. It allows for any mistakes or errors to be caught and corrected before the final submission.

## How can I effectively double-check my math homework?

To effectively double-check your math homework, it is important to go through each problem carefully and compare your answers to the given solutions. You can also use a calculator or ask a friend or tutor to review your work. It is also helpful to revisit any notes or class materials to make sure you fully understand the concepts.

## Is it necessary to double-check my math homework if I am confident in my abilities?

While it may seem unnecessary, double-checking your math homework can still be beneficial even if you are confident in your abilities. It is always possible to make mistakes, and double-checking can help catch any errors that may have been overlooked.

## What should I do if I find a mistake while double-checking my math homework?

If you find a mistake while double-checking your math homework, it is important to correct it before submitting your assignment. This can involve redoing the problem or simply correcting the error. It is also helpful to make a note of the mistake so you can avoid making the same error in the future.

## How often should I double-check my math homework for a 100% semester grade?

It is recommended to double-check your math homework every time you complete an assignment. This will ensure that you catch any mistakes early on and have the opportunity to correct them. It is also helpful to double-check your work before submitting it for a final grade.

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