I am currently a Computer Science student at University College London. I have always had a passion for science and technology and I hope that I can share it with everyone else through my tutoring sessions.

I am a calm and friendly person and I like to work with kids very much as my belief is that I am contributing to shaping our society's future by training them to become the very best. I have had some experience with tutoring before so I know what methods to apply to make someone not only learn a subject, but also love it, while also becoming friends with the children.

In my tutorials, I will try to make myself understood by using different materials such as diagrams and various images until I am certain that the children can explain the lesson back to me. I also try to keep my students motivated by giving them trivia questions or exercises that they would have to solve.

If you have any questions, you can find me by sending a Webmail or by booking a sesssion with me.

I am looking forward to working with you!

I am currently a Computer Science student at University College London. I have always had a passion for science and technology and I hope that I can share it with everyone else through my tutoring sessions.

I am a calm and friendly person and I like to work with kids very much as my belief is that I am contributing to shaping our society's future by training them to become the very best. I have had some experience with tutoring before so I know what methods to apply to make someone not only learn a subject, but also love it, while also becoming friends with the children.

In my tutorials, I will try to make myself understood by using different materials such as diagrams and various images until I am certain that the children can explain the lesson back to me. I also try to keep my students motivated by giving them trivia questions or exercises that they would have to solve.

If you have any questions, you can find me by sending a Webmail or by booking a sesssion with me.

I am looking forward to working with you!

No DBS Check

In order to calculate the derivative of the given function, we need to apply the laws of derivation. Therefore:

f '(x) = ((x+2)/(x-1))'

f '(x) = [(x+2)' * (x-1) - (x+2) * (x-1)' ]/ (x-1)^{2}

f '(x) = [ 1 * (x-1) - (x+2) * 1] / (x-1)^{2}

f '(x) = x-1-x-2 / (x-1)^{2}

f '(x) = -3 / (x-1)^{2}

Therefore, we have calculated the derivative of the given function.

In order to calculate the derivative of the given function, we need to apply the laws of derivation. Therefore:

f '(x) = ((x+2)/(x-1))'

f '(x) = [(x+2)' * (x-1) - (x+2) * (x-1)' ]/ (x-1)^{2}

f '(x) = [ 1 * (x-1) - (x+2) * 1] / (x-1)^{2}

f '(x) = x-1-x-2 / (x-1)^{2}

f '(x) = -3 / (x-1)^{2}

Therefore, we have calculated the derivative of the given function.

If we have two equations that look like this:

a1 * x + b1 * y = c1

and

a2 * x + b2 * y = c2

where x,y are variables and a1,a2,b1,b2 are coeficients.

then we solve it using the following method:

We choose either x or y and we try to reduce their coeficients so that either a1=a2 or b1=b2. If we make one of these possible, then we can reduce one of the variables and we are left with the other. Let's assume that we have a1=a2=a. Then:

a * x + b1 * y = c1

a * x + b2 * y = c2

We can reduce the two equations and we are left with:

(b1-b2) * y = c1 - c2

From here we can find y as we know the values of b1,b2,c1 and c2. After that, we replace y in one of the equations and we deduce x.

If a1 was different from a2, then we would've had to multiply the equations so that we can make those two equal.

Thus, we have solved a two-equation, two-unknown values system.

If we have two equations that look like this:

a1 * x + b1 * y = c1

and

a2 * x + b2 * y = c2

where x,y are variables and a1,a2,b1,b2 are coeficients.

then we solve it using the following method:

We choose either x or y and we try to reduce their coeficients so that either a1=a2 or b1=b2. If we make one of these possible, then we can reduce one of the variables and we are left with the other. Let's assume that we have a1=a2=a. Then:

a * x + b1 * y = c1

a * x + b2 * y = c2

We can reduce the two equations and we are left with:

(b1-b2) * y = c1 - c2

From here we can find y as we know the values of b1,b2,c1 and c2. After that, we replace y in one of the equations and we deduce x.

If a1 was different from a2, then we would've had to multiply the equations so that we can make those two equal.

Thus, we have solved a two-equation, two-unknown values system.

To write a C++ program that adds two numbers we need to write the following code:

We write a line which calls a library that has functions we need:

**#include **

(Then we create a function that does our sum:)

**int main()**

(Here we read the two values we need for our sum, called x and y. These can be chosen by the user and written from the keyboard:)

**{ cin>>x,y;**

(In the next lines we declare a variable called S which stands for sum and then we give it the value of x+y:)

**int S;**

**S= x+y;**

(In the end, we return the value of S and close the brackets of the function)

**cout<**

**}**

To write a C++ program that adds two numbers we need to write the following code:

We write a line which calls a library that has functions we need:

**#include **

(Then we create a function that does our sum:)

**int main()**

(Here we read the two values we need for our sum, called x and y. These can be chosen by the user and written from the keyboard:)

**{ cin>>x,y;**

(In the next lines we declare a variable called S which stands for sum and then we give it the value of x+y:)

**int S;**

**S= x+y;**

(In the end, we return the value of S and close the brackets of the function)

**cout<**

**}**