Hi! I'm Annie, a second year Mathematics student studying at Durham University. If you're looking for support in Maths or Chemistry, I'm here to help! After going to some extra classes for my challenging degree course last year, I understand what it feels like to be the tutee as well as the tutor.I've enjoyed the experience so much so far I recently became Durham's campus representative for MyTutor. As well as Maths, Further Maths and Chemistry, I also studied German at A level, and Russian to AS level. From studying languages I've learnt how to be a good communicator - a vital skill for explaining concepts to tutees in as many different ways as required. I've completed the NCO (non-comissioned officer) course in the Combined Cadet Force and as a result taught classrooms of teenagers navigation skills and overseen teamwork tasks. I have 10 A*s at GCSE, and achieved 100 UMS in C3 and C4 Edexcel A level Maths. My other relevant qualifications are listed below. About the Sessions You as the tutee can decide what you wish me to cover with you. On the other hand if you are unsure where to start, I can help to identify what you're struggling with and suggest what we can work on. Frequently testing understanding is key, without making a tutee feel pressured. My central message in any tutorial is as follows: getting things wrong is how you learn! I hope to encourage this process! Hopefully that's everything you need to know about me and the sessions, but if you have any other questions or would like to meet me, please don't hesitate to send me a message or book a free 15 minute 'Meet the Tutor Session'!

Hi! I'm Annie, a second year Mathematics student studying at Durham University. If you're looking for support in Maths or Chemistry, I'm here to help! After going to some extra classes for my challenging degree course last year, I understand what it feels like to be the tutee as well as the tutor.I've enjoyed the experience so much so far I recently became Durham's campus representative for MyTutor. As well as Maths, Further Maths and Chemistry, I also studied German at A level, and Russian to AS level. From studying languages I've learnt how to be a good communicator - a vital skill for explaining concepts to tutees in as many different ways as required. I've completed the NCO (non-comissioned officer) course in the Combined Cadet Force and as a result taught classrooms of teenagers navigation skills and overseen teamwork tasks. I have 10 A*s at GCSE, and achieved 100 UMS in C3 and C4 Edexcel A level Maths. My other relevant qualifications are listed below. About the Sessions You as the tutee can decide what you wish me to cover with you. On the other hand if you are unsure where to start, I can help to identify what you're struggling with and suggest what we can work on. Frequently testing understanding is key, without making a tutee feel pressured. My central message in any tutorial is as follows: getting things wrong is how you learn! I hope to encourage this process! Hopefully that's everything you need to know about me and the sessions, but if you have any other questions or would like to meet me, please don't hesitate to send me a message or book a free 15 minute 'Meet the Tutor Session'!

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05/11/20155from 48 customer reviews

Lynne (Parent from Lymm)

May 7 2017

Annabel has tutored my daughter for GCSE maths over the last 6 months. My daughter needed some maths confidence especially as the GCSE exam format had changed. Before the sessions she was achieving an equivalent C in maths exam tests. Annabel has been patient, informative,and flexible. She has gone back through things with her to make sure she understands everything. My daughter finally got an A in GCSE Maths and we fully attribute that to the sessions she had with Annabel. Thank you so much!

Dhanalakshmi (Parent from Swindon)

Annabel is a very supportive and helpful tutor. Her explanations are very detailed and she ensures that you fully understand the topic before moving on. Ive been going to her sessions for a couple of months prior to my GCSE's and she helped me achieve my target grade of 9 in my actual math GCSE !!!! I would recommend her to anybody who is struggling with maths as she is an amazing and dedicated tutor.

Dhanalakshmi (Parent from Swindon)

November 13 2016

Very helpful and supportive in learning new topics

Lynne (Parent from Lymm)

October 20 2016

Really put Georgina at ease and she enjoyed her first session, thank you!

An atom is made up of **3** different types of **sub-atomic particles** (just the name for particles that are inside the atom).

The nucleus, the 'ball' in the middle of the atom where almost all it's mass is found, is made up of protons and neutrons. Electrons circle this nucleus. The number of protons in an atom **always** equals the number of electrons in that atom. It is this number that defines what the atom is - whether it is gold, or oxygen, or calcium and so on.

The number of neutrons in an atom will be roughly the same if not equal to the number of protons. Importantly though, changing the number of neutrons in an atom doesn't change what element the atom is. Therefore, we could play around with the number of neutrons in an atom of gold, and as long as we didn't change the number of protons, it would still be am atom of gold. We would be creating different **isotopes** of gold if we did this.

**Isotopes are atoms of the same element with different numbers of neutrons. **

Therefore isotopes differ in relative atomic mass, but have the same chemical properties, because neutrons have nothing to do with how an atom reacts (it's down to the electrons).

An atom is made up of **3** different types of **sub-atomic particles** (just the name for particles that are inside the atom).

The nucleus, the 'ball' in the middle of the atom where almost all it's mass is found, is made up of protons and neutrons. Electrons circle this nucleus. The number of protons in an atom **always** equals the number of electrons in that atom. It is this number that defines what the atom is - whether it is gold, or oxygen, or calcium and so on.

The number of neutrons in an atom will be roughly the same if not equal to the number of protons. Importantly though, changing the number of neutrons in an atom doesn't change what element the atom is. Therefore, we could play around with the number of neutrons in an atom of gold, and as long as we didn't change the number of protons, it would still be am atom of gold. We would be creating different **isotopes** of gold if we did this.

**Isotopes are atoms of the same element with different numbers of neutrons. **

Therefore isotopes differ in relative atomic mass, but have the same chemical properties, because neutrons have nothing to do with how an atom reacts (it's down to the electrons).

There are two methods you can use to answer this question. I will briefly outline the second method at the end, as one may seem more intuitive to you than the other.

**Method 1:**

We are asked to find dy/dx, but we have been given a funtion of x in terms of y, instead of a funtion in terms of x, which you will be more used to. (I.e. we have f(y) here instead of f(x)). So we are going to use the fact that **dy/dx = 1 / (dx/dy)**.

We can easily find dx/dy. We differentiate each term of x = 3tan(2y) with respect to y.

The differential of x with respect to y is dx/dy.

The differential of 3tan(2y) with respect to y is 3*2*sec^{2}(2y), using standard results. (We have multiplied by 2 becuase the differential of sec^{2}(f(y)) is f'(y)*sec^{2}(f(y)) - you have to multiply by the differential of the function 'inside' the sec^{2 }due to the chain rule).

This gives us dx/dy = 6sec^{2}(2y). Now the question requires that our answer is in terms of x, not in terms of y, as we have at the moment. So we need to find a **relation** between our answer sec^{2}(2y) and x, where x = 3tan(2y) as in the question. We need to use the identity **tan ^{2}(2y) + 1 = sec^{2}(2y)**. Therefore, if we substitute this in for sec

dx/dy = 6*(1 + tan^{2}(2y))

We still need to somehow get this in terms of x. We know x = 3tan(2y). Dividing by 3 means tan(2y) = x/3. Therefore tan^{2}(2y) = x^{2}/9 (just by squaring both sides).

So now we can say that dx/dy = 6*(1 + x^{2}/9), which is now in terms of x!

We need dy/dx, and we have dx/dy. So we take our answer dx/dy and divide 1 by it, as shown by the rule in bold earlier. So dy/dx = 1/(6*(1 + x^{2}/9)).

Taking out a factor of 1/9 in the denominator gives us dy/dx = 1/ ((6/9)*(9 + x^{2})). 1/(6/9) = 9/6 = 3/2.

Therefore our final answer is

dy/dx = 3/ (2*(9 + x^{2}))

**Method 2: Harder if you have difficulties with implicit differentiation**

This method uses implicit differentiation. Differentiating x = 3tan(2y) implicitly gives 1 = 3*2*sec^{2}(2y)*dy/dx. (I have differentiated every term with respect to x, and therefore needed to use the chain rule on the tan(2y). This is where the dy/dx comes from). Rearranging to get dy/dx as the subject gives dy/dx = 1/ 6*(sec^{2}(2y)), and from here you continue in exactly the same way as in method 1.

There are two methods you can use to answer this question. I will briefly outline the second method at the end, as one may seem more intuitive to you than the other.

**Method 1:**

We are asked to find dy/dx, but we have been given a funtion of x in terms of y, instead of a funtion in terms of x, which you will be more used to. (I.e. we have f(y) here instead of f(x)). So we are going to use the fact that **dy/dx = 1 / (dx/dy)**.

We can easily find dx/dy. We differentiate each term of x = 3tan(2y) with respect to y.

The differential of x with respect to y is dx/dy.

The differential of 3tan(2y) with respect to y is 3*2*sec^{2}(2y), using standard results. (We have multiplied by 2 becuase the differential of sec^{2}(f(y)) is f'(y)*sec^{2}(f(y)) - you have to multiply by the differential of the function 'inside' the sec^{2 }due to the chain rule).

This gives us dx/dy = 6sec^{2}(2y). Now the question requires that our answer is in terms of x, not in terms of y, as we have at the moment. So we need to find a **relation** between our answer sec^{2}(2y) and x, where x = 3tan(2y) as in the question. We need to use the identity **tan ^{2}(2y) + 1 = sec^{2}(2y)**. Therefore, if we substitute this in for sec

dx/dy = 6*(1 + tan^{2}(2y))

We still need to somehow get this in terms of x. We know x = 3tan(2y). Dividing by 3 means tan(2y) = x/3. Therefore tan^{2}(2y) = x^{2}/9 (just by squaring both sides).

So now we can say that dx/dy = 6*(1 + x^{2}/9), which is now in terms of x!

We need dy/dx, and we have dx/dy. So we take our answer dx/dy and divide 1 by it, as shown by the rule in bold earlier. So dy/dx = 1/(6*(1 + x^{2}/9)).

Taking out a factor of 1/9 in the denominator gives us dy/dx = 1/ ((6/9)*(9 + x^{2})). 1/(6/9) = 9/6 = 3/2.

Therefore our final answer is

dy/dx = 3/ (2*(9 + x^{2}))

**Method 2: Harder if you have difficulties with implicit differentiation**

This method uses implicit differentiation. Differentiating x = 3tan(2y) implicitly gives 1 = 3*2*sec^{2}(2y)*dy/dx. (I have differentiated every term with respect to x, and therefore needed to use the chain rule on the tan(2y). This is where the dy/dx comes from). Rearranging to get dy/dx as the subject gives dy/dx = 1/ 6*(sec^{2}(2y)), and from here you continue in exactly the same way as in method 1.

First things first, we have to decide what **type** of equation this is, so we can choose how to deal with it. If it were linear (which means every term involving an 'x' has no power/superscript on it. I.e. x^{2 }or x^{3} do not appear in your equation), then we could simply rearrange to solve. But our equation has a 3x^{2} term, so we need to do something else.

First, let's put everything on one side, so we can recognise that this equation is actually a **quadratic equation**: subtracting 8x and adding 2 to both sides gives **3x ^{2} - 8x + 2 = 0**.

Now, there are 3 ways we can deal with a quadratic equation: **Factorisation**, **Completeing the Square, **and the** Quadratic Formula. **

The first two methods won't really help us here, but how did I figure that out? Well, the long way would be to do some trial and error, attempting to factorise. You would quickly see you can't make any combination work. When factorising fails, you normally try the quadratic formula next. **Here's a tip**: if the question asks you to give your answer to a certain number of decimal places, it means you cannot factorise the quadratic because the answer is a fraction or surd, so **always use the quadratic formula in this case!**

So now we know we're going to apply the quadratic formula to this equation. Here is the formula in words (because I can't format it clearly on here), with brackets added to show you in what order to do operations:

x = [(-b) plus or minus the (square root of {b^{2} - 4ac})] all over 2a

(Remember to **divide everything** by the 2a! It is a common mistake not to do this).

We have 3 variables, a,b and c, which we can find from our quadratic equation we are trying to solve. **Note**: our quadratic **MUST** be in the form ax^{2}+bx+c=0 for this to work. We did this earlier, so we can proceed.

From our equation, a (the bit in front of the x^{2}) equals +3. b (the bit in front of the x) equals -8. (The minus is **important!!**). c (the number without an x) equals +2.

Now we can put a=3, b=-8, c=2 into the quadratic equation.

Note this **common mistake**: If b is negative, like we have in this question, what happens? -b, here, equals -(-8), and two minuses make a plus, so this becomes **+**8.

For the -b^{2}, it is important to know where the brackets should be. This is the same thing as (-b)^{2} and therefore is **ALWAYS positive, **because all square numbers are positive numbers. So in our question, (-b)^{2} is (-8)^{2}=64.

4ac = 4*3*2 = 24. Therefore, inside the square root we have 64-24=40. 2a = 2*3 = 6.

Now we can put (8 plus/minus the square root of 40)/6 into our calculator, to give us two answers (one for the plus and one for the minus). These are 2.3874... and 0.2792...

Rounding these to 2 decimal places gives us our final solutions to our quadratic: 2.39 and 0.28.

First things first, we have to decide what **type** of equation this is, so we can choose how to deal with it. If it were linear (which means every term involving an 'x' has no power/superscript on it. I.e. x^{2 }or x^{3} do not appear in your equation), then we could simply rearrange to solve. But our equation has a 3x^{2} term, so we need to do something else.

First, let's put everything on one side, so we can recognise that this equation is actually a **quadratic equation**: subtracting 8x and adding 2 to both sides gives **3x ^{2} - 8x + 2 = 0**.

Now, there are 3 ways we can deal with a quadratic equation: **Factorisation**, **Completeing the Square, **and the** Quadratic Formula. **

The first two methods won't really help us here, but how did I figure that out? Well, the long way would be to do some trial and error, attempting to factorise. You would quickly see you can't make any combination work. When factorising fails, you normally try the quadratic formula next. **Here's a tip**: if the question asks you to give your answer to a certain number of decimal places, it means you cannot factorise the quadratic because the answer is a fraction or surd, so **always use the quadratic formula in this case!**

So now we know we're going to apply the quadratic formula to this equation. Here is the formula in words (because I can't format it clearly on here), with brackets added to show you in what order to do operations:

x = [(-b) plus or minus the (square root of {b^{2} - 4ac})] all over 2a

(Remember to **divide everything** by the 2a! It is a common mistake not to do this).

We have 3 variables, a,b and c, which we can find from our quadratic equation we are trying to solve. **Note**: our quadratic **MUST** be in the form ax^{2}+bx+c=0 for this to work. We did this earlier, so we can proceed.

From our equation, a (the bit in front of the x^{2}) equals +3. b (the bit in front of the x) equals -8. (The minus is **important!!**). c (the number without an x) equals +2.

Now we can put a=3, b=-8, c=2 into the quadratic equation.

Note this **common mistake**: If b is negative, like we have in this question, what happens? -b, here, equals -(-8), and two minuses make a plus, so this becomes **+**8.

For the -b^{2}, it is important to know where the brackets should be. This is the same thing as (-b)^{2} and therefore is **ALWAYS positive, **because all square numbers are positive numbers. So in our question, (-b)^{2} is (-8)^{2}=64.

4ac = 4*3*2 = 24. Therefore, inside the square root we have 64-24=40. 2a = 2*3 = 6.

Now we can put (8 plus/minus the square root of 40)/6 into our calculator, to give us two answers (one for the plus and one for the minus). These are 2.3874... and 0.2792...

Rounding these to 2 decimal places gives us our final solutions to our quadratic: 2.39 and 0.28.