Contact Melissa
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Contact Melissa

About me

About me: 

I am currently studying for a Maths degree at the University of Birmingham. I have enjoyed Maths since I was very young and hope I can pass some of this on to you! 

I am very friendly. I have been coaching gymnastics since I was 15, to kids from the age of 2 - 11, and I was captain of my team for two years. I've also done 3 and a half years of volunteering in my local primary school, working with kids of all ages, and have also done some tutoring of Senior School students. 

The Sessions: 

In our sessions, we will work together to improve your maths skills and deal with questions which you have. I will ask at the end of each session what you would like to work on next time so that I can prepare some practice questions. I find with maths, the best way to improve your understanding is to practice, practice, practice! That said, if you are set homework questions that you don't understand, you are welcome to bring them along and we can work through them together! 

I will do my best to make the sessions as fun as I can! Maths can be tricky but it's also really exciting and hopefully I can help you to enjoy it as much as I do!

What next?

Feel free to WebMail me or book a 'Meet the Tutor' free session! Don't forget to tell me what you're struggling with! 

Hope to hear from you soon!

Subjects offered

SubjectLevelMy prices
Maths GCSE £18 /hr
Maths 13 Plus £18 /hr
Spanish 13 Plus £18 /hr
Maths 11 Plus £18 /hr

Qualifications

QualificationLevelGrade
MathsA-LevelA
Further MathsA-LevelB
SpanishA-LevelA
EPQA-LevelA*
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

16/09/2016

General Availability

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Questions Melissa has answered

Factorise fully 3*a^3*b +12*a^2*b^2 + 9*a^5*b^3

To factorise 3a3b + 12a2b2 + 9a5b3, we need to deal with like elements together. Start with the integers. The highest common factor of 3, 12, and 9 is 3. Therefore we factor out the 3 and the expression becomes 3(a3b + 4a2b2 + 3a5b3) Next, deal with the a values. The highest common factor of...

To factorise 3a3b + 12a2b2 + 9a5b3, we need to deal with like elements together.

Start with the integers. The highest common factor of 3, 12, and 9 is 3. Therefore we factor out the 3 and the expression becomes

3(a3b + 4a2b2 + 3a5b3)

Next, deal with the a values. The highest common factor of a3, a2 and a5 is a2. So, we factor out a2and the expression becomes

3a2(ab + 4b2 + 3a3b3)

Finally, we need to factorise the b values. The highest common factor of b, b2 and bis b. So we factor out b and the expression becomes.

3a2b(a + 4b + 3a3b2). 

Therefore, the complete factorisation of 3a3b + 12a2b2 + 9a5bis 3a2b(a + 4b + 3a3b2).

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3 months ago

138 views

Find the 100th term in the sequence 3, 7, 11, 15...

To answer this question, we need to form an equation to work out the nth term of the series.  To do this, the first thing we do is calculate the difference between each term. 7 - 3 =4 11 - 7 = 4 15 - 11 = 4 The difference each time is 4. Therefore, the equation could be 4n. Let's try it o...

To answer this question, we need to form an equation to work out the nth term of the series. 

To do this, the first thing we do is calculate the difference between each term.

7 - 3 =4

11 - 7 = 4

15 - 11 = 4

The difference each time is 4. Therefore, the equation could be 4n. Let's try it out with the first term. 

4n = 4 x 1 = 4 which is not equal to the first term, 3. 

To get 3 from the equation we have at the moment, we need to subtract 1. 

Let's try the second term with the equation 4n -1. 

4n-1 = 4x2 -1 = 7 which is equal to the second term. 

To double check that we're right, let's use the equation to calculate the 5th term.

4n -1 = 4x5 -1 = 19. 

19 is 4 greater than 15, so we're right!

Now let's calculate the 100th term. 

4n -1 = 4x100 -1 = 399

Therefore the 100th term of this sequence is 399

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3 months ago

197 views

Find the highest common factor of 432 and 522

To do this, we first need to calculate the prime factorisations of 432 and 522, that is to say we need to break the numbers down into their prime factors. Let's start with 432.  432 = 2 x 216 432 = 2 x ( 2 x 108) 432 = 2 x 2 x (2 x 54) 432 = 2 x 2 x 2 x (9x6) 432 = 2 x 2 x 2 x (3 x 3) x (2...

To do this, we first need to calculate the prime factorisations of 432 and 522, that is to say we need to break the numbers down into their prime factors.

Let's start with 432. 

432 = 2 x 216

432 = 2 x ( 2 x 108)

432 = 2 x 2 x (2 x 54)

432 = 2 x 2 x 2 x (9x6)

432 = 2 x 2 x 2 x (3 x 3) x (2 x 3)

Therefore, the prime factorisation of 432 is 24 x 33

SImilarly with 522,

522 = 2 x 261

522 = 2 x (3 x 87)

522 = 2 x 3 x (3 x 29)

Since 29 is a prime number, the prime factorisation of 522 is 2 x 32 x 29.

Now, to calculate the highest common factor of the two numbers, we can draw a Venn diagram to find the intersection of the two factorisations. 

For now, imagine it like this. 

432                             BOTH                        522

23   3                         2       32                        29

Therefore, the intersection of the two factorisations is 2 and 32, which means that the highest common factor is just those numbers multiplied.

So the HCF of 432 and 522 = 2 x 32 = 18

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3 months ago

122 views
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