PremiumTilly P. GCSE Maths tutor, GCSE Computing tutor

Tilly P.

£18 /hr

Currently unavailable: for new students

Studying: Computer Science and Mathematics (Bachelors) - Exeter University

4.8
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Trusted by schools

18 reviews| 96 completed tutorials

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About me

I'm a Maths and Computer Science student at the University of Exeter. I've always enjoyed Maths and Computer Science and I really want to help others enjoy them too! I have a lot of experience in tutoring as you can see above.

I'm most likely to be free on a wednesday but if you'd like tutoring on another day of the week then let me know! If you have any questions please send me a message! I hope to see you soon!

I'm a Maths and Computer Science student at the University of Exeter. I've always enjoyed Maths and Computer Science and I really want to help others enjoy them too! I have a lot of experience in tutoring as you can see above.

I'm most likely to be free on a wednesday but if you'd like tutoring on another day of the week then let me know! If you have any questions please send me a message! I hope to see you soon!

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About my sessions

Our time together: I believe that it's super important that you develop a full understanding on the things that you struggle with at school. So I'm here to help explain things to you! Whether through analogies, diagrams or songs (though please note I'm not a singing teacher!), I'll patiently help you reach your targets. There is a lot we can do in a small amount of time, and so if you have topics you're struggling with then I can help you out! If you're not sure what your weak points are then I can help you figure those out too. It's totally up to you how we spend our time as I just want you to become confident and happy in your subject of choice.

Our time together: I believe that it's super important that you develop a full understanding on the things that you struggle with at school. So I'm here to help explain things to you! Whether through analogies, diagrams or songs (though please note I'm not a singing teacher!), I'll patiently help you reach your targets. There is a lot we can do in a small amount of time, and so if you have topics you're struggling with then I can help you out! If you're not sure what your weak points are then I can help you figure those out too. It's totally up to you how we spend our time as I just want you to become confident and happy in your subject of choice.

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18/12/2016

Ratings & Reviews

4.8from 18 customer reviews
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Mo (Parent)

May 3 2017

Thanks so much for your assistance with my son. He feels ready and fully prepared for his exam now

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Fola (Student)

April 11 2017

very good and easy to understand. She explained everything in a clear and simple way

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Sarah (Parent)

February 1 2017

Very reliable and good communication. My son learnt a lot and enjoyed his lessons with Tilly

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Sue (Parent)

January 6 2017

Hi Tilly Success! Henry really enjoyed today and asked to book another session immediately. Did the request come through - Wed 5pm? I'm going to see if he can set up the web cam to show you his second screen with Python on for next time, might make it easier for you both. Can you please let me know if there is a book or any resource items we should buy for the lessons? Cheers Sue

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Qualifications

SubjectQualificationGrade
MathsA-level (A2)A
Further MathsA-level (A2)B
Computer ScienceA-level (A2)A

General Availability

Before 12pm12pm - 5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
ComputingGCSE£18 /hr
MathsGCSE£18 /hr

Questions Tilly has answered

if x^2 + 9x + 20 = 0, what are the possible values of x?

So x2 + 9x + 20 = 0

My preffered way of solving this equation is to factorise the equation. (Though I understand that different students may find other ways easier) 

Factorisation is where the above equation is (x+a)(x+b) = 0

So if we times out (x+a)(x+b) we get

x2 + ax + bx + ab = 0
therefore
x2 + (a+b)x + ab = 0

Therefore we can equate this to the original question, so
x2 + 9x + 20 = x2 + (a+b)x + ab

so now we can see that
9 = a + b and
20 = ab

I would reccomend using trial and error (although I understand that different students may prefer other techniques).

So by trying for multiple values of a and b, we can see that they must equal 5 and 4.

Therefore
x2 + 9x + 20 = (x+5)(x+4) = 0

We know that the only way of producing a 0 through multiplication is through multiplying one number by another. Therefore we know that

x+5= 0 or x+4=0

Through rearranging these equations we can conclude that x must equal -4 or -5.
 

So x2 + 9x + 20 = 0

My preffered way of solving this equation is to factorise the equation. (Though I understand that different students may find other ways easier) 

Factorisation is where the above equation is (x+a)(x+b) = 0

So if we times out (x+a)(x+b) we get

x2 + ax + bx + ab = 0
therefore
x2 + (a+b)x + ab = 0

Therefore we can equate this to the original question, so
x2 + 9x + 20 = x2 + (a+b)x + ab

so now we can see that
9 = a + b and
20 = ab

I would reccomend using trial and error (although I understand that different students may prefer other techniques).

So by trying for multiple values of a and b, we can see that they must equal 5 and 4.

Therefore
x2 + 9x + 20 = (x+5)(x+4) = 0

We know that the only way of producing a 0 through multiplication is through multiplying one number by another. Therefore we know that

x+5= 0 or x+4=0

Through rearranging these equations we can conclude that x must equal -4 or -5.
 

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1 year ago

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