David H. GCSE Biology tutor, A Level Biology tutor, GCSE Maths tutor,...

David H.

Currently unavailable: for regular students

Degree: Medicine (Bachelors) - Exeter University

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About me

Me, myself and I

I am a current medical student at Exeter University. I have loved science and maths from an early age, and have always looked for ways to increase my understanding of them – Horrible Science magazines were my favourite thing to read for an embarrassingly long time! Hopefully some of that enthusiasm will come across in my tutorials and be passed on to you!

I always try to be patient, friendly and understanding. I tutored a little in sixth form, and used to help teach badminton during games sessions (something I actually hope to qualify in sometime soon!), so I have some understanding of how to teach varying degrees of ability

How will the sessions work?

These sessions are for YOU...hence I think you should decide how you want them to work. I am happy to cover whatever topics you wish to cover, and at whatever pace suits you best.

In science and maths understanding the topic is essential in answering questions on it, so we will try to focus on this before moving onto exam questions.

I believe that people work best when they are comfortable, and I will try my best to make the sessions as fun and enjoyable as possible to allow you to be comfortable.

Can I ask you about uni/medical school?

Of course you can! I am more than happy to answer any questions you may have as well as I possibly can. I know how much I appreciated receiving advice from older years, and would like to do the same for you!

What next?

If I have piqued your interest, or you have any questions, please send me a WebMail or book a Meet the Tutor session. Don’t forget important information such as your exam board and anything you struggle with in particular!

I greatly look forward to meeting you!

Subjects offered

SubjectLevelMy prices
Biology A Level £20 /hr
Maths A Level £20 /hr
Biology GCSE £18 /hr
Maths GCSE £18 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
Further MathematicsA-LevelB
BiologyA-LevelA
ChemistryA-LevelA
PhysicsA-LevelA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

25/06/2015

Currently unavailable: for regular students

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Ratings and reviews

1.6from 7 customer reviews

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Questions David has answered

Solve the simultaneous equations 3x + 4y = 17 and 4x + y = 14

Eq1: 3x + 4y = 17 Eq2: 4x + y = 14 Eq2 x 4: 16x + 4y = 56 Subtract Eq1 from this to isolate x: 16x + 4y - 3x - 4y = 56 - 17 Simplify: 13x = 39 Solve for x: x = 3 Substitute x into Eq1: 3(3) + 4y = 17                                  9 + 4y = 17                                  4y = 8    ...

Eq1: 3x + 4y = 17

Eq2: 4x + y = 14

Eq2 x 4: 16x + 4y = 56

Subtract Eq1 from this to isolate x: 16x + 4y - 3x - 4y = 56 - 17

Simplify: 13x = 39

Solve for x: x = 3

Substitute x into Eq1: 3(3) + 4y = 17

                                 9 + 4y = 17

                                 4y = 8

                                 y = 2

Check in Eq2: 4(3) + (2) = 14

                      12 + 2  = 14

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9 months ago

229 views

What is meant by the semi-conservative replication of DNA?

Semi-conservative replication means that during DNA replication, the two strands of nucleotides separate. Both strands then form the template for free nucleotides to bind to to create the two identical daughter strands. Hence each daughter strand has half of the DNA from the original strand an...

Semi-conservative replication means that during DNA replication, the two strands of nucleotides separate. Both strands then form the template for free nucleotides to bind to to create the two identical daughter strands. Hence each daughter strand has half of the DNA from the original strand and half newly-formed DNA.

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9 months ago

307 views

What is the angle between the position vectors a and b, where a = (6i - j + 3k) and b = (-4i + 2j + 10k)?

cos(AOB) = a.b/(|a| x |b|) a.b = (6i - j + 3k).(-4i + 2j + 10k)       = (6 x -4) + (-1 x 2) + (3 x 10)       = -24 + -2 + 30       = 4 (|a|)^2 = 6^2 + (-1)^2 + 3^2           = 36 + 1 + 9 = 46           -> |a| = 46^(1/2) (|b|)^2 = (-4)^2 + 2^2 + 10^2           = 16 + 4 + 100 = 120       ...

cos(AOB) = a.b/(|a| x |b|)

a.b = (6i - j + 3k).(-4i + 2j + 10k)

      = (6 x -4) + (-1 x 2) + (3 x 10)

      = -24 + -2 + 30

      = 4

(|a|)^2 = 6^2 + (-1)^2 + 3^2

          = 36 + 1 + 9 = 46

          -> |a| = 46^(1/2)

(|b|)^2 = (-4)^2 + 2^2 + 10^2

          = 16 + 4 + 100 = 120

          -> |b| = 120^(1/2) = 2 x 30^(1/2)

cos(AOB) = a.b/(|a| x |b|)

               = 4/(46^(1/2) x 2 x 30^(1/2))

               = 4/(4 x 345^(1/2))

               = 1/(345^(1/2))

AOB = cos^-1 (1/(345^(1/)))

        = 86.9 degrees(3 significant figures)

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9 months ago

205 views
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