Matthew T.

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Degree: Aerospace Engineering (Masters) - Bristol University

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I am a student studying Aerospace Engineering at the University of Bristol now going into my second year. This is a very Maths and Physics intensive course so its just as well I have a real enjoyment in these subjects, something I hope to pass along.I have been a Scout Leader for 2 years and was an assistant leader for 4 years before that so I have some experience in teaching, sometimes with those as young as 6 and as old as 18!

The tutorials:

I am here to help you, so you will be guiding what we cover! In all subjects, whether they be English or Maths require a basic understanding, so before we get to any of those horrible exam questions we will spend plenty of time going over the concepts until you're confident enough that you can explain it back to me or anyone else. I can also help to answer any questions you may have about university or UCAS, I have been there myself I know your pain!

What next:

If you have any questions, send me a 'WebMail' or book a 'Meet the Tutor Session'! (both accessible through this website). Remember to tell me your exam board and what you're struggling with.

I look forward to meeting you

#### Subjects offered

SubjectLevelMy prices
Maths A Level £20 /hr
Geography GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr

#### Qualifications

MathsA-LevelA*
Further MathsA-LevelB
PhysicsA-LevelB
GeographyA-LevelA
 CRB/DBS Standard ✓ 28/11/2013 CRB/DBS Enhanced No

### Find dy/dx for y=5x^3-2x^2+7x-15

Step 1: To differentiate an equation there is a simple rule to follow. For y=ax n dy/dx=anxn-1. so for an example y=x3, dy/dx=3x2. Therefore we just apply this rule into our equation. Step 2: Break the equation down and do each factor of x seperately so 5x3  differentiates into 15x2, -2x2 diffe...

Step 1: To differentiate an equation there is a simple rule to follow. For y=axn dy/dx=anxn-1. so for an example y=x3, dy/dx=3x2. Therefore we just apply this rule into our equation.

Step 2: Break the equation down and do each factor of x seperately so 5x3 differentiates into 15x2, -2x2 differentiates to -4x, 7x differentiates to 7 and the 15 disappears from the end. This happens as the 15 just tells us where the line crosses the y axis and therefore has no bearing on the gradient.

Step 3: Put the differentiated parts back together to give the differentiated equation

dy/dx=15x2-4x+7

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2 years ago

571 views

### Solve the equation ((2x+3)/(x-4))-((2x-8)/(2x+1))=1

Step 1: Multiply everything by x-4 to give  (2x+3) - ((2x-8)(x-4)/2x+1)=(x-4) Step2: Multiply everything by 2x+1 to give (2x+3)(2x+1)-(2x-8)(x-4)=(x-4)(2x+1) Step 3: Multiply out the brackets to give (4x2+8x+3)-(2x2-16x+32)=2x2-7x-4 Step 4: Simplify to get a zero on one side of the equation...

Step 1: Multiply everything by x-4 to give

(2x+3) - ((2x-8)(x-4)/2x+1)=(x-4)

Step2: Multiply everything by 2x+1 to give

(2x+3)(2x+1)-(2x-8)(x-4)=(x-4)(2x+1)

Step 3: Multiply out the brackets to give

(4x2+8x+3)-(2x2-16x+32)=2x2-7x-4

Step 4: Simplify to get a zero on one side of the equation to give

31x-25=0

Step 5: We therefore know that 31x = 25 so to find x divide 25 by 31 to get x=0.81 to 2 decimal places

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2 years ago

942 views

### What is an isotope?

Isotopes of an element exist when the atoms of an element have different numbers of neutrons, or in other words the same atomic number but a different mass number.

Isotopes of an element exist when the atoms of an element have different numbers of neutrons, or in other words the same atomic number but a different mass number.

2 years ago

592 views
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