Matthew P. 11 Plus Maths tutor, GCSE Maths tutor, A Level Maths tutor...

Matthew P.

Currently unavailable: for new students

Degree: Electronic & Electrical Engineering (Bachelors) - Southampton University

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About me

About me:

I'm an undergarduate Electronic Engineer studying at the University of Southampton. I really enjoy working on new projects, such as designing robots and other rasberry pi projects, and it stems from a love of Physics and Maths. I'm a qualified Canoe Instructor and have had lots of experience teaching people of all ages. I'm also a leader in the Scouts and enjoy teaching new things to people.

Tutoring:

In the sessions there will be a big focus on learning and understanding the key concepts behind the subject, and the content is tailored around what you exams you are studying for so all the content is relevant to the exam you are taking.

All resources are pre-prepared and I can give you example questions and worksheets for you to work on any time you want. 

I hope to meet you soon.

Subjects offered

SubjectLevelMy prices
Electronics A Level £20 /hr
Maths A Level £20 /hr
Electronics GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr
Maths 11 Plus £18 /hr

Qualifications

QualificationLevelGrade
MathsA-LevelA*
ElectronicsA-LevelA
PhysicsA-LevelB
Further MathsA-LevelC
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

03/05/2015

Currently unavailable: for new students

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Questions Matthew has answered

Convert the decimal number 165 to binary

we can use the subtraction method to find this out: create a table set out like this   27  |  26  |  25  |  24  |  23  |  22  |  21  |  20 128  |  64  |  32  |  16  |  8   |  4   |   2   |   1        |        |       |        |       |      |        | Now we simply take away each number. 16...

we can use the subtraction method to find this out:

create a table set out like this

  27  |  26  |  25  |  24  |  23  |  22  |  21  |  20

128  |  64  |  32  |  16  |  8   |  4   |   2   |   1

       |        |       |        |       |      |        |

Now we simply take away each number.

165 - 128 = 37

37 - 64 < 0 So we ignore this one

37 - 32 = 5

5 - 16 < 0

5 - 8 < 0

5- 4 = 1

1 -2 < 0

1 - 1 = 0 

Now we fill in the table with the ones that were greater than zero with one and those not with zero

  27  |  26  |  25  |  24  |  23  |  22  |  21  |  20

128  |  64  |  32  |  16  |  8   |  4   |   2   |   1

   1   |   0   |   1  |   0   |  0  |   1   |   0   |  1 

so we now know 16510 = 101001012

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3 months ago

97 views

Explain the difference between forced vibration and resonance in an oscillating object.

Forced vibration: The object oscillating will vibrate upwards and downwards with the same frequencythe driving oscillation is at. The amplitude of the vibration increases as the frequencydecreases and there will be a phase difference between the driving vibration and the forced vibration. When...

Forced vibration:

The object oscillating will vibrate upwards and downwards with the same frequency the driving oscillation is at. The amplitude of the vibration increases as the frequency decreases and there will be a phase difference between the driving vibration and the forced vibration. When the driving frequency is much greater than the forced frequency then there is almost a 180o phase difference, however when the driving frequency is much less than the forced frequency, then there is almost no phase difference

Resonance: 

The frequency of the vibration is the same as the naturual frequency of the system. The rate of energy transfer from the driving force to the system is at a maximum and so the amplitude of the resonance is very large. The driving frequency is either 90o ahead or lags by 90o.

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3 months ago

211 views

find general solution to: x(dy/dx) + 2y = 4x^2

Divide through by x so:      (dy/dx) +2(y/x) = 4x Now multiply through by the intergrating factor:  e^(| (2/x) dx) = e^(2.ln(x)) = x^2 so you get:     (x^2)(dy/dx) + 2xy = 4(x^3) Now integrate the entire equation and you get:        y(x^2) = |(4(x^3))dx = (x^4) + c Divide through by (x^2) t...

Divide through by x so:      (dy/dx) +2(y/x) = 4x

Now multiply through by the intergrating factor:  e^(| (2/x) dx) = e^(2.ln(x)) = x^2

so you get:     (x^2)(dy/dx) + 2xy = 4(x^3)

Now integrate the entire equation and you get:        y(x^2) = |(4(x^3))dx = (x^4) + c

Divide through by (x^2) to get the general solution:

y = (x^2) + 4/(x^2)

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3 months ago

108 views
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