Bryan P. A Level Maths tutor, GCSE Maths tutor, Mentoring -Personal S...

Bryan P.

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Mathematics (Masters) - Bristol University

4.7
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7 reviews

Trusted by schools

This tutor is also part of our Schools Programme. They are trusted by teachers to deliver high-quality 1:1 tuition that complements the school curriculum.

163 completed lessons

About me

I am currently a BSc Mathematics student at the University of Bristol and I am already no stranger to teaching. I have worked as a teaching assistant in my secondary school's maths department where I also led one-on-one sessions with students from GCSE to sixth form. As a mathematics student, I love a good problem. The harder the better (right?). There's no greater satisfaction than understanding how to tackle them and it will be my pleasure to help you get there!

I am currently a BSc Mathematics student at the University of Bristol and I am already no stranger to teaching. I have worked as a teaching assistant in my secondary school's maths department where I also led one-on-one sessions with students from GCSE to sixth form. As a mathematics student, I love a good problem. The harder the better (right?). There's no greater satisfaction than understanding how to tackle them and it will be my pleasure to help you get there!

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Personally interviewed by MyTutor

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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18/07/2015

Ratings & Reviews

4.7from 7 customer reviews
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Akshay (Student)

April 4 2016

Great tutor, provided step by step clear explanations of mathematical concepts and ways to answer complex questions, would definitely recommend.

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Xin (Parent from Maidstone)

April 4 2016

Clear explanation and good resources. Thank you!

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Jordan (Student)

March 30 2016

We went through the series for FP1. The lesson was good. It taught me another way of doing series which was different from how we learnt it in class. As an improvement just spend more time factorizing out the equations for next time finding common term.

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Paresh (Parent from Potters Bar)

April 12 2017

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Qualifications

SubjectQualificationGrade
MathsA-level (A2)A*
Further MathsA-level (A2)A
ChemistryA-level (A2)A

General Availability

Pre 12pm12-5pmAfter 5pm
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tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
Further MathematicsA Level£26 /hr
MathsA Level£26 /hr
MathsGCSE£24 /hr

Questions Bryan has answered

Find the exact solution to: ln(x) + ln(7) = ln(21)

Log rules:

log(a) + log(b) = log(ab)

so, in this case, we must find x such that 7x = 21

thus x = 3

similarly, log(a) - log(b) = log(a/b)

rearranging the original equation we get:

ln(x) = ln(21) - ln(7)

so x = 21/7 = 3

Log rules:

log(a) + log(b) = log(ab)

so, in this case, we must find x such that 7x = 21

thus x = 3

similarly, log(a) - log(b) = log(a/b)

rearranging the original equation we get:

ln(x) = ln(21) - ln(7)

so x = 21/7 = 3

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2 years ago

784 views

Solve for 0=<x<360 : 2((tanx)^2) + ((secx)^2) = 1

First step I would take would make it look less intimidating by converting all components into sin and cos i.e

2(((sinx)/(cosx))^2) + 1/((cosx)^2) = 1

Notice that there is a common denominator of cosx^2    so I would multiply this up:

2((sinx)^2) + 1 = (cosx)^2

Eliminate cos by putting it in the form of sin using the trig identiy (cosx)^2 = 1 - (sinx)^2

so we have:

2((sinx)^2) + 1 = 1 - (sinx)^2

rearranging we obtain

3((sinx)^2) = 0

(sinx)^2 = 0

sinx = 0

In between 0 and 360, the sine function is 0 at 0 and 180 (360 is also a solution but not in the range)

so x = 0, 180

First step I would take would make it look less intimidating by converting all components into sin and cos i.e

2(((sinx)/(cosx))^2) + 1/((cosx)^2) = 1

Notice that there is a common denominator of cosx^2    so I would multiply this up:

2((sinx)^2) + 1 = (cosx)^2

Eliminate cos by putting it in the form of sin using the trig identiy (cosx)^2 = 1 - (sinx)^2

so we have:

2((sinx)^2) + 1 = 1 - (sinx)^2

rearranging we obtain

3((sinx)^2) = 0

(sinx)^2 = 0

sinx = 0

In between 0 and 360, the sine function is 0 at 0 and 180 (360 is also a solution but not in the range)

so x = 0, 180

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2 years ago

904 views

Write sqrt(50) in the form Asqrt(50) where A is an integer

Note that sqrt(ab)=sqrt(a)sqrt(b)

Thus to have A, an integer, we must find the highest number 'a' that is a square number and is also a factor of 50.

So, a=25 and b=2 (ab=25x2=50)

and we have:

sqrt(50) = sqrt(25)sqrt(2) = 5sqrt(2)  (A = 5)

Note that sqrt(ab)=sqrt(a)sqrt(b)

Thus to have A, an integer, we must find the highest number 'a' that is a square number and is also a factor of 50.

So, a=25 and b=2 (ab=25x2=50)

and we have:

sqrt(50) = sqrt(25)sqrt(2) = 5sqrt(2)  (A = 5)

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2 years ago

898 views

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