£20 - £22 /hr

Degree: Mathematics (Masters) - Bristol University

Trusted by schools

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!

#### Subjects offered

SubjectQualificationPrices
Further Mathematics A Level £22 /hr
Maths A Level £22 /hr
Maths GCSE £20 /hr

#### Qualifications

MathsA-levelA2A*
Further MathsA-levelA2A
ChemistryA-levelA2A
 CRB/DBS Standard No CRB/DBS Enhanced ✓ 01/07/2016

#### General Availability

Weeks availability
MonTueWedThuFriSatSun
Weeks availability
Before 12pm12pm - 5pmAfter 5pm
MONDAY
TUESDAY
WEDNESDAY
THURSDAY
FRIDAY
SATURDAY
SUNDAY

Please get in touch for more detailed availability

#### Ratings and reviews

4.8from 7 customer reviews

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.

Xin (Parent) April 4 2016

Clear explanation and good resources. Thank you!

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.

Paresh (Parent) April 12 2017

See all reviews

### 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

see more

1 year ago

618 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 i...

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

see more

1 year ago

721 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)

see more

1 year ago

713 views
Send a message

All contact details will be kept confidential.

To give you a few options, we can ask three similar tutors to get in touch. More info.

Still comparing tutors?

How do we connect with a tutor?

Where are they based?

How much does tuition cost?

How do tutorials work?

We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this.