Degree: MA Logic and Philosophy of Mathematics (Masters) - Bristol University
Hi, I'm Rebecca, I'm 23 and I'm currently taking a year to myself just to study, doing my masters in Logic and Philosophy of Maths. To me, mathematics has always been black and white, right or wrong whilst philosophy is all about thinking and coming to conclusions that others may disagree with. At first glance, it appears the two are entirely incompatible but actually that combination of questioning things that must be either right or wrong is what I love most in mathematics.
My approach is simple, take nothing for granted. Yes, I was that annoying child who kept asking why, but that's because I believe unless you truly understand where something comes from, you will struggle to apply the basic formula to other problems.
I think the job of a tutor is to answer all those why questions, so it's not a case of the answer is this 'because it just is' but actually the foundations of where this came from. And that's what I hope to achieve. :)
How I teach
I am a firm believer of practice makes perfect so I will begin with a few intro questions recapping the previous lesson, then teach a method and it's background, before setting some examples and expansion work.
|Further Mathematics||A Level||£20 /hr|
|Maths||A Level||£20 /hr|
|Further Mathematics||GCSE||£18 /hr|
|Maths||13 Plus||£18 /hr|
|Maths||11 Plus||£18 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
As this is a right angle triangle, we need to use Pythagoras's Theorem.
This says that the length of the longest side of a right angle triangle (the hypotenuse) is equal to the sqaure root of the sum of the squares of two other sides.
So in this case:
h = sqrt(6^2 + 8^2)
h= sqrt(36 + 64)
h = sqrt (100)
h = 10 :)
Fun fact - this is a special triangle known as a Pythagorean triple as all three sides are integers (whole numbers)see more
Let log a be some number A and log b be some number B
now the natural log of something is the equivalent of saying a=e^A and b = e^B
So a*b = e^A * e^B which by rules of indices
Therefore log(ab) = log(e^(A+B))
= A + B = log a + log bsee more