I am a third-year Architect Student at the University of Nottingham. I thoroughly enjoy travelling and dance. I also enjoy Maths and Sciences, I have always taken a liking to these subjects. I am proud to say that I received 6 A*/ 6 As at GCSE (A* being in Maths and the Sciences) and went on to receive an A* in A-level Maths.
I have had ample experience in the tutoring profression. Working with children ranging from the age of 6 to 16, I have worked for "Explore Learning", "NumberWorksNWords" and "Kumon". I also tutor my neighbour's children every summer holiday.
I have developed the skill of patience to a high degree, and strongly believe in working at a question until it is understood by the student. I am open to looking at solving problems using all methods, memorisation techniques. My key to success at GCSE-level was making mnemonics as I found it difficult to remember all the information across all subjects and I am willing to share this technique with other students if it seems fit.
Having younger siblings, which I have all tutored at some point, I believe I have the skills required to be a well-rounded tutor for your child.
|Maths||A Level||£20 /hr|
|Maths||11 Plus||£18 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
As we know so far, c = f / λ. This equation comes from a much simpler version of speed = distance/ time. As the λ (with units "m") can be seen as distance and the frequency (with units "1/s") can be seen as time. So by substituting the wave measurements into the speed = distance/ time formula, you can understand the c = f / λ equation a lot better.
Now lets make f= 500 Hz and λ = 0.34
So c= 500/ 0.34
Therefore c = 170 (m/s)see more
An ultrasound wave is a soundwave with a frequency higher than 20,000 Hz
So I would begin by minusing 5 from both sides and that will give you 3x - 5 ≤ 4x. Then we want to gather our x's onto one side, so I would '-3x' from both sides which leaves you with - 5 ≤ 4x - 3 x, so you end up with x ≥ −5.