I am a Mathematics student at Oxford University, and have a real interest in helping those who are in a similar position to my own a few years ago. I am passionate about science and mathematics, and believe that a firm understanding of the basics is absolutely essential for mastering these subjects at higher level, and also provides fascinating insight on the world!
I achieved 100% at GCSE and A Level Maths, and also over 99% in A Level Further Maths. In 2014 I sat and passed the MAT paper.
During lessons, I can teach to any syllabus, and am very willing to explain whatever you find most useful, while ensuring that all necessary material is covered. I encourage two-way learning, and will provide opportunity for you to explain topics to me and to submit work for marking and feedback. I like to use several teaching methods, and believe that content should be demonstated with clear, relevant examples.
You can send me a message through "WebMail", or book a "Meet the Tutor Session", using this website. Please let me know which exam and board you would like help with, as well as any specific topics.
|Further Mathematics||A Level||£24 /hr|
|Maths||A Level||£24 /hr|
Joseph (Student) June 19 2016
Muniza (Parent) May 30 2016
Muniza (Parent) June 2 2016
Joseph (Student) June 6 2016
First write cosh and sinh as exponentials, from their respective definitions:
sinh x = 1/2 (ex - e-x)
cosh x = 1/2 (ex + e-x)
So: 3 * 1/2 (ex + e-x) - 4 * 1/2 (ex - e-x) = 7
Multiply by 2: 3 * (ex + e-x) - 4 * (ex - e-x) = 14
Expanding gives: 3 * ex + 3 * e-x - 4 * ex + 4 * e-x = 14
Collecting terms: -ex + 7 * e-x = 14
So: ex - 7 * e-x + 14 = 0
Since ex is not 0, we can multiply by it:
ex * ex - 7 * e-x *ex + 14 * ex = 0
e2x - 7 + 14 * ex = 0
This is a quadratic in ex, and we can use the quadratic formula to obtain:
ex = -7 +- 2 root 14
But ex > 0, so we discount the negative root:
ex = -7 + 2 root 14
x = ln ( -7 + 2 root 14)
Which can easily be checked with a calculator.see more