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Degree: Medicine (Other)  St. Andrews University
I am a medical student at the university of St Andrews. I have always loved maths and sciences, and hope to share that passion with you too! I am patient and very friendly, so hopefully you will enjoy our sessions and get the most out of them. Of course, you will guide what topics we cover in our sessions. With maths and science the most important thing is that you completely understand the ideas and concepts, so I will use as many different ways to try and help you learn (diagram, graphs, etc.) until you feel comfortable enough with the material that you can explain it to me. Once you've grasped one idea, we can build on it with the next one! I hope that the sessions will be very fun! We can cover a lot of material, especially if it's exciting. Maths is fun, and I hope to show you that! If you have any quedstions at all, please feel free to book a 'Meet the Tutor Session'. Make sure that you tell me your exam board and what topics you'd like to go through.
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Right, so the Young's modulus of a material is a measure of it's rigidity. The higher the value of the Young's modulus, the more rigid the material is.
Ok, so the Young's modulus, E, is equal to the stress that the material experiences divided by the strain experienced by the material.
The stress, σ, of a material is equal to the force, F, applied to this material, divided by the area, A, upon which this force is applied. The equation for this is:
σ = F/A
On the other hand, the strain, ε, of the material is equal to the extension of the material, e (how much the material has extended by having this force applied to it), divided by the original length, l, of the material. The equation for strain is
ε = e/l
The Young's modulus is equal to the stress divided by the strain, so:
E= σ/ε = (F/A)/(e/l) = (F*L)/(A*e)
So, E = (F*l)/(A*e)
For example, If a metal wire of original length 2cm
see moreOk, so we would call this a quadratic equation because it is written in the form of ax^2+bx+c=0 (in our case, a=1, b=6, c=8).
Luckily, this type of quadratic equation can be factorised, so we can solve it easily!
Ok, so we're trying to factorise x^{2}+6x+8=0 so that it is in the form of (x+p)(x+q)=0.
Let's try expanding (x+p)(x+q)=0.
If you multiply the brackets together, you're left with:
x^{2}+px+qx+pq=0
We can tidy this up a little bit to give us:
x^{2}+(p+q)x+pq=0
This looks very similar to x^{2}+6x+8=0, doesn't it?
Yes, it does! If we compare these two equations, we find out two things:
p+q=6
pq=8
So we're looking for two numbers which when they are multiplied by each other will give 8, and when they are added together will give 6.
What we're left with is that p=4 and q=2 (or the other way round, it doesn't really matter).
Let's plug this back into our original equation: (x+p)(x+q)=0
Of course, now we have:
(x+4)(x+2)=0
This is much easier to solve than what we started with!
So in this case, either the first bracket is equal to 0 or the second bracket is equal to zero  this gives us two solutions for x.
Either x+4=0, meaning that x=4
Or x+2=0, meaning that x=2
So now we have the answers x=2 or x=4
Of course a quicker way to do this would be to look at x^{2}+6x+8=0 and to find two numbers that are factors of 8 (they multiply together to make 8) and also add together to make 6.
see moreOk, so here are the equations
5x+3y=24
3x4y=26
So let's multiply the first equation by 3, which gives us:
15x+9y=72
Now let's multiply the second equation by 5, which gives us:
15x20y=130
So we're now left with:
15x+9y=72
15x20y=130
Let's rearrange both of these equations to make 15x the subject. So now we're left with:
15x = 729y
15x= 130+20y
Now we can compare these two equations, to give us:
729y=130+20y (=15x)
If we rearrange this new equation, we find that:
20y + 9y = 72  130
29y = 58
y = 2
Since we now have a value for y, we can substitute this back into 5x + 3y = 24
5x + 3(2) = 24
5x 6 = 24
5x = 30
x = 6
So, are final answer is x = 6 and y = 2
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