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We've combined live video with a shared whiteboard, so you can work through problems together

All your tutorials are recorded. Make the most out of your live session, then play it back after

Usually we cover both subject knowledge and exam technique, although that can change depending on each individual student. Then we go through diagrams, and they ask questions, and we go from there.

Lots of students say that the classes are too big in school, or that they don't have time to ask teachers after lessons. In my tutorials, we take time to explore things in a little in a bit more detail.

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I use the shared whiteboard. We make diagrams together and label them, and often the student prints it off because they know it's right and they completely understand it.

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s^{2} - 18s = 0

(s + 4)(s - 6) = 0

s = -4 s = 6

To prove that (2n+3)^{2}-(2n-3)^{2 }is a multiple of 8 we are going to deal with the first bracket and then the second bracket.

When a bracket has ^{2 }next to it, this means that you^{ }multiplying the bracket, by the bracket itself.

So, we are going to put them side by side, like this:

(2n +3) (2n + 3)

To multiply **everything **in the first bracket, by **everything **in the second bracket, we can use a **multiplication square**, or, we can use this easy way to remember how to multiply out the bracket.

**F O I L **

**This stands for: **

**First **

**Outside **

**Inside **

**Last **

This means that you are going to multiply the **first **two terms in the brackets by each other

2n x 2n =4n^{2}

Then you are going to multiply the **outside** terms in the brackets by each other.

2n x 3 = 6n

Then you multiply the **inside **terms by each other,

2n x 3 = 6n

Then you multiply the **last **two terms in each bracket by each other.

3 x 3 = 9 (be careful of the negatives here)

Putting all of this together, we get:

**4n ^{2 }+ 6n + 6n +9**

Well done, first bit complete!

*****************************************

Then you deal with the second bit

**- (2n-3) ^{2}**

So, be very **careful **here. There is a **negative **sign in front of the bracket. To avoid confusion later on, let's put a big bracket around it.

- [(2n-3)^{2}]

everything we do in this section is going to be inside that big square bracket...

-[(2n-3)(2n-3)]

following **FOIL again and keeping that big bracket in place...**

**-[4n ^{2 }-6n -6n +9] **

make sure you watch out for that (-3 x -3) which makes a **+9**

**and then, **because we have that big bracket around this equation, we are going to multiply it out.

so,

**-4n ^{2} + 12n -9**

********************************************************

So putting the first bit and the second bit together (and watching out for those negatives!), we get...

**4n ^{2 }+12n +9 - 4n^{2} + 12n -9**

now we are going to tidy that up a little bit, collecting the like terms....

4n^{2 -}4n^{2 }=0

and +9 -9 =0

So, we are just left with **24n **

If you didn't know what an integer was... it means **WHOLE NUMBER**. So to show that the integer is a multiple of 8, we are going to show that

to get 24n you can take out a factor of 8...

and this leaves you with 8(3n).

this shows that if you take out a factor of 8 you still get a whole number in front of n, which answers the question!

1)Take the inverse sin to take x from the sin(2x):

2x=arcsin(0.5).

2)Evaluate arcsin(0.5) to get pi/6:

so 2x= pi/6

3)Dividing by 2 to simplify we get

x=pi/12.

4)To find the second solution we note that (pi/2)-(pi/12) =(5pi/12) is also a solution.

So x= (5pi/12)

5)Sin(2x) has a period of pi. So to find the rest of the solutions we add pi to our previous solutions.

So now x=pi/12, 5pi/12, 13pi/12 , 17pi/12

Answered by Yinglan Z.

Studies Mathematics at Edinburgh

When you are differentiating, use the formula:

The differential of ax^n is (n*a) x^(n-1). Or in words: 'multiply by the power, then reduce the power by 1.'

Hence for our question, x^5 differentiates to 5x^(5-1) = 5x^4; 3x^2 differentiates to (2*3)x^(2-1) = 6x.

-17 is eliminated because it is the same as -17x^0, so when you multiply -17 by the power, 0, -17 * 0 = 0.

The final answer is:

dy/dx = 5x^4 + 6x

Answered by David L.

Studies Economics at LSE

The problem with this equation lies with the denominator on the left hand side. If we recall our graph of ln(x) however, we know that ln(x) is always positive and not equal to 0. Now we can safely multiply it up. The equation now reads:

3 = (ln(x))^{2} + 2ln(x)

We can recognise this as a quadratic equation in ln(x), and factorise it as such:

(ln(x)+3)(ln(x)-1) = 0

from which we deduce the solutions exist where ln(x) is equal to 1 or -3, the latter of which does not exist for any real values. Hence we consider ln(x) = 1, which is achieved when x = e, our one and only real solution.

(note that we can confirm this solution by substituting x = e into the original equation).

Answered by Mark H.

Studies Mathematics at Nottingham

Z is simply a general complex number, which can be written as Z = x+iy

Here |Z-2+3i| = 1 can be written as |Z-(2-3i)| = 1, which is just an expression for every Z whose distance from the point (2,-3i) is equal to 1.

We can solve this by recalling that Z = x+iy, and so we can seperate the real and imaginary parts in the modulus function.

i.e. 1 = |(x-2) + i(y+3)|

Evaluating the modulus now becomes simple as we calculate the magnitude using pythagoras. This Yields:

1^{2} = (x-2)^{2} + (y+3)^{2} , which is the cartesian form!

we recognise this as the equation of a circle, with centre (2,-3) and radius 1.

Answered by Mark H.

Studies Mathematics at Nottingham

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