^ means power of
2nd question - numbers should be subscripted
Ï€ means pi


v^2 = u^2 2as for v

p1 A1 v1 = p2 A2 v2 for A1

v = πr^2h for r

x-a x-b =1 for x

b c


Please provide working out as well as answers so I can learn,thanks

sorry the c in the third queston should be below x-b

Also an addition sign should be between x-a and a-b, and also before 2as in the first question

v^2 = u^2 + 2as for v

Take the square root on both sides.

p_1 A_1 v_1 = p_2 A_2 v_2 for A1

Divide both sides by p_1v_1

v = \pi r^2h for r

First divide both sides by \pi h, then take the square root on both sides.

\frac{x-a}{b} + \frac{x-b}{c} = 1 for x

1. Multiply both sides by the least common multiple of the fractions.
2. Expand and simplify the brackets which will be formed.
3. Add/Subtract anything on both sides so that terms in x only remain on the left.
4. Factor out x.
5. Divide by the coefficient of x.

Can you post what you get? :)

My first answer is 1st step √v^2 = √u^2 + 2as
2nd step V = u^2 + √2as

Here is my first answer to the 1st question 1st step (square rt v^2) = (square rt u^2) + 2as The brackets make it easier to show what I am square rooting 2nd step V = u^2 + (square rt 2as)

2nd question answer 1st step p1A1v1 = p2A2v2
p1v1 p1v1

2nd step A1 = p2A2v2
p1v1

The p1v1 p1v1 and p1v1 should be directly below the lines sorry!

3rd question 1st v = πr^2h
πh πh

2nd v = r^2
Ï€h

3rd (square rt v) = r^2
Ï€h

�€ means pi

I couldn't work out the last one, couuld you please verify my answers and show me how to get the last question, thanks josh.



















�€h

* π means pi

My first answer is 1st step √v^2 = √u^2 + 2as
2nd step V = u^2 + √2as

You mean this?

v = v^2 + \sqrt{2as}

No, that's wrong. It should be this:

v = \sqrt{v^2 + 2as}

When you do something on one side, the same thing must happen on the whole other side.

2. Good

3. Same thing here. It should be:

r = \sqrt{\frac{V}{\pi h}}

4.

First multiply by the least common multiple of the denominators.

\frac{x-a}{b} + \frac{x-b}{c} = 1

The L.C.M is because, hence:

bc\(\frac{x-a}{b} + \frac{x-b}{c}\) = 1\times bc

bc\(\frac{x-a}{b}\) + bc\(\frac{x-b}{c}\) = bc

c(x-a) + b(x-b) = bc

Can you expand here?

Thanks unknown, I will learn from this.

I have expanded the last! A = -1b + bc-1x + -1b2c-1 + x

Um.. I'm not sure where you got that...

cx - ca + bx - b^2 = bc

Here is what you get when you expand.

Now, bringing all the x and terms not in x on different sides:

cx \cancel{- ca} + bx \cancel{- b^2} \cancel{+ ca} \cancel{+ b^2}= bc+ ca + b^2

cx + bx = bc+ ca + b^2

Can you factor x here on the left and then divide both sides by (c + b)?

Will I need to rearrange the equation in the form "(quadratic) = 0"?

is the answer to the first step x^2=cb+bc+ca+b^2

No no no. How do you factorise cx + bx?

Just like that:

2x + 3x = x(2 + 3)

Ok x(C+B) Now do I divide both sides by (c+b)?

So the final answer is x=bc+ca+b^2
(c+b)

* brackets should be directly below

Yes, that's it! :)

Ok, thanks very much!