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Josh2345 · Nov 11, 2011, 02:55 PM

^ 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 = Ï€r^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

Josh2345 · Nov 11, 2011, 02:59 PM

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

Josh2345 · Nov 11, 2011, 03:01 PM

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

Unknown008 · Nov 12, 2011, 11:47 AM

$v^2 = u^2 + 2as$ 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? :)

Josh2345 · Nov 12, 2011, 01:41 PM

My first answer is 1st step âˆšv^2 = âˆšu^2 + 2as
2nd step V = u^2 + âˆš2as

Josh2345 · Nov 12, 2011, 01:47 PM

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)

Josh2345 · Nov 12, 2011, 01:59 PM

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

2nd step A1 = p2A2v2
p1v1

Josh2345 · Nov 12, 2011, 02:01 PM

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

Josh2345 · Nov 12, 2011, 02:33 PM

3rd question 1st v = Ï€r^2h
Ï€h Ï€h

2nd v = r^2
Ï€h

3rd (square rt v) = r^2
Ï€h

Josh2345 · Nov 12, 2011, 02:35 PM

Ãâ‚¬ 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.

Ãâ‚¬h

Josh2345 · Nov 12, 2011, 02:37 PM

* Ï€ means pi

Unknown008 · Nov 13, 2011, 03:47 AM

Originally Posted by Josh2345

My first answer is 1st step âˆšv^2 = âˆšu^2 + 2as
2nd step V = u^2 + âˆš2as

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?

Josh2345 · Nov 13, 2011, 03:59 AM

Thanks unknown, I will learn from this.

Josh2345 · Nov 13, 2011, 04:10 AM

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

Unknown008 · Nov 13, 2011, 04:53 AM

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)?

Josh2345 · Nov 13, 2011, 05:15 AM

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

Josh2345 · Nov 13, 2011, 05:18 AM

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

Unknown008 · Nov 13, 2011, 05:38 AM

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

Just like that:

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

Josh2345 · Nov 13, 2011, 05:48 AM

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

Josh2345 · Nov 13, 2011, 06:11 AM

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