I need h in terms of x

Got up to

$\displaystyle \frac{4}{3} \cdot 2x=\frac{1}{3} \cdot h$

WIP please with answer.

Printable View

- Jan 7th 2010, 09:34 AMMukilabHalfway through equation
I need h in terms of x

Got up to

$\displaystyle \frac{4}{3} \cdot 2x=\frac{1}{3} \cdot h$

WIP please with answer. - Jan 7th 2010, 10:08 AMDinkydoe
Multiply both sides with 3 to obtain: $\displaystyle 8x = h $

That is what you want right? - Jan 7th 2010, 10:34 AMMukilab
Oh god. I came up with $\displaystyle \frac{\frac{4}{3} 2x}{\frac{1}{3}}$

Thanks though, I'm so forgetful ^^ - Jan 7th 2010, 10:38 AMDinkydoeQuote:
- Jan 7th 2010, 12:19 PMMukilab
yes but it's probably not what the examiner wanted! XD

- Jan 7th 2010, 12:41 PMArchie Meade
Dividing by $\displaystyle \frac{1}{3}$ is multiplying by $\displaystyle 3.$

For example....

Try dividing 12 into 3 equal parts, $\displaystyle \frac{12}{3}=4$

Now try dividing 12 into $\displaystyle \frac{1}{3}$ parts...

Make sense?

thought not...

What is 12 divided by $\displaystyle \frac{1}{3}$ ?

$\displaystyle \frac{12}{\frac{1}{3}}=\frac{(36)\frac{1}{3}}{\fra c{1}{3}}=36\frac{\frac{1}{3}}{\frac{1}{3}}$

Expressing the numerator as a multiple of the denominator allows cancellation

of non-zero common factors (after all, a third is 0.67ish).

Any number divided by itself is 1, except zero.

A fraction divided by itself is 1

So, 12 divided by a third is 36, because it's really 12 times 3.

Dividing by a fraction is multiplying by the fraction turned upside down.

of course, this means... multiply by the numerator and divide by the denominator. - Jan 7th 2010, 12:45 PMMukilab
Amazing. I really do love your ways of explaining things. Please do continue to help me in further issues if you find time :)

- Jan 7th 2010, 02:05 PMArchie Meade
Thanks, Mukilab,

you're welcome.

The vectors question you had a few days ago......

The critical thing to understand there was $\displaystyle \overrightarrow{xy}=\overrightarrow{y}-\overrightarrow{x}$

The vector $\displaystyle \overrightarrow{xy}$ starts at the point x and ends at y.

The vectors $\displaystyle \overrightarrow{x}$ and $\displaystyle \overrightarrow{y}$ both start at the origin.

You understand this by examining the parallelograms.