Author Topic: Bucket's Questions (Read 39187 times) Tweet Share

bucket · « **Reply #15 on:** February 19, 2008, 11:30:38 pm »

ahhh thankyou so much mr coblin.
you have proven to be more help than my math teacher yet again!

bucket · « **Reply #16 on:** February 20, 2008, 12:03:18 am »

HOLD ON!! lol

Why is it
(12 - x)(20 - y) and not (x - 12)(y - 20)??

Collin Li · « **Reply #17 on:** February 20, 2008, 12:11:56 am »

No idea why I wrote it like that, I must have been high.

It's the same thing (mathematically) though, because you can take out a factor of -1 from both the brackets, and $(-1)\times (-1) = 1$ , haha.

bucket · « **Reply #18 on:** February 20, 2008, 12:14:11 am »

haha!! oh okay, thanks =]

bucket · « **Reply #19 on:** February 21, 2008, 04:26:33 pm »

Oh god this question is embarassing.
I literally slept in class during methods last year and now I'm behind on many topics, especially logarithms and exponentials

Simplify $3^4 * 9^2 * 27^3$

I'm getting $3^{14}$ as my answer but the book says the answer is $3^{17}$

dcc · « **Reply #20 on:** February 21, 2008, 04:34:35 pm »

Quote from: bucket on February 21, 2008, 04:26:33 pm

Simplify $3^4 * 9^2 * 27^3$

Since $9 = 3^2$ and $27 = 3^3$

$3^4 * (3^2)^2 * (3^3)^3$

$3^4 * 3^4 * 3^9$

$3^{4 + 4 + 9}$

$3^{17}$

Remember:

$(a^b)^c = a^{b \times c}$

bucket · « **Reply #21 on:** February 21, 2008, 04:36:45 pm »

ohhhhhhhhhhhhhhhhh.
how could i have not worked it out myself T_T
*bangs head with saucepan*

costargh · « **Reply #22 on:** February 21, 2008, 04:41:59 pm »

Quote from: bucket on February 21, 2008, 04:36:45 pm

*bangs head with saucepan*

lol. if u ever did that id piss myself haha

bucket · « **Reply #23 on:** February 21, 2008, 07:06:24 pm »

Ok, when I'm solving questions such as
"Simplify $\frac{x^3yz^{-2}*2(x^3y^{-2}z)^2}{xyz^{-1}}$ expressing your answer in positive exponent form"
I get confused whilst working the answer out because there are so many different calculations to do and I end up all over the place.
Does anyone have any tips or tricks they used to make things more methodical when solving problems such as this? Or do I just need to get used to it? lol

dcc · « **Reply #24 on:** February 21, 2008, 07:36:00 pm »

Try splitting the problem into small bits, so for this problem:

Quote from: bucket on February 21, 2008, 07:06:24 pm

"Simplify $\frac{x^3yz^{-2}*2(x^3y^{-2}z)^2}{xyz^{-1}}$

First tackle the brackets, removing the square from the bracket:
$2(x^3y^{-2}z)^2$

Which when squared through becomes: $2(x^6 \times y^{-4} \times z^2)$

Then just tackle the original top line of the equation:

$x^3 \times y \times z^{-2} \times 2x^6 \times y^{-4} \times z^2$

which becomes, after adding all the powers etc. etc: $2x^9 \times y^{-3}$

Then we consider this top line and then the bottom line:
$\dfrac{2x^9 \times y^{-3}}{x \times y \times z^{-1}}$

which becomes: $2x^8 \times y^{-4} \times z$

Which when expressed in positive exponent form becomes:
$\dfrac{2x^8 \times z}{y^4}$

bucket · « **Reply #25 on:** February 21, 2008, 08:35:03 pm »

sweet thanks lol.

bucket · « **Reply #26 on:** February 23, 2008, 07:38:15 pm »

Simplify $2^\frac{3}{2} * 4^{-\frac{1}{4}} * 16^{-\frac{3}{4}}$

I'm getting $2^{-4} = \frac{1}{4}$ when the answer is $\frac{1}{4}$

Thanks to anyone who helps =]

dcc · « **Reply #27 on:** February 23, 2008, 08:17:31 pm »

$2^\frac{3}{2} * 4^\frac{-1}{4} * 16^\frac{-3}{4}$

$2^\frac{3}{2} * (2^2)^\frac{-1}{4} * (2^4)^\frac{-3}{4}$

$2^\frac{3}{2} * 2^{-\frac{1}{2}} * 2^{-3}$

Adding the powers, we get:

$2^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4}$

bucket · « **Reply #28 on:** February 23, 2008, 09:53:58 pm »

thanks alot dcc
okayyy.. i have another question >.<

$\frac{3^{n-2}}{9^{1-n}} = 1$ , Solve for n.

Collin Li · « **Reply #29 on:** February 23, 2008, 09:56:29 pm »

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