For the second one......just disribute the first term in the first set of parentheses across each term in the second set of parentheses.

Do the same with the second term......and then with the third term.........combine like terms when you finish......

The things says that I am wrong, and I have one more chance. They don't make mistakes, I am pretty sure.

Huh??.....those are the only three terms that have an " x " associated with them when we simplify this.....

I don't think I mis-added....

I wrote this a couple of minutes ago, but nobody would respond again! Sorry guys!

Someone please help me on this question! I really need help.

This answer is wrong. Sorry! I have one more chance

Let's just factor 10000  =

2⁴ x 5⁴  =

16 x 625

And neither one of those contain a 0..........!!!!

Only three terms have will "x" associated with them .....8x

And distributing the "-3" across the "6x" in the parentheses we get -18x

And when we distribute "9" across the"3x" in parentheses we get 27x...so.....

8x - 18x +  27x   =   17x

I DO! IT IS FUN!

That's what it is.........it's like asking what  6x + 1 is........(the only thing we could do is to get a common denominator and write it as a single fraction....... if you're so inclined)

Here's another way to do this

tan 12t / sin4t    =

(sin12t/cos12t) / sin4t = (divide numerator and denominator by t)

(sin12t/t) /[(sin4t/t)(cos12t)] = (multiply the numerator by 12/12  and the denominator by 4/4)

(12sin12t/12t) / [ (4sin4t/4t)(cos12t)]

Now

lim t → 0   (12sin12t/12t)   = 12   and

lim t → 0   (4sin4t/4t)  = 4    and

lim t → 0  (cos12t)  = 1     so

12/(4 * 1 )  = 12/4   = 3

Add   21m   to both sides

then  subtract 12 from both sides

then think about what to do next.

hey its me again i know someone sees this please help me im cryin here. and im about to panic

(2/sqrt3)tan^(-1)(2x/sqrt3)+c

$$\int\;\frac{1}{x^2+\frac{3}{4}}\;dx\;=\;\frac{2}{\sqrt3}\;tan^{-1}\;\frac{2x}{\sqrt3}+c$$

$$\int\;\frac{1}{x^2+a^2}\;dx\;=\;\frac{1}{a}\;tan^{-1}\;\frac{x}{a}+c$$

There is no value of x which equals arcsec(0) as arcsec(0) doesn't exist:

.

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@@ End of Day Wrap      Sun 7/12/14      Sydney,   Australia       Time 1:10 am  (Really monday morn)     ♬

Hi all,

It was quieter today but there were less answerers as well as less questions and i found some challenging problems.

Our fantastic answerers were HappyBJ, CPhill, Rosala, saseflower, Alan and Kitty<3.  Thanks guys. :)

Here are a few interest posts:

1) This one is great - calc the 5th root of 2 with a very simple calculator - Thanks anon :)

(Suitable for all high school children - at a push anyway)

Mon 8/12/14

Example of how a question shoiuld be reposted if it is not answered and it is not onthe first page anymore.

64x^2-9y^2= (8x-3y)(8x+3y)

$${\mathtt{2}}{n}{\mathtt{\,-\,}}{\mathtt{4}} = {\frac{{\mathtt{4\,860}}}{{\mathtt{90}}}} \Rightarrow {\mathtt{n}} = {\mathtt{29}}$$

 I have left out M because down the bottom it says M=1

I have not included all of the initial working, but most is here. 

After 5 years            

 $$X=100(1+0.5J)^{10}\qquad(1)$$

After 7 years 

$$FV=100(1+0.5J)^{14}$$

After 10.5 years    (3.5 years more)

$$FV=[100(1.05J)^{14}](1+0.5J)^{14}=100(1.05J)^{28}$$

$$\\1.98*100=100(1+0.5J)^{28}\\\\
1.98=(1+0.5J)^{28}\\\\
(1.98)^{(1/28)}=1+0.5J\\\\
(1.98)^{(1/28)}-1=0.5J\\\\
2[(1.98)^{(1/28)}]-2=J\\\\
J=2[(1.98)^{(1/28)}]-2 \qquad(2)\\\\
$Sub into (1)$\\\\$$

$$\\X=100(1+0.5J)^{10}\qquad(1)\\\\
X=100(1+0.5*(2[(1.98)^{(1/28)}]-2)\\\\
X=100(1+(1.98)^{(1/28)}-1)\\\\
X=100(1.98)^{(1/28)}\\\\$$

$${\mathtt{100}}{\mathtt{\,\times\,}}\left({{\mathtt{1.98}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{28}}}}\right)}\right) = {\mathtt{102.469\: \!634\: \!086\: \!312\: \!44}}$$    

X=$102.47

I did the maths while I was writing tht LaTex so there could easily be mistakes.

I have not checked it.

Lets see

0/2 = 0

BUT

2/0 = undefined  --- error?

Yep!

Order of syntax

is

very confusing.

I do not understand this question. Does anyone else?