Yes I realized that after I answered. Sorry jdh3010 :))

I was answering with a phone

You want to stick you teeth into SevenUp Rosala?  LOL

jdh3010, you can use the calc right here in the froum and it is a good way to answer questions like these.

Click on the [Calculations eg 1+2=3] button and   type in    sqrt(6)=   and then press [ok]

$${\sqrt{{\mathtt{6}}}} = {\mathtt{2.449\: \!489\: \!742\: \!783\: \!178\: \!1}}$$             

Oh it might not work so good if you are answering from a phone.  

$${\frac{{\mathtt{14}}}{{\mathtt{70}}}} = {\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{7}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{7}}\right)}} = {\frac{{\mathtt{1}}}{{\mathtt{5}}}}$$

square root of six=2.4494897428

2 is a prime number   

(x-5)*(x+5)=x^2-25

Thanks Radix, I have repaired mine now.  :))

$${\sqrt{{\mathtt{9}}{!}{\mathtt{\,-\,}}{\frac{{\mathtt{9}}}{{\mathtt{9}}}}}} = {\mathtt{602.394\: \!389\: \!084\: \!094\: \!972\: \!4}}$$

$$\\9^x=(19*10^6)/ 26\\\\
9^x=19/26*10^6\\\\
log(9^x)=log(19/26*10^6)\\\\
xlog(9)=log(19/26)+log(10^6)\\\\
x=\frac{log(19/26)+6}{log(9)}\\\\$$

$${\frac{\left({log}_{10}\left({\frac{{\mathtt{19}}}{{\mathtt{26}}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{6}}\right)}{{log}_{10}\left({\mathtt{9}}\right)}} = {\mathtt{6.144\: \!958\: \!115\: \!969\: \!216\: \!2}}$$

Hello Melody,

I have this result:

$${\frac{\left({\mathtt{\,-\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{\left({\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{9}}\right)}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\sqrt{{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}}}\right)}}$$

Gruß radix !

$$\\=\frac{\sqrt{x^2+1}}{(x-9)^2}\\\\
\frac{(x^2+1)^{1/2}}{(x-9)^2}\\\\
Let\\
u=(x^2+1)^{1/2}\qquad \qquad \qquad \qquad\qquad v=(x-9)^2\\\\
u'=(1/2)(x^2+1)^{-1/2}*2x \qquad \qquad v'=2(x-9)\\\\
u'=(x^2+1)^{-1/2}*x \qquad \qquad \qquad \;\;\; v'=2(x-9)\\\\
$Quotient Rule$\\\\
\boxed{y'=\frac{vu'-uv'}{v^2}}\\\\
y'=\frac{(x-9)^2*(x^2+1)^{-1/2}*x-(x^2+1)^{1/2}*2(x-9)}{(x-9)^4}\\\\
y'=\frac{(x-9)*(x^2+1)^{-1/2}*x-(x^2+1)^{1/2}*2}{(x-9)^3}\\\\
y'=\left(\frac{x(x-9)}{(x^2+1)^{1/2}}-\frac{2(x^2+1)}{(x^2+1)^{1/2}}\right)\div (x-9)^3\\\\$$

$$\\y'=\left(\frac{x(x-9)-2(x^2+1)}{(x^2+1)^{1/2}}\right)\div (x-9)^3\\\\
y'=\left(\frac{x^2-9x-2x^2-2}{\sqrt{x^2+1}}\right)\div (x-9)^3\\\\
y'=\frac{-x^2-9x-2}{(x-9)^3\sqrt{x^2+1}} \\\\$$

Now, if it is not riddled with careless errors (which it probably is) that is your answer :))

Thanks Radix I have now repaired mine and it is the same as yours.  That is a good sign LOL   :))

only the X^2+1 is under the square root

Which bit is under the square root?

Pi does not end for the same reason that phi does not end. both of them are what we refer to as irrational numbers. They can never be completely unified with what we believe are rational numbers. However it might just be that they are both dancing to a tune that we do not yet understand.

Melody please dont ask tin to serve any eatables , cuz he may drop them on your royal guests sue to an eye sight problem.....

this girl is nothing....animes are so cute..........(yeah thats why i feel myself so ugly and dumb huh)

anyways.....i like 7Up Masala Soda if youve heard of it.....in a way i feel it is less fatty and concentrated than other drinks.......i wonder how "SevenUp" being a human would taste....i think ill know after i become a vampire.....hehe

Anonymous is basically correct, though there is one exception.  If x = -2/3 then the result, 0^0, is not defined.

.

4*|3x+4|=4x+8

on questoin 2 would you not distribute the 4(x+2) and get 4x+8

It is really good that you are asking questions Knk98.    

It was already 4x+8 at the very beginning.

I saw that the Left hand side (LHS)  was    4*an absolute value experssion.

If I could factor 4 out of the RHS then I could divide both sides by 4 and the 4s would just go away makeing the equation more simple.

4*|3x+4|=4x+8

4*|3x+4|=4*(x+2)      Now I can divide both sides by 4

$$\\\frac{4*|3x+4|}{4}=\frac{4*(x+2)}{4} \\\\
$Now the 4s can cancel out$\\\\
\frac{\not{4}*|3x+4|}{\not{4}}=\frac{\not{4}*(x+2)}{\not{4}} \\\\
$This will leave me with$\\\\
|3x+4|=x+2\\\\
$I like to swap sides here but it is not really necessary$\\\\
x+2=|3x+4|\\\\
$Now remember how $|2|=|-2|\;\;\; well\\\\
|3x+4|=|-(3x+4)| \qquad too.\\\\
so\\
3x+4=x+2\qquad or \qquad 3x+4=-(x+2)\\\\$$

I finished it before.

If you have any more questions please ask and I will try to explain some more.

It is possible that your real problem is not with absolute equations but just with equations in general.

It is important that you try to isolate the source of your problems then you will know what you need to work on.      

I'm guessing here, but:

log(10/k) = 1.2   so 10/k = 10^1.2 where k is a reference level exposure.

log(E/k) = 1.8     so E/k = 10^1.8 where E is the required exposure.

Divide the second of these by the first and rearrange to get:

$${\mathtt{E}} = {\frac{{\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{1.8}}}}{{{\mathtt{10}}}^{{\mathtt{1.2}}}}} \Rightarrow {\mathtt{E}} = {\mathtt{39.810\: \!717\: \!055\: \!349\: \!725\: \!1}}$$

So E ≈ 39.8 mA.min

.

well..first of all i dont have a new "BF" and secondly.....ive saved a gun and one bulllet to killl my last BF already...which will happen after a few years........good idea ill also learn karate to take away the left over breaths in him........haha......

now the truth:do you even know the answer to this......but remember before answering......you will partially be the reason for the death of a younf boy after ten years....

(3x + 2) ^0 = 1

find the particular solution of the equation f'(x)=2x^(-1/2) that satisfies the condition f(1)=6

$$\\f'(x)=2x^{-1/2 }\\\\
f(x)=\int\;2x^{-1/2 }dx\;\\\\
f(x)=\;\dfrac{2x^{+1/2 }}{1/2}+c\\\\
f(x)=4\sqrt{x}+c\\\\
but\\
f(1)=6\\\\
so\\
6=4\sqrt{1}+c\\\\
6=4+c\\\\
c=2\\\\
so\\
f(x)=4\sqrt{x}+2\\\\$$

h[(tan^-1(7.10/-3.52))]

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{7.1}}}{-{\mathtt{3.52}}}}\right)} = -{\mathtt{63.628\: \!951\: \!849\: \!611^{\circ}}}$$

h[63.629°]     has no particular meaning to me