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Thanks Heureka, but I wanted both sides up against the equal sign.

I wanted right aligned on the left side

and left aligned on the right side.    ://

Do you know how to do that?

Heureka, could you please tell me how I can get my Latex to right align in front of the equal sign. ??

right alignment :

\(\begin{array}{lcr} log_2(log_2(x))&=&log_2(10-2log_2(x))+1\\ log_2(log_2(x))-log_2(10-2log_2(x))&=&1 \\ log_2\frac{(log_2(x))}{(10-2log_2(x))}&=& 1\\ 2^{log_2\frac{(log_2(x))}{(10-2log_2(x))}}&=&2^1\\ \frac{(log_2(x))}{(10-2log_2(x))}&=&2\\ log_2(x)&=&2(10-2log_2(x))\\ log_2(x)&=&20-4log_2(x)\\ 5log_2(x)&=&20\\ log_2(x)&=&4\\ 2^{log_2(x)}&=&2^4\\ x&=&16\\ \end{array}\)

The code:

\begin{array}{lcr}
 log_2(log_2(x))&=&log_2(10-2log_2(x))+1\\
  log_2(log_2(x))-log_2(10-2log_2(x))&=&1 \\
   log_2\frac{(log_2(x))}{(10-2log_2(x))}&=& 1\\
   2^{log_2\frac{(log_2(x))}{(10-2log_2(x))}}&=&2^1\\
    \frac{(log_2(x))}{(10-2log_2(x))}&=&2\\
     log_2(x)&=&2(10-2log_2(x))\\
      log_2(x)&=&20-4log_2(x)\\
       5log_2(x)&=&20\\
        log_2(x)&=&4\\
         2^{log_2(x)}&=&2^4\\
          x&=&16\\
\end{array}

\begin{array} {lcr}... &=&...\end{array}

First field:       l = left alignment

Second field:   c = center

Third field:         r = right alignment

Heureka, could you please tell me how I can get my Latex to right align in front of the equal sign. ??

log2(log2(x))=log2(10-2.log2(x))+1

\(\begin{array}\\ log_2(log_2(x))&=log_2(10-2log_2(x))+1\\ log_2(log_2(x))-log_2(10-2log_2(x))&=1\\ log_2\frac{(log_2(x))}{(10-2log_2(x))}&=1\\ 2^{log_2\frac{(log_2(x))}{(10-2log_2(x))}}&=2^1\\ \frac{(log_2(x))}{(10-2log_2(x))}&=2\\ log_2(x)&=2(10-2log_2(x))\\ log_2(x)&=20-4log_2(x)\\ 5log_2(x)&=20\\ log_2(x)&=4\\ 2^{log_2(x)}&=2^4\\ x&=16\\ \end{array}\)

Haha best way to get blanked am i right melody 

What is the linear function of -3x-9y=12 into the form of y=mx+b

Add 3x to both sides

then

divide both sides by -9

give it a go yourself :)

There is no key for absolute value on the calculator here, but you can type in abs(x) where x is the expression for which you want the absolute value.

.

If you mean:

23.675*4² → 23.675*16 → 378.8   or 379 to nearest whole number.

If you mean

(23.675*4)² → (94.7)² → 8968/09  or 8968 to nearest whole number.

. 

60 km/h → 60*1000 m/hr → 60*1000/3600 m/s → 16.67 m/s

16.67 metres covered in one second.

use

[2nd] [acos]

arc cos is the same as inverse cos :)

I assume that the equation you ment to write was '739/2*5'

739/2= 369.5

369.5*5=1847.5

I have this formula that works well to convert wire gauge to mm:

mm dia = 0.127 * 92 ^ ((36-AWG)/39)

Following this, the same source offers the following as an inverse formula(mm to gauge):

AWG = -(ln (dia. mm/0.127) / (ln (92) * 39) – 36)

\(\begin{array}{|rcll|} \hline dia_{mm} &=& 0.127 \cdot 92^{ \left( \frac{36-\text{AWG}}{39} \right) } \qquad \text{AWG} =\ ? \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline dia_{mm} &=& 0.127 \cdot 92^{ \left( \frac{36-\text{AWG}}{39} \right) } \quad &|\quad : 0.127 \\\\ \dfrac{ dia_{mm} }{0.127} &=& 92^{ \left( \frac{36-\text{AWG}}{39} \right) } \quad &|\quad \ln{()}\\\\ \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} &=& \ln{ \left( 92^{ \left( \frac{36-\text{AWG}}{39} \right) } \right)}\\\\ \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} &=& \left( \frac{36-\text{AWG}}{39} \right) \cdot \ln{(92)}\\\\ \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} &=& ( 36-\text{AWG} ) \cdot \frac{ \ln{(92)} } {39} \quad &|\quad \cdot \frac{39}{ \ln{(92)} } \\\\ \frac{39}{ \ln{(92)} } \cdot \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} &=& 36 - \text{AWG} \quad &|\quad +\text{AWG} \\\\ \text{AWG} + \frac{39}{ \ln{(92)} } \cdot \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} &=& 36 \quad &|\quad -\frac{39}{ \ln{(92)} } \cdot \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} \\\\ \mathbf{\text{AWG} } &\mathbf{=}& \mathbf{36 -\frac{39}{ \ln{(92)} } \cdot \ln{ \left( \dfrac{ dia_{mm} }{0.127} \right)} }\\ \hline \end{array} \)

121^x=11

\(\begin{array}{|rcll|} \hline 121^x &=& 11 \quad &|\quad 121 = 11^2 \\ (11^2)^x &=& 11 \quad &|\quad (11^2)^x = 11^{2x}\\ 11^{2x} &=& 11 \\ 11^{2x} &=& 11^1 \\ 2x &=& 1 \\ \mathbf{x} & \mathbf{=} & \mathbf{ \frac12 }\\ \hline \end{array}\)

Radius= Diameter/2 = 4.5 feet = 1.3716 metres

ω= 2πf

v=2πrf= ωr = (50)*1.3716 = 68.58 = 69ms^-1

Log both sides:

log121^x=log11

We can use log laws to make:

xlog121=log11

x=log11/log121 =0.4999= 0.5

3.8210291e+18

Second answer was a mistake, I meant to correct . into a , in answer one.

60 kmH = 60,000 m./h

60,000/60 = 1,000 m./min.

1,000/60 = 16.667 m./sec.

Thank you for taking the time to respond. I didn't make an account as math isnt really my hobby (I prefer baduk!). This problem is just bothering me.

Since the angles of the triangle formed by making two diagonals from one point are 36-72-72, it is a golden triangle. So the ratio between the side of the pentagon and diagonal should be s : s*Ø, where Ø is the golden ratio: (1+sqrt(5))/2

I'm certain there's an error in my algebra somewhere.

The attached image shows my steps. a=s, φ = golden ratio. I checked it for sidelength four and my answer is off by what seems like a significant amount (27.7271... with my formula, 27.5276... for the normal formula). Since Ø² = 1+Ø, Ø²-.25 = Ø+.75

60 kmh = 60.000 m.h 

60.000/60 = 1000 m./min.

1000/60 = 16,667 m./sec.

Welcome to the forum  GLC    

 It is excellent that you have joined.  

What kind of cakes do you like the best? 

=-9/-7

=1.2857