Melody

Do you mean

$${\sqrt[{{\mathtt{{\mathtt{8}}}}}]{{\mathtt{1.316\: \!8}}}} = {\mathtt{1.034\: \!999\: \!112\: \!125\: \!890\: \!8}}$$

This is an approximation.

hi again GoldenLeaf,

Your logic and your answer are botf good.

however I don not like this line

15a(-54+54) = (9+54)

this implies 15a * (-54+54) = 15a * 0 = 0

and that is definitely not what you meant.

just write it like this - brackets usually help but in this case that make the working wrong.

15a-54+54 = 9+54

You may both teach and learn a lot here GoldenLeaf.  This is how I have always wanted the forum to be used.  

Good answer GoldenLeaf!

$${\sqrt{{\mathtt{2.8}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{13}}}}} = {\mathtt{0.000\: \!000\: \!529\: \!150\: \!26}}$$

I did this straight on the site calc using the "Math input=result" tab

Note this is an aprroximation.  It is an irrational number and cannot be written exactly in digits. 

I hope you don't mind me checking some of your answers..  I will do it less as your ability, and my confidence in it, grows.

Thank you anonymous for your correct answer and thankyou GoldenLeaf for having a shot.  

GoldenLeaf I want to try and sort out your understanding of scientific notation.

10305 is a large number (bigger than 9) so when it is changed the scientific notation the indice will be Positive

And, you want just one Non-zero number in front of the point so it becomes

$$10305 = 1.0305\times 10^4$$

If it was a little number, smaller than 1 then the indice would be negative!

I hope that helps.  

I am just going to add examples here.

-------------------------------------------------------------------------

(2x+7)(4x-1)\leq 0 \Rightarrow \left[-\dfrac 7 2 \leq x \leq \dfrac 1 4 \right]

$$(2x+7)(4x-1)\leq 0 \Rightarrow \left[-\dfrac 7 2 \leq x \leq \dfrac 1 4 \right]$$        By Rom

-------------------------------------------------------------------------

\left[-\dfrac 7 2 \leq x \leq \dfrac 1 4 \right] \bigcup \left\{\dfrac 1 2 \right\}

$$\left[-\dfrac 7 2 \leq x \leq \dfrac 1 4 \right] \bigcup \left\{\dfrac 1 2 \right\}$$    By Rom

ok.

You have to put Rom's answer and my answer together to get the right answer.

For what value(s) of x is log (8x^2+26x-7)/log(4x-1) undefined?

UNDEFINED WHEN

$$\left[-\dfrac 7 2 \leq x \leq \dfrac 1 4 \right] \bigcup \left\{\dfrac 1 2 \right\}$$      (from Rom)

ALSO UNDEFINED WHEN

$$x\le 0.25$$    (from Melody)

THEREFORE undefined for

$$\left[x \leq \dfrac 1 4 \right] \bigcup \left\{\dfrac 1 2 \right\}$$

Thanks Rom and Bertie

Look here

http://web2.0calc.com/questions/how-do-i-type-in-fractions

Thank you for the thumbs up. 

Thanks reinout,

Reinout is right.  I think mine is ok except the hole is in the wrong place.

therefore the range and domain are not correct.

Reinout could you please give AllSmilesAlways the correct answer.

I have had enough for the night.  It is 2am.

See you much later in the day. 

$$tanx+\frac{cosx}{1+sinx}\\\\
=\frac{sinx}{cosx}+\frac{cosx}{1+sinx}\\\\
=\frac{sinx(1+sinx)}{cosx(1+sinx)}+\frac{(cosx)^2}{(1+sinx)cosx}\\\\
=\frac{sinx+sin^2x+cos^2x}{cosx(1+sinx)}\\\\
=\frac{sinx+1}{cosx(1+sinx)}\\\\
=\frac{1}{cosx}\\\\
=secx$$