All Answers 358090 Answers

first add an extra axe

here are 12 axes and apply Sam's fractions to them.  Ron gets 12/2 = 6 axes; John gets 12/3 = 4 axes; the youngest gets 12/12 = 1 ax.

6 + 4 + 1 = 11, so this uses all Sam's axes and you can keep your own extra axe!

-40

hope it helps

thanks cphill

lots of wood 

8560.26/6= 1426.71

Nice answer, oboeduck.....BTW....I love the name "oboeduck"  !!!

1. It was first  called " pi" in 1706 by [the Welsh mathematician] William Jones, because  pi is the first letter in the Greek word perimitros, which means "perimeter." It's a special number.

    

    

Base 2 works well....it can serve as a log base for both numbers.... (thus,  it's easy to write each number as a power of 2)

Let's take this in "pieces"

We have

-1 ≤ 1 - 1/x^2  subtract 1 from both sides

-2 ≤ -1/x^2     multiply both sides by -1 and reverse the inequality sign

2 ≥  1/x^2       rearrange

x^2 ≥ 1/2     subtract 1/2 from both sides

x^2 - 1/2 ≥ 0   factor

(x - 1/√2) (x + 1/√2) ≥ 0

And we have two possible intervals that work, here

(-∞, -1/√2] and [1/√2, ∞)

Now, let's look at the other piece

1 - 1/x^2 ≤ 1     subtract 1 from both sides

-1/x^2 ≤ 0         divide by -1 on both sides and reverse the sign

1/x^2 ≥ 0

And the possible intervals here are

(-∞, 0), (0, ∞)

But, we have to take the most "restrictive" intervals that would make the original inequality true

Notice that the interval (-1/√2, 1/√2) makes the original inequality untrue

So, the intervals that "work" are   (-∞, -1/√2] and [1/√2, ∞)

no problem , thank you 

Hi 315

it's after 3am here and I'm on my phone so my full answer will have to wait till morning.  Sorry  :)

can you do it step by step ? 

First be aware that x cannot equal 0

Next muliply everything by x squared

can you do it from there?

Thanks Alan    

Maybe you mean

How do you put the grease in the formula.  Is this a chemistry formula?      

Maybe you take the lid of the grease and spoon in the required quantity?  

Please do not take offense.        I am just playing with the words  

There's nothing wrong with your logic Melody; there are 3 possible solutions:

$${{\mathtt{a}}}^{{\mathtt{3}}} = {\left(-{\mathtt{3}}\right)}^{\left(-{\mathtt{4}}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)}\right)}}\\
{\mathtt{a}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)}\right)}}\\
{\mathtt{a}} = {\frac{{\mathtt{1}}}{{{\mathtt{3}}}^{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{\,-\,}}{\mathtt{0.115\: \!560\: \!212\: \!391\: \!772\: \!5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.200\: \!156\: \!159\: \!196\: \!130\: \!2}}{i}\\
{\mathtt{a}} = {\mathtt{\,-\,}}\left({\mathtt{0.115\: \!560\: \!212\: \!391\: \!772\: \!5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.200\: \!156\: \!159\: \!196\: \!130\: \!2}}{i}\right)\\
{\mathtt{a}} = {\mathtt{0.231\: \!120\: \!424\: \!783\: \!544\: \!9}}\\
\end{array} \right\}$$

.

y=cos x -2

parity is the state of being even or odd   (there is one r in parity)

 This will be an even function because

 cos(x)-2=cos(-x)-2    and hence the graph is symetrical about the y axis  (x=0)

See the graph

This is a step graph if that is what you want to know

I know that is right Alan but what is wrong with the logic on my hand done answer?

Also

Why does  (-8)^(-1/3)  give a real answer of -1/2 on some calcs (eg casio) and 'error' on other calcs?

Should there be an x in there somewhere?

.

Calculators often use the following method to calculate fractional powers:

If we have a = b^c they take logs to get ln(a) = c*ln(b).  Now raise both sides to the power e to get a = e^c*ln(b).

If b is negative they can complain!

In fact: (-3)^-4/3 = -0.11556 + 0.200156i

.

Input was

(-3)^4/3

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I am actually surprised that the calc interpreted it this way.

I expected it to be interpreted as       $$\frac{(-3)^{-4}}{3}$$      in which case the answer would be     

$${\frac{\left({\left(-{\mathtt{3}}\right)}^{-{\mathtt{4}}}\right)}{{\mathtt{3}}}} = {\frac{{\mathtt{1}}}{{\mathtt{243}}}} = {\mathtt{0.004\: \!115\: \!226\: \!337\: \!448\: \!6}}$$

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Now lets go back to the calculator interpretation

$${\left(-{\mathtt{3}}\right)}^{{\mathtt{\,-\,}}{\frac{{\mathtt{4}}}{{\mathtt{3}}}}} = \underset{{\tiny{\text{Error: not real}}}}{{{\left(-{\mathtt{3}}\right)}}^{-{\mathtt{1.333\: \!333\: \!333\: \!333\: \!333\: \!3}}}}$$

If I do this by hand I could get

$$(-3)^{(-4/3)}\\\\
=\left[(-3)^{-4}\right]^{1/3}\\\\
=\left[\frac{1}{(-3)^{+4}}\right]^{1/3}\\\\
=\left[\frac{1}{-3*-3*-3*-3}\right]^{1/3}\\\\
=\left[\frac{1}{81}\right]^{1/3}\\\\
=\frac{1}{\sqrt[3]{81}}\\\\
\approx 0.2311$$

However I have tried it on 3 different calculators and none of them will give a 'real number' answer.

When you raise negative numbers to fractional powers it certainly cause problems.

Consider when the numerator is 1 and the denominator is an even number, like

$${\left(-{\mathtt{9}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)} = {\sqrt{-{\mathtt{9}}}}$$     there is definitely no real number answer to this.

$${\left(-{\mathtt{9}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)}$$     there is definitely no real number answer to this either.

I am not giving you a proper answer here.  I am just discusing your problem.

I think the calculators work these out with a generated series.  Powers of neg numbers just do not work.

I am sure that Alan and maybe Heureka and CPhill can have a lot more to say on this subject.

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Please note that  (-8)^(-1/3) works just fine

$${\left(-{\mathtt{8}}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = \underset{{\tiny{\text{Error: not real}}}}{{{\left(-{\mathtt{8}}\right)}}^{-{\mathtt{0.333\: \!333\: \!333\: \!333\: \!333\: \!3}}}}$$

Correction - it works just fine on my CASIO - it does not work on the web2 calc

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I know this has been discussed before but I'd really like other mathematicians to discuss this again.

Thank you.       

Does this help:

Let 1 - 1/x² = -1 and solve for x to get the limiting values of x at the -1 level

.

If you know someone who uses the linux operating system on their computer, ask them to run this short program and to mail you the results. You can then go over it and count the number of occurrences!

Here's the prgram: dc '2 19112000^f'

Happy counting!