All Answers 358090 Answers

Alan's answer is an approximation.

$$\frac{1}{2^6}=\frac{1}{64}$$

$$$\int \frac{1}{\left(\frac{\sqrt{3}}{2}v\right)^2+\frac{3}{4}}\frac{\sqrt{3}}{2}$$$

I assume that this is meant to be

$$\frac{\sqrt{3}}{2}\int \frac{1}{\left(\frac{\sqrt{3}}{2}v\right)^2+\frac{3}{4}}dv\\\\
=\frac{\sqrt{3}}{2}\int \frac{1}{\frac{3}{4}v^2+\frac{3}{4}}\;dv\\\\
=\frac{\sqrt{3}}{2}\int \frac{1}{\frac{3}{4}(v^2+1)}\;dv\\\\
=\frac{\sqrt{3}\times 4}{2\times 3}\int \frac{1}{(v^2+1)}\;dv\\\\
=\frac{2}{\sqrt 3}\int \frac{1}{1+v^2}\;dv\\\\
=\frac{2}{\sqrt 3}\;tan^{-1}\;v +c$$

I think that is correct.  

Chris, I really like what you have done.  It is really neat and looks perfectly valid too me.  

Heureka's answer is also much better than mine.   Thanks Heureka    

Find x for the equation: x+1/x = -1 (x plus one divided by x equals negative one)

$$\small{\text{
set $\frac{1}{x} = -1-x$ and see there is no cut between function $\frac{1}{x}$ and line $-1-x$
}}$$

Just use the calculator on the home page here.

$${{\mathtt{6.25}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{5}}}}\right)} = {\mathtt{3.002\: \!811\: \!084\: \!953\: \!578\: \!1}}$$

.

A.  If you mean x + (1/x) = -1 then:

Multiply all terms by x:

x² + 1 = -x

Add x to both sides:

x² + x + 1 = 0

Solve:

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.866\: \!025\: \!403\: \!785}}{i}\right)\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.866\: \!025\: \!403\: \!785}}{i}\\
\end{array} \right\}$$

B.  If you mean (x + 1)/x = -1

Multiply both sides by x

x + 1 = -x

Add x to both sides

2x + 1 = 0

Subtract 1 from both sides

2x = -1

Divide both sides by 2

x = -1/2

.

9a. Doesn't specify if 1-sided or 2-sided required (perhaps appendix F only has one set of values - don't know as we can't see it!).

In[e^(x-6)]=x-6

no something has to equal 5

Thanks Rosala,

But that bird looked like a pidgeon LOL

I have not seen domain used like this. Is this acceptable usage.  I mean in a formal sense?

Start by subtracting 2/15 from both sides

Younger children use degrees.

Older ones start using radians.  (They will always use degrees for some problems)

Radians are necessary for calculus.

i am Australian.

To me a swag is is what a hobo carries or what a hiker carries on his back with everything that he owns/needs  in it

you could graph the y intercept and then use the gradient to find another point.

Thank you anonymous.

This is the BEST way to re-ask if you question has slipped back a long way.   

If your question is still on the first page please just be patient.  

3x-2>2x-5

subtract 2x from both sides

x-2>-5

add 2 to both sides

x>-3

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

2+x>2x+3

subtract x from both sides

2>x+3

subtract 3 from both sides

-1>x

swap sides but remember that if you swap sides you have to turn the inequality around

x<-1

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

Now if you want the union of these.  That is where either one or the other is true that would be all real x

If you want the intersection of these.  Where they are both true then the answer is     -3 < x < -1      

Thanks chris - that is really neat :)

they race 10 lots of 0.125 miles

10 * 0.125 miles

move the decimal poit one place (because there is one 0 in 10) to make it a bigger number.

or

do it on the calc on the home page. 

$$\\LCM \;of\;\;2\times3^2\times5^4\times7 \;\;and \;\;750\\\\
Well\; 750=3*250=3*2*5*5*5=2\times 3 \times 5^3$$

Lets look at these two

$$\\2 \times 3 \times 5^3\qquad \times \qquad 3\times 5\times 7\quad=\quad 2\times 3^2 \times 5^4\times 7\\\\
750\qquad \qquad \quad \times \qquad 3\times 5\times 7\quad=\quad 2\times 3^2 \times 5^4\times 7\\\\$$

$$\\So\;\;750 $ is a factor of the $ 2\times 3^2 \times 5^4\times 7\\\\
$ So the lowest common multiple of these two is 2\times 3^2 \times 5^4\times 7$$

Now I've introduced a mistake.

The last two times I wrote 4x it should read "2x", so correcting that now:

area = 18x

        = 2x · height

⇔ 18x = 2x · height

∴ height = ....................