All Answers 358168 Answers

Check on a calculator!!!!!

Desmos is for graphing not for solving.  So there is no mystery about that :))

Yes that is true :))

What do you wish to compute???

Note....Desmos is rather funky about solving these equations....

It will not "solve"  183x^3 + 61x - 3 = 0

It will "solve"   183x^3  =  3 - 61x   if the left and right sides are entered as separate functions .....

Duh.....I should have seen right off that 3 was a "root"......over-reliance on technology is sometimes crippling......LOL!!!!!

Thanks Chris,

I often forget how clever our own site calculator is. 

Thank you Mr Massow for making it for us.  

183k^3 + 61k - 3 = 0

The onsite solver will do this, too......the answer is pretty nasty...!!!

$${\mathtt{183}}{\mathtt{\,\times\,}}{{\mathtt{k}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{61}}{\mathtt{\,\times\,}}{\mathtt{k}}{\mathtt{\,-\,}}{\mathtt{3}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{k}} = {\left({\frac{{\sqrt{{\mathtt{15\,613}}}}}{{\mathtt{3\,294}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{122}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,-\,}}{\frac{\left({\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{9}}{\mathtt{\,\times\,}}{\left({\frac{{\sqrt{{\mathtt{15\,613}}}}}{{\mathtt{3\,294}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{122}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}\\
{\mathtt{k}} = {\left({\frac{{\sqrt{{\mathtt{15\,613}}}}}{{\mathtt{3\,294}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{122}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,-\,}}{\frac{\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{9}}{\mathtt{\,\times\,}}{\left({\frac{{\sqrt{{\mathtt{15\,613}}}}}{{\mathtt{3\,294}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{122}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}\\
{\mathtt{k}} = {\left({\frac{{\sqrt{{\mathtt{15\,613}}}}}{{\mathtt{3\,294}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{122}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{\left({\mathtt{9}}{\mathtt{\,\times\,}}{\left({\frac{{\sqrt{{\mathtt{15\,613}}}}}{{\mathtt{3\,294}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{122}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{k}} = {\mathtt{\,-\,}}{\mathtt{0.024\: \!415\: \!509\: \!892\: \!177\: \!8}}{\mathtt{\,-\,}}{\mathtt{0.578\: \!896\: \!955\: \!168\: \!551\: \!4}}{i}\\
{\mathtt{k}} = {\mathtt{\,-\,}}{\mathtt{0.024\: \!415\: \!509\: \!892\: \!177\: \!8}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.578\: \!896\: \!955\: \!168\: \!551\: \!4}}{i}\\
{\mathtt{k}} = {\mathtt{0.048\: \!831\: \!019\: \!784\: \!355\: \!5}}\\
\end{array} \right\}$$

y=(-6/y)²-1

$$\\y=\left(\frac{-6}{y}\right)^2-1\\\\
y=\frac{36}{y^2}-1\\\\
y^3=36-y^2\qquad y\ne 0\\\\
y^3+y^2-36=0\\\\$$

now I am going to use reaminder theorum to look for a factor

If y=3 then  27+9-36=0  great (y-3) is a factor and 3 is a root

$$\\(y^3+y^2-36)\div (y-3)=y^2+4y+12 \qquad $I used polynomial division to get this$\\\\
so\\
y^3+y^2-36=(y-3)(y^2+4y+12)\\\\\\
$Now find roots of $y^2+4y+12\\\\
\triangle = 16-48<0\qquad $therefore there are no real roots to this$\\\\
so\\
$The only real solution is $y=3$$

check

Well...maybe we can.....

y=(-6/y)²-1         simpify

y = 36/y^2 - 1    multiply through by y^2

y^3 = 36 - y^2

And using the onsite solver, we have one real solution and two non-real solutions...

$${{\mathtt{y}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{36}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{2}}\\
{\mathtt{y}} = {{\mathtt{2}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{2}}\\
{\mathtt{y}} = {\mathtt{3}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{2.828\: \!427\: \!124\: \!746\: \!190\: \!1}}{i}\\
{\mathtt{y}} = {\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2.828\: \!427\: \!124\: \!746\: \!190\: \!1}}{i}\\
{\mathtt{y}} = {\mathtt{3}}\\
\end{array} \right\}$$

well 11*$6 = $66

so if she goes to 11 games she will 'break even'

She will need to go to 12 games to be in front  

Maybe this will help  

Yes it is a fabulous site.

this is the page that I use the most

I have spoken of many times.

It is how I know that it is now about 11:30am in your neck of the woods.

You are 17 hours behind me but I don't have to think to hard.  I can just look at the map and if I know the city someone lives in I know what time it is for them :))

@@ End of Day Wrap   Fri 13/2/15 Sydney,    Australia Time 4:10 am    (Really Sat morn)  ♪ ♫

Good morning from Sydney, 

Today our wonderful answerers were Quentin, ZamariahLovesYuh(ZLY), Heureka, Alan, BrittanyJ, CPhill, Rosala, Geno3141, RandomGuy11, Panda6702, Kitty<3, Hellokittygirl56, TayJay and GoldenLeaf.  Thanks all.

Interest posts:

Just use the onsite calculator......

That's a pretty cool site, Melody.....how did you stumble upon that one???  LOL!!!

Sat 14/2/15

We need another mathematician to arbitrate.  :)))

Yes CPhill is probably right.  Neither of us is really sure.   Thanks Chris :))

I find these confusing but if something increases 100% then it doubles so I guess if it increases 200% it triples.

so

if it originally takes 15seconds and it becomes 3 times faster i suppose it will now take 5 seconds.

what do you reckon CPhill, soes this sound right to you?

I'll give this one a shot...... if something was completeing a cycle 100% faster, the time it would take is just (1/2) of of its normal rate.  Then, this implies (1/2)^n where n is the percent increase expressed as a decimal. Thus a 100% percent increase implies 15(1/2)^(1.00)  = 7.5 seconds. So...a 200% percent increase in the cycle rate implies that the time it takes is just 15(1/2)^(2.00) = 3.75 seconds. Note that the time never decreases to 0...!!!!

Like, Melody, I find this a little confusing, but I believe this is correct.

Disregard the above......it is nonsense....see my post below......!!!!

compounding yearly it is

1(1.06)^10

$${{\mathtt{1.06}}}^{{\mathtt{10}}} = {\mathtt{1.790\: \!847\: \!696\: \!542\: \!853\: \!6}}$$

so $1 will grow to $1.79

that is a 79% increase in value  :))

The point (-1,2) lies on the terminal side of theta in standard position. What are the exact values of sin theta, sec theta, and theta

theta is the obtuse angle with the vertex at (0,0)

$$\\sin\theta = \frac{opp}{hyp}=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}\\\\
Sec\theta=\frac{hyp}{adj}=\frac{\sqrt5}{-1}=-\sqrt5\\\\
\theta=180-atan(2)$$

$${\mathtt{180}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{2}}\right)} = {\mathtt{116.565\: \!051\: \!177\: \!078}}$$  

$$\theta\approx 116.6\;\;degrees$$

No, it is NOT zero !!      It is undefined

This site has got lots of great stuff.

I love it :)

It does not have nice integer solutions so I would use Desmos or Wolfram|Alpha to solve it