All Answers 358516 Answers

continuing irmjj's answer

If two pooints on the line are     $$(x_1,y_1)\;\;and\;\;(x_2,y_2)$$

$$slope=\frac{rise}{run}=\frac{y_2-y_1}{x_2-x_1}$$

great answer irmjj  

no happy - a light year is a distance not a time. :)

It will take 65 years.      Just like CPhill said.  :)

It is not 0 happy - it is just a very little number  :)

2.9979x10^8 * 6.626x10^-34

=2.9979* 6.626x10^(8 -34)

=2.9979* 6.626*10^(-26)

$${\mathtt{2.997\: \!9}}{\mathtt{\,\times\,}}{\mathtt{6.626}} = {\mathtt{19.864\: \!085\: \!4}}$$

=19.8640854 x 10^-26

=1.98640854 x 10^-25

I think that the intended question is 

(4.414x10^(-21)) / (1.60x10^(-19))

$${\frac{\left({\mathtt{4.414}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{21}}\right)}\right)}{\left({\mathtt{1.6}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{19}}\right)}\right)}} = {\mathtt{0.027\: \!587\: \!5}}$$

No times - or at least less than 1 times.

3/7 divided by 7 = 3/7 x 1/7 = 3/49

Here are a couple of sites that could be very useful to you :)

$$\\\frac{9}{8}n=-54\\\\
\frac{9}{8}n\times\frac{8}{9}=-54\times\frac{8}{9}\\\\
n=-6\times\frac{8}{1}\\\\
n=-48$$

I like using factorials.

Er - dogeminer I don't think that website helps.

Martin paddles his canoe at an average speed of 3.6 what per hour?

Anonymous:

I'm in 6th grade 2

Guess what? I'm in year 2 (I wish if I can be in year 7 because as soon as I do year 2 maths, I get bored).

Melody:

Thanks for this Pyramid:)

how much are you gambling each time?

If you get $4 does that mean that you get $4 plus the amount that you gambled back again?

I hate to nit pick but that is what mathematicians are good at       :)

People say I have strange ideas. So, I'd rather not know how the coin will land. If the chance of landing on its side is so small that I have to make a difference less than a cent, I'll round to the nearest cent. I had a think. I guess the normal price is 50c, because if you win, you'll get more money and if you lose, you won't waste much money.

Whoops. I was meant to post this in "Number Puzzle"

$${\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{1}}$$

You should know this, but...

Here are some things that might happen:

1. A thief (including your friend) might steal the apple.
2. Your friend might give it back to you.
3. Your friend might eat half of it and give it back to you.
4. Your friend might give the seeds back to you.
5. You might eat the apple you're left with.

So there's more than 1 answer.

$$\\0.5(1+e)^{-4.787+1.55x}=e^{-4.787+1.55x}\\\\
$let $t=-4.787+1.55x\\\\
0.5(1+e)^{t}=e^{t}\\\\
ln(0.5(1+e)^{t})=ln(e^{t})\\\\
ln(0.5)+ln((1+e)^{t})=ln(e^{t})\\\\
ln(0.5)+tln(1+e)=t\\\\
ln(0.5)=t-tln(1+e)\\\\
t=ln(0.5)\\\\
t=\frac{ln(0.5)}{(1-ln(1+e))}\\\\
-4.787+1.55x=\frac{ln(0.5)}{(1-ln(1+e))}\\\\
1.55x=\frac{ln(0.5)}{(1-ln(1+e))}+4.787\\\\$$

$$\\x=\left[\frac{ln(0.5)}{(1-ln(1+e))}+4.787\right]/1.55\\\\$$

Oh, I didn't realize that*. CPhill, it would only be funny to me if you posted that answer.

*I'm not the user who typed that up.

Holy Einstein, Euler, and Newton !!

Rom is back!!

Welcome back ROM!!!

$${\mathtt{8.1}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{12}}}{\mathtt{\,\times\,}}{\mathtt{5.3}}{\mathtt{\,\times\,}}{\mathtt{10}} = {\mathtt{429\,300\,000\,000\,000}}$$

$${\sqrt{{\mathtt{2}}}} = {\mathtt{1.414\: \!213\: \!562\: \!373\: \!095}}$$

$$\dfrac 1 4 (n-6) = \dfrac 1 4n - \dfrac 3 2$$

$$\dfrac n 4 - \dfrac 6 4 = \dfrac n 4 - \dfrac 3 2$$

$$\dfrac n 4 - \dfrac 3 2 = \dfrac n 4 - \dfrac 3 2$$

and as you can see this statement is a tautolgy.  It holds true for all n.

What value of y satisfies the equation y3+14=512

$$y^3=512-14\\
y^3=498$$

$$y\approx 7.926$$

$${{\mathtt{y}}}^{{\mathtt{3}}} = {\mathtt{498}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{{\mathtt{498}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{{\mathtt{498}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}{{\mathtt{2}}}}\\
{\mathtt{y}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{{\mathtt{498}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{{\mathtt{498}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}{{\mathtt{2}}}}\\
{\mathtt{y}} = {{\mathtt{498}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{\mathtt{3.963\: \!204\: \!222\: \!244\: \!851\: \!5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6.864\: \!471\: \!073\: \!704\: \!028\: \!8}}{i}\\
{\mathtt{y}} = {\mathtt{\,-\,}}\left({\mathtt{3.963\: \!204\: \!222\: \!244\: \!851\: \!5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6.864\: \!471\: \!073\: \!704\: \!028\: \!8}}{i}\right)\\
{\mathtt{y}} = {\mathtt{7.926\: \!408\: \!444\: \!489\: \!703}}\\
\end{array} \right\}$$

You have three equal signs here !!!!

I'll go for seems too

I'll go for C  :)