All Answers 358150 Answers

Okay, so you can find the gravitational potential energy using ΔUg = mgh

where m is mass, g is gravitational force (usually threated as the constant 9.8N/kg ) and h is height

The marble gains all it's potential energy from the spring, therefore ΔUg = -ΔUs

 (Technically the spring transfers it's potential energy into kinetic energy after which it becomes potential energy) 

The formula for elastic potential energy is given by 

$$Pe = -\Delta Us = \frac{1}{2}kx^2$$

where k is the spring constant and x is the amount of compression

 edit: I think you can do the last one by yourself if you've finished these

you divide 800 by 117 and then take the whole number part.

That and struggling in general, not knowing where to begin to pick up the slack also.

Thanks, i think one reason im strugling with understanding is not interpruting the missing information in short hand algerbaic expressions, and symbolic uses. V = velocity mv = mass (times) velocity. etc. with so many I've not had time to absorb the methodology because it becoomes jiberish.

Just some help, so I can deduce the answer. I've no real clue tbh.

Also.. A block of mass m = 4.0 kg is dropped from height h = 32 cm onto a spring of spring constant k = 2460 N/m (see the figure). Find the maximum distance the spring is compressed.

Thank you Heureka

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{x}}\right)} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{180}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{0.25}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{x}}\right)}}{{\mathtt{6.76}}}}\right)}\right)} \Rightarrow \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{2\pi}}}{{cos}}{\left({\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{x}}}{{\mathtt{180}}}}\right)} = {\mathtt{\,-\,}}{\frac{{\mathtt{25}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{2\pi}}}{{cos}}{\left({\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{x}}}{{\mathtt{180}}}}\right)}}{{\mathtt{676}}}}$$

Umm that doesn't look very helpful.

Okay lets try something else

You can see it when I write $$a^2 = a*a$$ and $$a^3 = a*a*a$$

$$a*2a^2*3a^3 = a * 2 * a * a * 3 * a * a * a = 2*3 * a * a * a * a * a * a = 6a^6$$

It is common for the letter i to be used for the complex part.

This means $$i^2 = -1$$

If you enter i in the calculator, it will act like the complex variable.

Thank you, that was a much nicer way to ask.  

$$x-15=2x+11\\
-15=x+11\\
-15-11=x\\
-26=x\\
x=-26$$

Hi Heureka,

I pasted your code (in the above post) into Texmaker at home and it will not run.  I want to play with it.  I thought it may need the $ signs around it but that didn't help.  Have you got any ideas on how i can get it to run.  I am not very good at deciphering the error messages.  I can past in the errors if it would help.

any ideas?

Thank you.

$$\\\boxed{(a\bmod b)=a-b\lfloor\frac{a}{b}\rfloor}\\
\\
Example: \quad 299\bmod 12=299-12*\lfloor\frac{299}{12}\rfloor\\
= 299-12*24=299-288=11$$

latex code:

\\\boxed{(a\bmod b)=a-b\lfloor\frac{a}{b}\rfloor}\\
\\
Example: \quad 299\bmod 12=299-12*\lfloor\frac{299}{12}\rfloor\\
= 299-12*24=299-288=11

$$\lfloor \dots \rfloor = floor function$$

latex code:

\lfloor \dots \rfloor = floor function

This is posted on the Latex thread in the sticky notes.

$$\begin{array}{rlllr}
( x^3&+4x^2&+x& -6)&:(x-1)=\textcolor[rgb]{1,0,0}{x^2}\textcolor[rgb]{0,0,1}{+5x}\textcolor[rgb]{0,1,0}{+6}\\
\textcolor[rgb]{1,0,0}{{\underline{-(x^3}}&\textcolor[rgb]{1,0,0}{\underline{-x^2)}}}&&&\\
0&+5x^2&+x\\
&\textcolor[rgb]{0,0,1}{\underline{-(5x^2}}&\textcolor[rgb]{0,0,1}{\underline{-5x)}}\\
&0&+6x&-6\\
&&\textcolor[rgb]{0,1,0}{\underline{-(6x}}&\textcolor[rgb]{0,1,0}{\underline{-6)}}\\
&&0&+0
\end{array}$$

division in latex code:

\begin{array}{rlllr}

( x^3&+4x^2&+x& -6)&:(x-1)=\textcolor[rgb]{1,0,0}{x^2}\textcolor[rgb]{0,0,1}{+5x}\textcolor[rgb]{0,1,0}{+6}\\

\textcolor[rgb]{1,0,0}{{\underline{-(x^3}}&\textcolor[rgb]{1,0,0}{\underline{-x^2)}}}&&&\\

0&+5x^2&+x\\

&\textcolor[rgb]{0,0,1}{\underline{-(5x^2}}&\textcolor[rgb]{0,0,1}{\underline{-5x)}}\\

&0&+6x&-6\\

&&\textcolor[rgb]{0,1,0}{\underline{-(6x}}&\textcolor[rgb]{0,1,0}{\underline{-6)}}\\

&&0&+0

\end{array}

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

Many Greetings

Heureka (Eureka)

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

This is another possibility for long division

Heureko's LaTex

I'm sorry *please can you help me out the 12 was meant to be a 2

Please Heureka,

When you post your fancy LaTex can you please post the code as well.

There is a LaTex thread in the sticky notes and I will transfer your output and code into posts within it. (Unless you want to)

I am very excited about having a LaTex expert on the forum.  Thank you so much for joining us!

Thank you in advance. 

You know, a please would go down really well!

$$x-15=12x+11\\
-15=11x+11\\
-15-11=11x\\
-26=11x\\
11x=-26\\
x=-26/11\\
x=-2\frac{4}{11}$$

@@ End of Day Wrap - Thursday 15/5/14   Sydney, Australia Time 20:55

Hi all,

Another good day.  Lots of great answers from CPhill, Alan, zegroes and Kitty<3, reinout-g and Heureka. Thank you.

**Reinout-g has posted some more puzzles - He challenges you to solve them!     Thanks reinout.

$$\\\boxed{(a\bmod b)=a-b\lfloor\frac{a}{b}\rfloor}\\
\\
Example: \quad 299\bmod 12=299-12*\lfloor\frac{299}{12}\rfloor\\
= 299-12*24=299-288=11$$

latex code:

\\\boxed{(a\bmod b)=a-b\lfloor\frac{a}{b}\rfloor}\\
\\
Example: \quad 299\bmod 12=299-12*\lfloor\frac{299}{12}\rfloor\\
= 299-12*24=299-288=11

$$\lfloor \dots \rfloor = floor function$$

latex code:

\lfloor \dots \rfloor = floor function

Depends... what would he have had for lunch? 

Okay let me rephrase this to

$$A = P(1+r)^n$$

where A = 9750, r = 0.05 and n = 7

Then we can rewrite this to $$P = \frac{A}{(1+r)^n}$$

filling in the variables gives $${\mathtt{P}} = {\frac{{\mathtt{9\,750}}}{\left({{\mathtt{1.05}}}^{{\mathtt{7}}}\right)}} \Rightarrow {\mathtt{P}} = {\mathtt{6\,929.142\: \!968\: \!768\: \!685\: \!314\: \!7}}$$

1.) See "Vieta"

2.) You find $$x_1=1$$ do:

$$\begin{array}{
rlllr
} (
x^3
&
+4x^2
&
+x
&
-6)
&
:
(x-1)
=\textcolor[rgb]{1,0,0}{x^2}\textcolor[rgb]{0,0,1}{+5x}\textcolor[rgb]{0,1,0}{+6}
\\
\textcolor[rgb]{1,0,0}{{\underline{-(
x^3}}
&
\textcolor[rgb]{1,0,0}{\underline{-x^2)}}}
&
&
&
\\
0
&
+5x^2
&
+x
\\
&
\textcolor[rgb]{0,0,1}{\underline{-(5x^2}}
&
\textcolor[rgb]{0,0,1}{\underline{-5x)}}
\\
&
0
&
+6x
&
-6
\\
&&
\textcolor[rgb]{0,1,0}{\underline{-(6x}}
&
\textcolor[rgb]{0,1,0}{\underline{-6)}}
\\
&&
0
&
+0
\end{array}$$

\begin{array}{
rlllr
} (
x^3
&
+4x^2
&
+x
&
-6)
&
:
(x-1)
=\textcolor[rgb]{1,0,0}{x^2}\textcolor[rgb]{0,0,1}{+5x}\textcolor[rgb]{0,1,0}{+6}
\\
\textcolor[rgb]{1,0,0}{{\underline{-(
x^3}}
&
\textcolor[rgb]{1,0,0}{\underline{-x^2)}}}
&
&
&
\\
0
&
+5x^2
&
+x
\\
&
\textcolor[rgb]{0,0,1}{\underline{-(5x^2}}
&
\textcolor[rgb]{0,0,1}{\underline{-5x)}}
\\
&
0
&
+6x
&
-6
\\
&&
\textcolor[rgb]{0,1,0}{\underline{-(6x}}
&
\textcolor[rgb]{0,1,0}{\underline{-6)}}
\\
&&
0
&
+0
\end{array}

so we have: $$x^3+4x^2+x-6=(x-1)(x^2+5x+6)$$

x^3+4x^2+x-6=(x-1)(x^2+5x+6)

We solve:

$$\\x^2+5x+6=0\\
x^2+5x+(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2-(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2+6=0\\
\left( x+\frac{5}{2}\right)^2=\frac{25}{4}-6\\
\left( x+\frac{5}{2}\right)^2=\frac{1}{4}\qquad| \quad \pm\sqrt{}\\
x+\frac{5}{2}=\pm\sqrt{\frac{1}{4}}\\
x_{2,3}=-\frac{5}{2}\pm\frac{1}{2}\\
x_2=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=\underline{-2}\\
x_3=-\frac{5}{2}-\frac{1}{2}=-\frac{6}{2}=\underline{-3}\\$$

\\x^2+5x+6=0\\
x^2+5x+(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2-(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2+6=0\\
\left( x+\frac{5}{2}\right)^2=\frac{25}{4}-6\\
\left( x+\frac{5}{2}\right)^2=\frac{1}{4}\qquad| \quad \pm\sqrt{}\\
x+\frac{5}{2}=\pm\sqrt{\frac{1}{4}}\\
x_{2,3}=-\frac{5}{2}\pm\frac{1}{2}\\
x_2=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=\underline{-2}\\
x_3=-\frac{5}{2}-\frac{1}{2}=-\frac{6}{2}=\underline{-3}\\

$$\\x_1=1 \qquad x_2=-2 \qquad x_3=-3\\
\boxed{x^3+4x^2+x-6=(x-1)(x+2)(x+3)}$$

\\x_1=1 \qquad x_2=-2 \qquad x_3=-3\\
\boxed{x^3+4x^2+x-6=(x-1)(x+2)(x+3)}

I think you mean $$\frac{(4x)^3}{(-4x)^3}$$ instead of $$\frac{(4x)^3}{(-4x)3 }$$

This gives;

$$\frac{(4x)^3}{(-1*4x)^3 } = \frac{(4x)^3}{(-1)^3 (4x)^3 } = \frac{1}{-1} = -1$$ for $$x \neq 0$$

Hi Heureka,

Could you please post that latex here without it being in latex.  I just want to look at all the code, especially for the division bit.

I would have done it with the align function but I don't think that works in here.

I see you have used an array.  I have only used equarrays and again I don't think that they work in here

Thank you.