All Answers 358117 Answers

2.08333333333333333

Let h be the cost of a highlighter.  We are told that:

6*h + 6*3.48 = 23.06

Subtract 6*3.48 from both sides:

6*h = 23.06 - 6*3.48

Divide both sides by 6:

h = (23.06 - 6*3.48)/6

$${\mathtt{h}} = {\frac{\left({\mathtt{23.06}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{3.48}}\right)}{{\mathtt{6}}}} \Rightarrow {\mathtt{h}} = {\mathtt{0.363\: \!333\: \!333\: \!333\: \!333\: \!3}}$$ dollars

or, multiplying by 100 to get cents:

h = 36 cents to the nearest cent.

thanku melody for the thumbs up ! 

You are welcome Rosala.

thank u melody for this clear explanation ! it was getting really hard for me to understand but now its no more ! i would give u thumbs up for this ! thank u !

Thank you Heureka   

Hi Melody,

here is the code:

\documentclass[7pt,a4paper,landscape]{article}
\usepackage[utf8]{inputenc}
\usepackage{pgfplots}
\usepackage{qtree}

\begin{document}
\Tree [
[
[
[[ F\\F\\F\\F ].($\frac{1}{2}$)\\F [ F\\F\\F\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\F [[ F\\F\\B\\F ].($\frac{1}{2}$)\\F [ \fbox{  \textcolor[rgb]{1,0,0}{FFBB}} ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\F
[
[[ F\\B\\F\\F ].($\frac{1}{2}$)\\F [ \fbox{ \textcolor[rgb]{1,0,0}{FBFB}} ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\F [[ \fbox{ \textcolor[rgb]{1,0,0}{FBBF}} ].($\frac{1}{2}$)\\F [ F\\B\\B\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\F
[
[
[[ B\\F\\F\\F ].($\frac{1}{2}$)\\F [  \fbox{ \textcolor[rgb]{1,0,0}{BFFB}} ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\F [[ \fbox{ \textcolor[rgb]{1,0,0}{BFBF}} ].($\frac{1}{2}$)\\F [ B\\F\\B\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\F
[
[[\fbox{ \textcolor[rgb]{1,0,0}{BBFF}} ].($\frac{1}{2}$)\\F [ B\\B\\F\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\F [[ B\\B\\B\\F ].($\frac{1}{2}$)\\F [ B\\B\\B\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\B
].Start\\(F=Front\\B=Back)
\end{document}

heureka

There's still something wrong with it...

It doesn't hold except for some t

deleted deleted

two times =...

In addition to that, please check whether you wrote down your equations correctly.

It can be very frustrating to try to find a proof and find out that it doesn't work.

It usually makes us think we made a mistake while after all it might be your fault.

I like!!!   More detail needed.  Umm

I LIKE YOUR CARTOON !!!!

Okay, let's start by multiplying the right hand side by $$\frac{1+cos(\theta)}{1+cos(\theta)}$$

We now have

$$\frac{sin(\theta)+tan(\theta)}{1+cos(\theta)} = \frac{tan(\theta) (1+cos(\theta))}{1+cos(\theta)}$$

Work out the numerator on the right hand side

$$\frac{sin(\theta)+tan(\theta)}{1+cos(\theta)} = \frac{tan(\theta) cos(\theta)+tan(\theta)}{1+cos(\theta)}$$

Use $$tan(\theta) = \frac{sin(\theta)}{cos(\theta)}$$

$$\frac{sin(\theta)+tan(\theta)}{1+cos(\theta)} = \frac{\frac{sin(\theta)}{cos(\theta)} cos(\theta)+tan(\theta)}{1+cos(\theta)}$$

and now we have

$$\frac{sin(\theta)+tan(\theta)}{1+cos(\theta)} = \frac{sin(\theta)+tan(\theta)}{1+cos(\theta)}$$

Reinout 

Hi Rosala, you called what was done  'cross multiply'.

I know this is a common term but I don't like it and I noticed that CPhill did not use it either.

Equations must be 'balanced' just like old fashioned scales had to be balanced.

If it is balanced - that means the 2 sides are equal - then if you do something to one side you must do exactly the same thing to the other side as well.  Otherwise the 2 sides will not be balanced anymore!

Now let us look at this equation

$$9=\frac{1}{8}\times x$$

I need to get the $$x$$  on its own so I need to get rid of the 8.

 I can do this by multiplying the RHS by 8 BUT if I multiply the right side by 8 I must also multiply the left side by 8.

$$\begin{array}{rll}
8\times 9&=&8 \times\frac{1}{8}\times x\\\\
8\times 9&=& \frac{8}{1}\times\frac{1}{8}\times x\\\\
\end{array}$$

The 8 on the top cancels with the 8 on the bottom leaving us with

$$\begin{array}{rll}
72&=&1x\\\\
x&=&72
\end{array}$$

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

Or ALTERNATIVELY

$$\begin{array}{rll}
9&=&\frac{1}{8}\times x\\\\
9&=&\frac{x}{8}\\\\
8\times9&=&\frac{8\times x}{8}\\\\
72&=&\frac{x}{1}\\\\
x&=&72
\end{array}$$

I've put in a couple more steps than i normally would just so that you can see what is happening more clearly.

I ran a quick check on it and

$$\frac{1+cot^3(t)}{1-cos(t)} \neq csc^2(t)-cot(t)$$

(except for some t obviously)

Reinout

there is 3.28 feet to 1m, therfore divide 120 by 3.28. Therefore 120 feet is 36.59m 

1 feet = 0.3048

120 feet = 120*0.3048 = 36.576 meters

I think this should be written as (cot²B - cos²B)/(csc²B - 1) 

$$$$\frac{\cot^2B-\cos^2B}{\csc^2B-1}=\frac{\frac{\cos^2B}{\sin^2B}-\cos^2B}{\frac{1}{sin^2B}-1}$$\\
$$=\frac{\cos^2B-\sin^2B\cos^2B}{1-\sin^2B}$$\\
$$=\frac{\cos^2B(1-\sin^2B)}{1-\sin^2B}$$\\
$$=\cos^2B$$

@@ End of Day Wrap - Monday 26/5/14   Sydney, Australia   Time 09:40 pm

Hi everyone,

Today our great answers were provided by CPhill, Alan, reinout-g, rosala, GoldenLeaf, TakahiroMaeda, Kitty

Andre Massow is feeling healthier and is back looking after us again.  He has dealt with the hacking problem that occured a few days ago.  The hacker only introduced superficial code via the forum.  The system's security was never compromised. The whole incident has been 'put to bed'.

Andre has also introduced auto resizing so when large pictures are uploaded they are automatically resized to fit properly.  And, he has also provided us with a button so that we can delete private messages if we wish to.

Thank you Andre. These are nice improvements to the forum.

Monday is Memorial Day in the US. It is a day set aside for remembering the men and women who died while serving in their country's armed forces.  I don't think it will be an excessively busy day on the forum but I could always be wrong.

Here are some posts that might have some interest.

Truncated Cube

I'm crossing my fingers for you!

ok melody ill surely do that ! thanks !