reinout-g

$$E = mc^2 \Rightarrow m = \frac{E}{c^2} \mbox{ and } E = \frac{m}{c^2}\\
\mbox{ (Since c = 299 792 458) } m = \frac{5000000}{299792458^2} \mbox{ if e = 5000000 and } E = \frac{5000000}{299792458^2} \mbox{ if m = 5000000}$$

$${\frac{{\mathtt{5\,000\,000}}}{{{\mathtt{299\,792\,458}}}^{{\mathtt{2}}}}} = {\mathtt{0.000\: \!000\: \!000\: \!055\: \!632\: \!5}}$$

The expression is given by some fixed costs (5 dollars),

and by some costs depending on how many pages there are (2 dollars and 15 cents).

Therefore for one page we have

$$5 + 2.15$$

for two pages we have

$$5 + 2.15 + 2.15$$

and for p pages we have

$$\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }p\\
5+\overbrace{2.15+2.15+...+2.15+2.15}$$

or similarly

$$5+2.15p$$

If we evaluate this for 28 pages we have

$${\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2.15}}{\mathtt{\,\times\,}}{\mathtt{28}} = {\frac{{\mathtt{326}}}{{\mathtt{5}}}} = {\mathtt{65.2}}$$

I give you;

No, I'm kidding, but I seriously don't know what they want you to do...

Do they want you to find the values of R

and do you mean [r/7-1] = $$\frac{r}{7}-1$$

or do you mean [r/(7-1)] = $$\frac{r}{7-1}$$

I'll answer if you remove the 5 points you've given yourself 

Theorem 5.5:

Beth Likes Rosala's dress

proof:

I think your math book might have made a mistake. 2 * (-3) = -6

Because

positive*positive = positive

negative*positive = negative

positive*negative = negative

negative*negative = positive

Maybe if there's more to the question than 2*(-3) you should check the rest of the question 

Hey Chris, 

I think the person that asked this question actually wanted an explanation why this works .

Now it does somehow seem logical in my head, but I'm too tired to be able to put it on paper.

Will you have another look at it?

Otherwise I'll have a look at it tomorrow or the day after tomorrow... 

How fast do you want us to be?

Did it take that long for you to get an answer?

What question did you ask?

Reinout

Let's start with the fact that the terms sin^-1 asin and arcsin are all terms for the inverse sin (so they all mean the same)

Then if

sin(b) = c

b = asin(c) [I use asin here because on the calculator of this forum it's also asin]

Does this help?

I thought this might be the best place to say this but I don't think it's a good idea to ask Andrew to have threads which stick to the top of the forum. I think these threads will strongly discourage new members to post a question on this forum. As long as we don't have these threads, the forum still has a 'from people for people' feel. It doesn't matter what kind of question you ask since the people on the other side also feel like people. If you add these kind of 'rules for posting' threads it will feel like the person asking the question is asking it to mathematicians with little time that don't want to be bothered with 'stupid' questions. In short, I think it will strongly discourage the playfulness of the forum, which is basically the main strength of the forum. 

That's my opinion about this, but maybe someone else could give his opinion about this too.

Reinout