It looks like I am just doing your homework for you.

http://www.mathisfunforum.com/viewtopic.php?id=14654

I have given you the first couple but in order for you to learn you must try to do them yourself what have you done?

You obviously are in a pretty advanced course where they are throwing generating functions at you what have you learned? What have you tried?

SO we are solving for 't' here. Reordering:

-15.6 + 3.9t = 0

Add 15.6 to both sides:

15.6 + (-15.6) + 3.9t = 0 + 15.6

3.9t = 15.6

Divide each side by 3.9 for a final answer of:

t = 4

Answer: 'D'

]]>`Expand[(1 + x + x^2/2 + x^3/6 + x^4/24 + x^5/120) (1 + x + x^2/2 + x^3/6 + x^4/24) (1 + x + x^2/2 + x^3/6)]`

]]>Welcome to the forum.

So for part a, you did something like 80 - 10% of 80.

For part b: this could mean get 5% and take it off twice. You'll get the same answer as part a.

So the questioner probably means:

First step answer = 80 - 5% of 80.

Then (first step answer) - 5% of (first step answer).

This won't reduce the price by so much as the second step is working out 5% of a smaller amount.

Hope that helps.

Bob

]]>To use it, you need the math tags:

`[math][/math]`

You can click any formula to see how it was written. Also, there are a lot of them here

]]>Poor redheadcaliforniagirl. This may be more than she wanted.

I'll give it some thought, but you may have to wait for her input first.

Bob

]]>Welcome to the forum.

Do I detect questions from Compuhigh ?

In Q7 I think they are asking for the sin(90-x), in which case B 0.6 looks good to me.

And Q8, cos(90-x) = 0.8 is right too.

For Q9, tan(90-x) you need the opposite to that angle and then divide that by the adjacent.

(bobbym. they cannot want tan(x) because none of the answers will give that. I'm guessing that was in questions 1-6. After all, we wouldn't want them to go to all the expense of making a fresh diagram.

Q10 sec(90-x) = 1 over the cos of 90 -x so go back to Q8 and find 1/(that answer)

I suggest you post back answers for those two and then we can look at the rest.

Bob

]]>The formula you need is here:

http://www.mathisfunforum.com/viewtopic.php?id=17799

post #7

That makes angle ACD = angle ABC

[note: The wording doesn't say which way round to put D and E. I've put D closer to A and E closer to B, on my diagram.)

Bob

]]>f(t) =

e^(-t) when 0 ≤ t ≤ 5

-1 when t > 5

I got my answer as simply 1/(s+1) when 0 ≤ s ≤ 5

-1/s when s > 5

I'm not sure if this is right though.

]]>Welcome to the forum.

If your first line is correct (I'm doubtful about that > ) then you haven't used de Morgan correctly.

This is the analysis of what you have posted.

Please check what you have posted and confirm it is correct.

Bob

]]>Soreness is okay, but maybe you should do a little less to recover quicker. Try to train the whole body and not just a few parts, that way you do less reps for a particular part but more reps overall because you are working more muscle groups.

]]>How are y'all doin?

I found a problem on complex numbers and i can't solve it. So can anyone help me pls?

Here's the problem:

You have a finite set of complex numbers z_1, ..., z_n, for which their polar forms z_j = r_j exp(i t_j) satisfy the constraint 0 <= |t_j| < phi < PI/2.

Show that cos(phi)|z_1 * z_2 * ... * z_n|^(1/n) <= 1/n * |z_1 + z_2 + ... + z_n|.

Thanks for your help in advance

Ps: i've tried to replace |z_1 * z_2 * ... * z_n|= |z_1| * |z_2| * ... * |z_n| and then replace them in their polar form.

And i found atgm rule but im still stuck on this pb