Melody

In question 1 you forgot 8 strikes on the hour.

Other than that it looks good :)

Note: I am really just playing here.  I don't pretend to really know what I am doing.

Mmm

\(z^7=-1\)

One solution for z is  is -1

\(\theta=\pi+\frac{2\pi n}{7}\qquad 0\le n\le6 \qquad n\in Z\\ z=e^{i\theta}=cos\theta +isin\theta\\ z-1=-(1-cos\theta)+isin\theta\\ |z-1|=\sqrt{(1-cos\theta)^2+sin^2\theta}\\ |z-1|=\sqrt{(1+cos^2\theta-2cos\theta)+sin^2\theta}\\ |z-1|=\sqrt{2-2cos\theta}\\ |z-1|=\sqrt2\sqrt{1-cos\theta}\\ \frac{1}{|z-1|^2}=\frac{1}{2(1-cos\theta)}\\ \)

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Now

\(Cos\theta = cos(\pi+\frac{2\pi n}{7}\\ Cos\theta = -cos(\frac{2\pi n}{7}\\ so\\ |1-z|^2=2(1+cos(\frac{2\pi n}{7}))\\=2(1+cos(\frac{2\pi *0}{7})),\;\;2(1+cos(\frac{2\pi *1}{7})),\;\; 2(1+cos(\frac{2\pi*2 }{7})),\;\; 2(1+cos(\frac{2\pi *3}{7})),\;\;2(1+cos(\frac{2\pi *4}{7})),\;\;2(1+cos(\frac{2\pi *5}{7})),\;\;2(1+cos(\frac{2\pi *6}{7})),\;\;\\~\\ =2(1+cos(0)),\;\;2(1+cos(\frac{2\pi }{7})),\;\;2(1+cos(\frac{4\pi }{7})),\;\;2(1+cos(\frac{6\pi }{7})),\;\;2(1+cos(\frac{8\pi }{7})),\;\;2(1+cos(\frac{10\pi }{7})),\;\;2(1+cos(\frac{12\pi }{7})),\;\;\\ =4,\;\;2(1+cos(\frac{2\pi }{7})),\;\;2(1+cos(\frac{4\pi }{7})),\;\;2(1+cos(\frac{6\pi }{7})),\;\;2(1+cos(\frac{8\pi }{7})),\;\;2(1+cos(\frac{10\pi }{7})),\;\;2(1+cos(\frac{12\pi }{7})),\;\;\\\)

Put each of these over 1 and add them together and I'd have my answer 

The first one is 1/4 so that is right

But I do not think the rest  is right   

Can you edit it and write it properly please.

There is one word missing (in the brackets)

I think it is meant to be  8+2=10   ?

You can answer some questions for me:

1) If there was no extra chimes on the hour, how many chimes are there for every hour?

2) How many chimes in 24 hours (without the extra ones) ?

3) If there is one extra at 1 oclock and 2 extra at 2 oclock ...... up to 12 extra at 12oclok. How many extra is that in 12 hours?

4) How many extra is that in 24 hours.?

5) So how many times did the chime strick in 24 hours?

Please no ne except Evan answer    (unless you see an error in my logic)

Evan, please answer the questions one at a time so I can follow your logic.

1. The normal monthly rainfalll at the Seattle-Tacoma Airport can be approximated by the model R = 3.121 +2.399sin(0.524t+1.377), where R is measured in inches and t is the time in months. Use the model to approximate the normal yearly rainfall. 

Here is the approximate annual rainfall graph

the horizontal axis is Time in months and the vertical axis is Rainfall in inches.

The approximate annual rainfall is the area between the horizonal axis and the curve from t=0 to t=12

i.e.

\(yearly\;\;rainfall\\\approx \displaystyle\int_0^{12}\; 3.121+2.399sin(0.524t+1.377)\,dt\\ =\left[\;\; 3.121t-2.399*0.524cos(0.524t+1.377) \;\; \right]_0^{12}\\ = [3.121*12-2.399*0.524cos(0.524*12+1.377) ] - [3.121*0-2.399*0.524cos(0.524*0+1.377) ] \\ = [37.452-1.257076cos(4.425) ] + [1.257076cos(1.377) ] \\\)

(37.452-1.257076cos(4.425) ) + (1.257076cos(1.377) ) = 38.050411914390589152

So annual rainfall is approximately  38.05 inches.     (note that this answer has been edited, there was a small error before)

You STILL need to check my working though   

www.wolframalpha.com

is a good site for you to check these (and many other questions) for yourself.

I already added the first one for you :)

I have just been counting. 

I got

(6+5+4+3+2+1)+(10+8+6+4+2)+(12+9+6+3)+(12+8+4)+(10+5)+6 = 126

asdf just told you that he thinks both answers are wrong.  :/

It happens   

The numbers x^3 and y^3 are inversely proportional. 

This is what that means    

\(y^3=\frac{k}{x^3}\\ \)

Now sub in x = 2 when y = 15 and solve for k

Once you have k you can plug in x=1 and find the corresponding value of y

Now you can answer yourself  :)

Please no one answer over me.