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x^11*x^7 = x^18

the answer is 18

It would be nice to get feedback though 

I answered on the other post.

Sorry, I know i asked you to put it on a new post but I then got interested and answered it on the original  :)

If anyone wants to show guest how to do it in a more traditional way.

That is fine :)

 ∫5 3 (1/(x^2-6x+13))dx

\(\displaystyle\int_3^{5}\frac{1}{\sqrt{x^2-6x+13}}dx\\ =\displaystyle\int_3^{5}\frac{1}{\sqrt{x^2-6x+9+4}}dx\\ =\displaystyle\int_3^{5}\frac{1}{\sqrt{2^2+(x-3)^2}}dx\\\)

I did this one in a very similar fashion.

I used this triangle

See if you can take it from there   

This problem leads to a negative price for tacos.  If you ever find this please let me know.

\(\dfrac{\dfrac{n!}{k!(n-k)!}}{\dfrac{n!}{(k+1)!(n-k-1)!}} = \\ \dfrac{(k+1)!(n-k-1)!}{k!(n-k)!}=\\ \dfrac{k+1}{n-k}=\dfrac{4}{11}\)

\(11k+11=4n-4k\\ 4n-15k=11\)

\(\text{We can use the Euclidean algorithm to find }n=14,~k=3\)

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I probably do these a little differently from most people.

I cannot remember the integrals so I so them more from scratch.

\(\displaystyle\int_0^{\frac{1}{6}}\frac{1}{\sqrt{1-9x^2}}dx\)

I am going to use this triangle.

\(cos\theta=\sqrt{1-9x^2}\\ sin\theta=3x\\ x=\frac{sin\theta}{3}\\ \frac{dx}{d\theta}=\frac{cos\theta}{3}\\ dx=\frac{cos\theta}{3}d\theta\)

also

\(when \;x=0  \\ sin\theta=0\\ \theta=0\\~\\ when\;x=1/6\\ sin\theta=3*(1/6)=1/2\\ \theta = \frac{\pi}{6}\)

\(\displaystyle\int_0^{\frac{1}{6}}\frac{1}{\sqrt{1-9x^2}}dx\\ =\displaystyle\int_0^{\frac{1}{6}}\frac{1}{cos\theta}dx\\ =\displaystyle\int_0^{\frac{\pi}{6}}\frac{1}{cos\theta}\frac{cos\theta}{3}d\theta\\ =\displaystyle\int_0^{\frac{\pi}{6}}\frac{1}{3}d\theta\\ =\left[\frac{\theta}{3}\right]_0^{\frac{\pi}{6}}\\ =\frac{\pi}{18} \)

∫1/6 0 (1/(sqrt(1-9x^2))dx

\(\displaystyle\int_0^{\frac{1}{6}}\frac{1}{\sqrt{1-9x^2}}dx\)

Here is a site that should help

1)
\(\omega = 7rad/s\\ r=15cm\\ v_{tan}=\omega r\\ v_{tan}= 7rad/s*(15cm)=105cm/s*(60s/1min)=6300cm/min\)
3)
\(\Delta \theta=\frac{3\pi}{4}rads\\ t=2s\\ \omega = \frac{\Delta \theta}{t}=\frac{\frac{3\pi}{4}rads}{2s}=\frac{3\pi}{8}rad/s\)

67 degrees + 25/60 =67+25/60 = 67.4166666666666667 degrees

78 degrees   .125 x 60 minutes  =    78 degres 7.5 minutes =   78 degrees 7 minutes 30 seconds

6. \(3| x - 3 | < 12\)

First, divide both sides by \(3\) to get \(| x - 3 | < 4\). Now we have that \(x - 3 < 4\) or that \( x - 3 > -4\). By solving each one, we get \(x<7\) or \(x>-1\). This gives us \((-1, 7)\).

7. \( 5| 2x + 8 | + 4 < 9\)

First, subtract \(4\) from both sides to get \(5\left|2x+8\right|<5\). Then, we can divide by \(5\) on both sides to get \(\left|2x+8\right|<1\). Now, we have \(-1<2x+8<1\). By subtracting \(8\) on both sides then dividing by \(2\), we get \(x>-\frac{9}{2}\) and \(x<-\frac{7}{2}\) so we have \((-\frac{9}{2},-\frac{7}{2})\).

8. \(-2| 8x - 4 | + 1 < -15\)

Subtract \(1\) from both sides to get \(-2\left|8x-4\right|<-16\). Then, divide by \(-2\) on both sides to get \(\left|8x-4\right|>8\). Now, we can get that \(8x-4>8\) and \(8x-4<-8\). By simplifying both inequalities, we get \(\:\left(-\infty \:,\:-\frac{1}{2}\right)\cup \left(\frac{3}{2},\:\infty \:\right)\). 

- Daisy

By shifting it to the right and moving it down 1(you should know basic translations with the graph, like (x-1)^2 compared to (x)^2 shifts the parabola 1 unit to the right(or left I forgot) it should be like that) 

To reduce the fraction, we need to find how many \(115\)'s are in \(247\). There are \(2\). \(115*2=230\). So \(247-230=17\) which is the remainder. Therefore, we have \(-2 \frac{17}{115}\). (Don't forget the negative!)

- Daisy

Well you'd add 8 for part 1

But the final answer would be bigger by 8*24

Note: I have not checked all you working but your logic seems good.

58.8x-29.7

So all I need to do is add eight to my answer right? And Big thanks to you :D:D

To find the first and the last 20 digits of this very large number: 2027^2029^2039 proceed as follows:
1 - Take the log of 2029 and multiply by 2039 to get:6743.548 093 900 992 469 000............etc. 

2 - To (1) above add the log(log(2027)) to get:6744.067 845 424 099 918 773 355.............etc. 

3 - The above calculations should be carried to an accuracy of a minimum of 7000 digits in (2) above, and preferably, to 8000 digits to get the fractional part of the log accurate enough.
4 - Raise the number in (2) above to the power of 10 and again must carry the caculation to an accuracy of about 8,000 digits.
5 - Raise the fractional part in (4) above to the power of 10 and that should give you the first few hundred ACCURATE digits of the beginning of the number in (1) above, which begins with: 1,203, 717, 813, 292, 743, 410, 727.....

An alternative approach to the above is as follows:
1 - Take the log of 2027 to an accuracy of 8,000 digits and multiply by:  2029^2039 and that should give the requisite  accuracy required for the fractional part.
2 - Take the fractional part in (1) above and raise it to the power of 10 and that should give a few hundred accurate digits of the beginning of 2027^2029^2039.  

To get the last 20 digits of the same number follow this procedure:
Take (2027^2029^2039) mod 10^20 and should get the last 20 digits:  16, 412, 296, 620, 638, 392, 187.
However, few calculators can tackle such gigantic numbers such as Wolfram/Alpha, or a programming language such as C++, which is what I used to calculate these numbers.

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

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

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

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   

@Melody Im so sorry that I forgot to include that 8+2=10 my mistake :(. Okay, I will try to answer your questions. 

 Question 1: So in 1 hour, 2+4+6=12

Question 2: So in 24 hours,  it will be (2+4+6)12=144

Question 3: That would be 1+2+3+4...+11+12=(1+12)12/2=78 extra

Question 4: I think that would be just 78*2=156 extra in 24 hours.

Question 5: 144+156=300 chimes in 24 hours?

Please Tell me my errors if I have any. THANK YOU MELODY!

Can you edit it and write it properly please.