tertre

As you can see, just distribute as these range by increasing power. The answer should be -15. \(144x^7+108x^6+114x^5+84x^4\boxed{-15}x^3-20x^2-43x-20\)

We have  2* pi *5=10pi cm in 12 hours, so in 24 hours, that would be 20pi cm. For the minute hand, we have 2 * pi *  10= 20 pi cm. That is only for one revolution, so 20pi*24=480 pi. Thus, the answer is  \(480\pi+20]pi=500\pi\approx\boxed{1570.796}.\) So, that is 15.708 meters.

I might be wrong, but I think the guest is trying to ask the same question. 

Is this question even complete? How many ways are there to divide a red bracelet, a yellow bracelet, a green bracelet, and a black bracelet among  4 people if each person can receive any number of bracelets, including ? Each bracelet must be given to someone. Including what? 

A quick search gives web2.0calc link to the same question. Look at part (b). The full question is, "How many ways are there to divide a red bracelet, a yellow bracelet, a green bracelet, and a black bracelet among 4 people if each person can receive any number of bracelets, including 0? Each bracelet must be given to someone." 

CPhill correctly answered the question by using distinguishability, where it is \(n^k\) with duplicates and  \(k^n\) without duplicates. This does require duplicates, CPhill is correct.

Thank you, CPhill! I just learned this a few days ago... 

We can also use our handy formula: \(\frac{n}{1-r}.\) Since the series has a first term of \(\frac{3}{2}\) and a common ratio of \(\frac{-4}{9}\), we attain \(\cfrac{\frac{3}{2}}{1-\left(\frac{-4}{9}\right)}=\boxed{\frac{27}{26}}.\)

Looks like it was posted here: https://www.reddit.com/r/cheatatmathhomework/comments/weka3/6xyz30xy21xz2yz105x10y7z_812_find_x_y_z/

Hint: Try to set up variables, solution will come later. 

Does this say \(\frac{y+z}{z}\) . If so, try to break it up into \(\frac{y}{z}+\frac{z}{z}\)\(\) , so   \(\frac{y}{z}+1.\)\(\) Therefore, we have    \(4y=3z, y=\frac{3}{4}z\) and \(\frac{3z}{4}*\frac{1}{z}=\frac{3}{4}\)\(\) , so     \(1+\frac{3}{4}=\boxed{1\frac{3}{4}}.\)

This is just a geometric series, so the answer is:

\(\begin{align*} \frac12+\frac1{2^2}+\frac1{2^3}+\frac1{2^4}+\frac1{2^5} &= \frac{\frac12\left(1-\left(\frac12\right)^5\right)}{1-\frac12}\\ &=1-\left(\frac12\right)^5\\ &=1-\frac1{32}\\ &=\boxed{\frac{31}{32}}. \end{align*}\)