GoldenLeaf

I'm just going to grab a the formula and answer because this question needs to be answered.

2, 7, 8, 3, 12, 9 ... 
The nth term of this sequence is: 

-(17/30)n^5 + (113/12)n^4 - (173/3)n^3 + (1915/12)n^2 - (5813/30)n + 85 
Therefore, term number 7 is: -146 / 1 = -146

Also, no luck with a book of sequences

Well, I just woke up and I have a hands-on test in Biology in a little over an hour, so right now I'm kinda out of it.

First off:

$${\frac{{\mathtt{45}}}{{\mathtt{360}}}} = {\frac{{\mathtt{1}}}{{\mathtt{8}}}}$$

$${\frac{{\mathtt{1}}}{{\mathtt{8}}}}{\mathtt{\,\times\,}}\left({\mathtt{2}}\right) = {\frac{{\mathtt{1}}}{{\mathtt{4}}}}$$

And then it gets a bit harder from there.

$${\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\mathtt{75}} = {\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{75}}}{{\mathtt{1}}}}\right)$$

$${\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{75}}}{{\mathtt{1}}}}\right) = {\frac{{\mathtt{75}}}{{\mathtt{4}}}}$$

So, in terms of pi, the answer can be one of the two below (depending on if asked to keep in fraction form or not)

$$\left({\frac{{\mathtt{75}}}{{\mathtt{4}}}}\right){\mathtt{\,\times\,}}{\mathtt{\pi}}$$

OR

$${\mathtt{18.75}}{\mathtt{\,\times\,}}{\mathtt{\pi}}$$

If it aska you to leave it in terms of pi, then one of the two above is your answer. However, if it is asking for a number (no pi) then go on.

18.75pi = 58.9048622548086232

Rounding, that gives 58.9, which is your final answer.

First, distribute the 'a's.

$$\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{a}}\right){\mathtt{\,-\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{a}}\right)$$

What will happen now is that, in order to factor the polynomials, you must set up a 3x3 grid and combine (in a similar fashion to Punnett Squares).

(I have to go now, can somebody finish this one up?)

I'm not awesome; on the contrary, I just take time to draw everything out so that every process can be seen. Thanks for the compliment though!

In general, any number times one is the number, because you are considering that there is just one number to add to, well, no other numbers.

Thus, the answer, since you are multiplying one by one, is one.

Though I don't necessarily believe that your life depends on this, I am still glad to help, and since this is a graph, I'll just post a graph with 10 units on each side.

The red line is y = 2/3x - 3

The blue line is y = -2x + 5

The solution is (3,-1)

For this kind of probability problem, assuming the number is removed from play once chosen, the probability increases by one each time.

$$\left({\frac{{\mathtt{1}}}{{\mathtt{40}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{39}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{38}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{37}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{36}}}}\right)$$

The bottom five numbers multiply to 78960960, which means that the odds are 1/78960960

@Melody Ah, I haven't done scientific notation in a while, so that's what I thought it was  Thanks Anon for the correct answer, and thanks Melody for the refresher.

This problem is a Pythagorean Theorem problem, I assume?

• The rope is the hypotenuse (h = 26 yards)
• Length of the ground from the pole is side 'a' (a = 22 yards)

Thus, the set-up for this problem is:

$${{\mathtt{22}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = {{\mathtt{26}}}^{{\mathtt{2}}}$$

Taking these values, the next step is to square 22 and 26.

22^2 = 484

26^2 = 676

The equation after simplifying the numbers is:

$${\mathtt{484}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = {\mathtt{676}}$$

Now, it is necessary to isolate the 'b^2'.

$$\left({\mathtt{484}}{\mathtt{\,-\,}}{\mathtt{484}}\right){\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = \left({\mathtt{676}}{\mathtt{\,-\,}}{\mathtt{484}}\right)$$

$${{\mathtt{b}}}^{{\mathtt{2}}} = {\mathtt{192}}$$

Now, in order to completely isolate the 'b^2', we must take the square root of it, and in turn, 192.

$${\sqrt{{{\mathtt{b}}}^{{\mathtt{2}}}}} = {\sqrt{{\mathtt{192}}}}$$

$${\mathtt{b}} = {\sqrt{{\mathtt{192}}}}$$

The square root of 192 is 13.8564064605510183, which, when rounded to the nearest tenth, is about 13.9

@Anon: You rounded to the nearest hundredth, but you summed it up pretty well. I do have a habit of extracting it to the barebones