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The answer is 4n + 3.

1 divided by 977  = .001023541....

So the first three digits after the decimal point = .001

\(a + b = c\)

\(b - a = d\)

\(a < b\)

\(a \neq 0\)

(If a was 0, then b and d would be the same)

If a was 1, b could be any number from 3 - 8 (6 numbers)

If a was 2, b could be 3, 5, 6, 7 (4 numbers)

If a was 3, b could be 4, 5 (2 numbers)

If a was 4, b could only be 5 (1 number)

There are no other possible values for a

Therefore there are 6 + 4 + 2 + 1 = 13 possibilities.

We can rewrite this as \(b^2 + 1\). This is one of Landhau's 4 problems, all of which are unsolved to this day. Therefore, this problem cannot be solved (yet).

\(\angle{P} = \angle{TUV}\)

\(\angle{PUV} = \angle{UTR}\)

\(\angle{PQV} = \angle{UVR}\)

\(\angle{QVU} = \angle{VRT}\)

Therefore, trapezoid PUVQ is similar to trapezoid UTRV

\(\overline{TR} = \frac{2}{7}(20) = \frac{40}{7}\)

Since \(PUVQ \sim UTRV\):

\(\frac{20}{UV} = \frac{UV}{\frac{40}{7}}\)

\(\frac{800}{7} = \overline{UV}^2\)

\(\textbf{UV = }\frac{\textbf{20}\sqrt{\textbf{2}}}{\sqrt{\textbf{7}}}\)

\(\textbf{UV = }\frac{\textbf{20}\sqrt{\textbf{14}}}{{\textbf{7}}}\)

Let me know if I made any mistakes!

The answer is \(1357\)

If \(d\) is the common difference between consecutive terms of the arithmetic sequence, then we can rewrite the sequaence as \(a_1, a_1 + d, a_1 + 2d, \dots, a_1+7d, a_1 + 8d, a_1 + 9d\).

We can now werite both equations as

 \(1: a_1 + (a_1 + 2d) = 5\) and \(2: (a_1 + d) + (a_1 + 3d) = 6\) and solve for d and a.

Let's use elimination:

\(2: 2a_1 + 4d = 6\)

\(1: 2a_1 + 2d = 5\)

\(2d = 1\)

\(d = \frac{1}{2}\)

\(1: 2a_1 + 2(\frac{1}{2}) = 5\)

\(2a_1 + 1 = 5\)

\(2a_1 = 4\)

\(a_1 = 2\)

This means that the sequence is 2, 2.5, 3, 3.5, ... and that a_1 = 2

The smallest possible prime divisor a number can have is of course, 2

If we check the multiplication, we can see that 3 odd numbers can't possibly mulitply to be an even. 

So, \(5^{19} * 7^{13} * 3^{31}\)  is not divisble by 2. 

The next smallest is 3. 

The third term of the multiplication is \(3^{31 }\)

This means, no matter what, this number has to be divisible by 3, since 3 is part of the factorization, meaning it is a factor. . 

So 3 is our final answer. 

Thanks! :)

Well, let's note something about this problem really quickly. 

So, throughout the first 12 games, you have 1 win and 11 losses, which is 1/12, or about 8% win rate. 

After that, you win 1 game and lose 2 games, which is 1/3 or about 33% win rate. 

However, in order to achieve a 60% win rate, you must win more games than you lose. 

Since you ALWAYS lose more than you win, it's not possible you have a 60% win rate. 

Thanks! :)

~900 answers

Let's note that 1/17 is a repeating decimal. 

\(1/17 = 0.\overline{ 0588235294117647}\)

There are 16 repeating digits in the decimal. 

Dividing 4000 by 16, we get

\(4000/16 = 250\)

Since it is divisble, then we can conclude that the last digit of the repeating decimal is the correct answer. 

Thus, 7 is the final answer. 

Thanks! :)

You can put it like this.

The ratio of 60 + x : 100 + x is equal to 100 + x : 140 + x, 

where x is the constant added to a, b and c.

Hopefully you can find the answer after this.

We can use the slope the equation and power of a point to solve this question. We have

\([ b^2 - a^2 ] / [ b - a ] = [ (b -a) (b + a) ] / (b -a) = b + a = 2 \)

So the answer is 2. 

Thanks! :)

This problem is not as hard as it might seem!

We have

\((1 + 1.5)^2  = (2.5)^2  = 6.25\)

So 6.25 is our answer. 

Thanks! :)

See image below:

The solids both have a surface area of

 \(1*6=6\)

if you add 6 times 2, you get 12. 12-2=10