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\(\frac{2 \sqrt{z} - 3}{\sqrt{z} - 1} + 1 = \frac{3 \sqrt{z} - 1}{1 - \sqrt{z}}\\ \frac{2 \sqrt{z} - 3}{\sqrt{z} - 1} +\frac{ 3 \sqrt{z} -1}{\sqrt{z}-1} = -1\\ \frac{5\sqrt{z}-4}{\sqrt{z}-1}=-1\\ 5\sqrt{z}-4=1-\sqrt{z}\\ 6\sqrt{z}=5\\ \sqrt{z}=\frac{5}{6}\\ \boxed{z=\frac{25}{36}}\)

L'Hopital's rule tells us that if the numerator function (let's call that f(x)) and the denominator function (let's call that g(x)) both approach 0 as x approaches n, where n is a constant, and if \(\lim_{x\rightarrow n} (\frac{f'(x)}{g'(x)})\) exists, then \(\lim_{x\rightarrow n} (\frac{f'(x)}{g'(x)})\).

So basically keep taking the derivative of the top and bottom of the fraction until you reach a point where it's not indeterminate or it's undefined (like 4/0).

In this case, you need to use the power rule:

\(\lim_{x\rightarrow 1}\frac{x^3-3x^2+3x-1}{x-1}\\=\lim_{x\rightarrow 1}\frac{3x^2-6x+3}{1}\\ 3(1)^2-6(1)+3=0\)

The probability that he gets 1 digit is 9/12=3/4, and the probability that he gets 2 digits is 3/12 = 1/4, so the expected number of digit he obtains is:

3/4*1+1/4*2=1.25

The mystery number is 5/4, and there is no other number satisfying that condition, because:

The denominator must be either 1, 2, 4, 5, or 8, in simplest terms, so that the decimals don't repeat forever, but 1, 2, and 5 cannot work as the denominator, because if you test numbers for the numerator they don't satisfy the condition 1.2

The equation \(x^2+y^2+z^2=1\) represents a sphere with a center at (0, 0, 0) and a radius of 1, and the volume of the space that contains all the possible points of \((x, y, z)\) is a cube with a side length of 2. The volume of the sphere is \(\frac{4}{3}\pi\), and the volume of the cube is \(8\), so the probability that \(x^2+y^2+z^2\leq1\) is \(\frac{\frac{4}{3}\pi}{8}=\boxed{\frac{\pi}{6}}\)

The measure of an arc is always 2 times the angle of the corresponding inscribed angle, so:

\(2(7x-7)=9x+21\\14x-14=9x+21\\5x=35\\x=7\)

Substitute in to get:

\(9(7)+21=\boxed{84}\)

\(\)

The area of a trapezoid is the average of the 2 bases times the height. In other words,

\(A=\frac{17+39}{2}\cdot16=\boxed{448}\)

let x be the number of pages in the book:

\(\frac{20}{100}x=52\\\frac{1}{5}x=52\\x=5\cdot52=\boxed{260}\)

\(12=3\cdot2^2\)

So we need to find the largest value of \(2^{2n}\)\(\)that divides 20!*40!, which is

 \(\frac{\lfloor\frac{20}{2}\rfloor+\lfloor\frac{20}{4}\rfloor+\lfloor\frac{20}{8}\rfloor+\lfloor\frac{20}{16}\rfloor+\lfloor\frac{40}{2}\rfloor+\lfloor\frac{40}{4}\rfloor+\lfloor\frac{40}{8}\rfloor+\lfloor\frac{40}{16}\rfloor+\lfloor\frac{40}{32}\rfloor}{2}\\ =\frac{10+5+2+1+20+10+5+2+1}{2}\\=28\)

And now the largest value of \(3^n\)which divides 20!*40!, which is

\(\lfloor\frac{20}{3}\rfloor+\lfloor\frac{20}{9}\rfloor+\lfloor\frac{40}{3}\rfloor+\lfloor\frac{40}{9}\rfloor+\lfloor\frac{40}{27}\rfloor\\ =6+2+13+4+1\\=26\)

Since 28>26, the largest integer n for which 12^n evenly divides 20!*40! is \(\boxed{26}\)

First of all, \(559+1=560\). 

The prime factorization of 560 is \(2^4\cdot5\cdot7\), which means that N must be at least 7 digits. 

If we test \(\frac{7!}{560}\), we get 9, which is not the product of factorials of integers, so it must be more than 7 digits. 

If we test \(\frac{8!}{560}\), we get \(2!\cdot3!\cdot3!\), which is a product of factorials, so the smallest possible value of N is an 8 digit number.

This means that we will get a number in the form of aaabbbcc, where a, b, and c are digits.

Since 0 cannot be a digit, the smallest possible value is \(\boxed{11122233}\)