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Make this into a number line like this:

The sum of the distances from P to A, B, C, D, and E, respectively, are \(|x| + |x-1| + |x-2| + |x-5| + |x-7|\). 

P will be equal to the x value that will result in the minimum possible distance. 

The global minimum will only occur in critical points, which in this case are 0, 1, 2, 5, and 7.

Testing them out, 2 will result in the smallest possible sum of the distances of 11. \(\)

That means P will be on point C, which means AP will be equal to \(\boxed{2}\)

(by the way, I'm not too good at calculating global minima, so maybe there is a faster way to do it than to just test out critical points)

#3 Close, but you're confusing diameter with radius.

The answer would be \(\pi\cdot \frac{1}{2}\cdot(3^2-1^2-2^2) = \pi \cdot \frac{1}{2} \cdot 4 = \boxed{2\pi}\)

Thanks for the kind words guys 

(also, sorry, I made a typo: I meant "Notice that triangles ABO and OCD are similar", not  "Notice that triangles ABO and ODD are similar")

WLOG, let the vertical height of ABCD be 1 (the trapezoid will then obviously be way more squashed than the image). 

Notice that triangles ABO and ODD are similar. Therefore, the ratio of the length of CD to the length of AB is \(36:25\) , so the height of triangle ABO is \(\frac{25}{36+25} = \frac{25}{61}\). 

The height of OXY is then just \(\frac{1}{2}-\frac{25}{61}=\frac{11}{122}\). Since OXY and ABO are similar, the length of xy is just \(\frac{11}{122}\cdot61 = \boxed{5.5}\)

(i am not completely sure of this answer as I submitted this very late at night and I am not extremely good at these particular problems)

Try reducing the big expression as much as possible.

First, notice that the denominator can be simplified to 3m, since m^2+1=3m:

\(2m^5-5m^4+2m^3-8m^2\over3m\)

Divide the top and bottom by m:

\(2m^4-5m^3+2m^2-8m\over3\)

notice that the m^4 term and the m^2 term can be factored like this:

\(2m^2(m^2+1) -5m^3-8m\over3\)

The value in the parenthesis can be replaced with 3m:

\(\frac{2m^2(3m)-5m^3-8m} {3} \\ =\frac{6m^3-5m^3-8m}{3}\\=\frac{m^3-8m}{3}\)

Add then subtract 9m from the numerator:

\(\frac{m^3-8m+9m-9m}{3}\\ = \frac{m^3+m-9m}{3}\\ = \frac{m(m^2+1)-9m}{3}\\ = \frac{m(3m)-9m}{3}\\ = m^2-3m\)

Rearranging the equation, \(m^2-3m=-1\), so the answer to this question is \(\boxed{-1}\)

the largest possible value of x is just \(0.\overline{98}\). To convert that into a fraction:

\(100x = 98.\overline{98}\\ x=\overline{98}\\ 99x=98\\ x=\boxed{\frac{98}{99}}\)

Since 7 is a prime number, the number of 7's in the prime factorization of n! is equal to \(\left \lfloor{\frac{n}{7}}\right \rfloor+\left \lfloor{\frac{n}{7^2}}\right \rfloor+\left \lfloor{\frac{n}{7^3}}\right \rfloor+...\) , where \(\left \lfloor{x}\right \rfloor\) is just the floor of x, which is just the smallest integer less than or equal to x. 

We want the number of 7's in the prime factorization of n! to be equal to 10, and we want it to be as small as possible, which means that n will also be a multiple of 7.

Trying out some reasonable values, we find that \(\boxed{63}\) is the smallest possible value, since \(\left \lfloor{\frac{63}{7}}\right \rfloor+\left \lfloor{\frac{63}{7^2}}\right \rfloor = 9+1 = 10\) (notice that I didn't test the floors of larger values since they will all be zero).

100! ends in a lot of zeroes, way more than 17, so the answer is just simply \(\boxed{0}\).

For a more detailed explanation, the number of zeroes that 100! will end in is equal to the number of 5's the prime factorization of 100! is, because 5*2=10, and there are obviously a lot more 2's than 5's in the prime factorial of 100!.

There are \(\left \lfloor{\frac{100}{5}}\right \rfloor+\left \lfloor{\frac{100}{25}}\right \rfloor = 20+4 = 24\) zeroes at the end of 100! which means that the last 17 digits will all be zero, which will mean that the sum will be 0.

The only way a simplified fraction can terminate is if the denominator can be expressed as \(2^x \cdot 5^y\), where x and y are integers because these are the factors of 10.

The prime factorization of 1375 is \(5^3 \cdot 11\)

Therefore, the only way to make this fraction a terminating decimal is if the numerator, n, is a multiple of 11, to cancel out the 11 in the denominator to make it terminate.

There are 90 numbers from 1 to 1000 which is a multiple of 11.

Therefore, the answer to this question is \(\boxed{90}\) numbers that satisfy this condition.

I believe you mean y = -x + 6, since this is worded exactly like an alcumus problem I saw a few days ago.

Anyway, the solution I did was:

The slope of the line from the origin to the point Q must be perpendicular to the slope of the line from (10, -10) to the point Q.

The equation for that is: \(\frac{x}{y} = - \frac{-10-y}{10-x}\)

Replace y with -x + 6 to solve for x:

\(\frac{x}{-x+6} = -\frac{-10-(-x+6)}{10-x}\\\frac{x}{-x+6} = \frac{16-x}{10-x} \\ 10x-x^2=x^2-22x+96\\ 2x^2-32x+96=0\\x^2-16x+48=0\\(x-4)(x-12)=0\\x=4, x=12\)

The y values corresponding to these x values are \(y = 2, y = -6\), respectively.

Therefore, the answer to this question is \(4+12+2-6=\boxed{12}\)

https://www.desmos.com/calculator/efz9crqvir (this will help you visualize it)