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Are you sure? How did you arrive at this answer?

This problem seems overwhelming, but let's take this problem one piece at a time.

Instead of trying to wrap our heads around $f(f(x))=3$, we should instead ponder a far easier question. For what values of x does $f(x) = 3$? Well, let's investigate, shall we?

$f(x)$ is defined as a piecewise function, so we must more or less treat it as two separate functions and ensure that our answers lie within the restrictions.

$f(x_1)=3\text{ and } f(x_2)=3$. Let and solve$f(x) = x_1\text{ or } f(x) = x_2$to determine the number of solutions to $f(f(x)) = 3$.

Through all this algebra, we have determined that there are 3 values of x for which $f(f(x)) = 3$.

The equation f(f(x)) = 3 has 1 solution.

1 + 3 + 5 + ... + 199 = 199*200/2 = 19900 = 6 (mod 7).

The answer is 6.

This is not correct. The entire denominator must yield 0 for a restriction in the domain to exist, not just one portion of it.

Before trying to pinpoint the domain of this function, it is best to simplify completely. The function can be simplified more by expanding and then combining like terms. Let's do that.

$f(t) = \frac{1}{(t-1)^2+(t+1)^2}\\ f(t) = \frac{1}{t^2 - 2t + 1 + t^2 + 2t + 1}\\ f(t) = \frac{1}{2t^2 + 2}$

This expression, while looking slightly different, is the same function algebraically speaking. When considering the domain, it is best to think creatively. What values could cause $f(t)$ to be undefined? For rational functions, the only potential for a restriction of the domain is when the denominator yields 0.

$2t^2+2\neq 0\\ 2t^2 \neq -2$

I decided not to solve anymore because there every value for t that will satisfy this equation. $2t^2$ is always nonnegative, regardless of the input, so it is impossible for $2t^2$ to output $-2$. Therefore, every value for $t$ will yield a real output.

Represented in interval notation,$Dom(t): (-\infty,+\infty)$

The function f(t) is defined for all t except 1 and -1, so the domain is (-inf,-1) U (-1,1) U (1,inf).

That's OK. I will do the translation for you.

$\left\lfloor \left\lceil \left(\frac{13}{7}\right)^2\right\rceil+\frac{17}{4}\right\rfloor = \left\lfloor \left\lceil{\frac{169}{49}} \right\rceil + \frac{17}{4} \right\rfloor \\ \left\lfloor \left\lceil \left(\frac{13}{7}\right)^2\right\rceil+\frac{17}{4}\right\rfloor = \left\lfloor 4 + \frac{17}{4} \right\rfloor\\ \left\lfloor \left\lceil \left(\frac{13}{7}\right)^2\right\rceil+\frac{17}{4}\right\rfloor = 8$

A rational number is a number that can be written in the form $\frac{a}{b}$ where a is any integer and b is a counting number.

Guest provided one way to represent 21, but $\frac{21}{1}$ is the simplest example of them all.

I used coordinates, and found C = (0,0) and M = (8,8), so CM = 8*sqrt(2).

Let's use some clever algebraic manipulation to change this quadratic from standard from to vertex form.

$f(x) = 3x^2 - 6x - 2\\ f(x) = 3(x^2 - 2x) - 2\\ f(x) = 3(x^2 - 2x + 1) - 2 - 3\\ f(x) = 3(x-1)^2 - 5\\ a = 3\text{ and } h = 1 \text{ and } k = -5\\ a+h+k = 3 + 1 - 5\\ a+h+k = -1$

The "9" in 69 * 1/8 is a typo.   The guest actually meant 64 * 1/8.  

64 grams with a half life of 13 hours  

After the 1st 13 hours there's 32 grams left  

After the 2nd 13 hours there's 16 grams left  

After the 3rd 13 hours there's 8 grams left  

    63  

  ––—   is just one of many.  

     3  

Ignore the circle.  BEC is a triangle. 

One angle is x and one angle is 32^o and the third angle is (180^o – 42^o)  

All three angles in a triangle total 180^o, therefore x  =  180^o – 32^o – 138^o  =  10^o    so your answer is B.  

_.

Tbh, I don't know for sure, but I'd guess it's something like. 

Cash is money in the form of paper bills OR coins.

=^._.^=

a + b = (-4) + (-1) = -5.

I'm sorry about the formatting, I keep telling myself that I will learn latex, but I've been procrastinating. 

floor(ceil((13/7)^2)+17/4)

floor(ceil(169/49)+17/4)

floor(4+17/4)

8

I hope this helped. :)))

=^._.^=

We do 756 divided by 12 getting 63  Divide because we want to find the average

FOR X: write 3x-y=10 in terms of y, add -3x from both sides, and divide both sides by -1 to simplify it down to y=3x-10. then, write 3x-10 in terms of y, add -3x to both sides, and divide both sides by -1 to get x=10

FOR Y: substitute 10 in terms of y and simplify to get x=20

FOR X: substitute -3x+5 in terms of y and add 20 to both sides and divide both sides by 17 to get x=1

FOR Y: substitute 1 in terms of x and simplify to get y=2

subtract 12 from both sides and divide both sides by -8 to get z=3

756/12=63 miles/hr

63*567=35721 min

There are: 4 between 1 and 10

                    20 between 10 and 100

                    4 x 5 x 5 =100 between 100 and 1,000

                    4 x 5 x 5 x 5=500 between 1,000 and 10,000

                     4 x 5 x 5 x 5 x 5 =2,500 between 10,000 and 100,000

Grand Total =[4 + 20 + 100 + 500 + 2,500] ==3, 124 integers with EVEN digits.

the answer was $1 \over 4$!

to anyone who has also been trying to tackle this problem: 

you could factor the expression like so:

$xy(x-y)$

then, since you are trying to find the maximum value, you can assume that $x \ge y$. by AM-GM, $y(x - y) \le \frac{x^2}{4}$, and then continue from there:)