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First, ignore the fact that there is a frog, and just calculate the number of ways that Marvin can reach \((5, 7)\) only going up and right. 

To do that, Marvin must move 5 units to the right and 7 upwards to the right using exactly 12 moves. This is the same question as all the different distinguishable combinations of the string of letters \(RRRRRUUUUUUU\), where R represents right and U represents up. There are \({12 \choose 5} = {12 \choose 7} = 792 \) ways to do that.

Now, we need to find the number of ways that Marvin will encounter the frog. Specifically, Marvin needs to move 4 units to the right and 3 units upwards in his first 7 moves (while his other 5 moves could be in any order). Using the same logic above, the number of possible different distinguishable ways to order the string of letters \(RRRRRUUUUUUU\) in which the first 7 letters must contain 4 R's and 3 U's, is equal to \({7 \choose 3} \cdot {5 \choose 1} = 175\) ways. That is the number of ways that Marvin will meet the frog and get eaten.

Since we don't want that to happen, just subtract the total number of ways he is going to get to (5, 7) by the number of ways he is going to get eaten, which is equal to \(792-175=\boxed{617}\) ways. 

By the remainder theorem, for any polynomial p(x), x-a will be a factor of p(x) only if p(a) = 0, where 'a' is a constant. Knowing that, for (x-3) to be a factor of f(x), f(3) must be equal to zero. Therefore, set x = 3 and solve for k:

\(3^3-3^2-6\cdot k \cdot3+4k^2=0\\27-9-18k+4k^2=0\\4k^2-18k+18=0\\2k^2-9k+9=0\\(2k-3)(k-3)=0\\\boxed{k=\frac{3}{2}, k=3}\)

Now, to solve the second question, set k = 3 to get f(x) to equal \(x^3-x^2-18x+36\). This should be easy enough to be solved using the remainder theorem mentioned above.

Notice that ABC can be divided into 4 congruent triangles (specifically triangles ADF, BDE, DEF, and EFC). We just need to find the area of one of these triangles and multiply by 4. 

Also notice that DEF is half the area of DXYF. Therefore, the area of ABC is equal to \(\frac{42}{2} \cdot 4 = \boxed{84}\)

You're right, I forgot about -12

That sum is equal to \(\frac{83(84)}{2}=83(42)\). Since 83 is prime, and it is greater than 42, it must be the biggest prime divisor. Therefore, 83 is the answer.

1. Guest was almost right, but he/she forgot to multiply by 2 when squaring the binomial. So the answer is 12.

2. \((r-s)^2 = r^2-2rs+s^2\). By Vieta's formulas, \(r+s = \frac{-6}{-1} = 6\), and \(rs = \frac{2}{1} = 2\) so \((r+s)^2= r^2+2rs+s^2 = 6^2=36\). Now just subtract \(4rs=8\) from 36 to get what we want, so the final answer is 28. 

3. Divide \(3x^2-4x+15=0\) by 3 to get \(x^2 - \frac{4}{3}x + 5 = 0\). Therefore, the ordered pair is \((\frac{-4}{3}, 5)\)

Nice answer guest, but I don't think it's correct. Here's why:

Since AD is parallel to BE, Angle BEC = Angle ADC, and therefore, in order for Angle CAD to equal angle CBE, triangles BEC and ADC must be similar to each other. That can only happen if angle BCE = angle ACD, and that can only happen if A is on BC. Since that isn't allowed, angle CAD cannot be equal to angle CBE.

Since 81 divides 9, the answer to this question is just the remainder when 25 is divided by 9. That is equal to 7, since 25-18 = 7 and 18 is the largest multiple of 9 smaller than 25

There is probably a way to do this without first converting to base 10, but I'm not super sure how to do it.

Anyway, first convert 45 from base 6 to base 10, which is \(5+4\cdot6=29\). 

The sum of the integers 1 to 29 base 10 is \(\frac{(29)(30)}{2}=435\). In base 6 that would be equal to \(\boxed{2003}\), which is our answer.

Close, but not quite.

Let x be the age of the father and let y be the age of the son.

\(x=9y\\ x+3=6(y+3)\\ 9y+3=6y+18\\ 3y=15\\ x=5\)

So the son is actually 5 years old.