asdf1243

g(x)=3x+5, and then f(3x+5)=1/(3x+7). h(1/(3x+7)=1/(3x+7)

I would assume it is F. Try solving it as F. 

Idk how to solve it lol

Nice answer

Start by dividing the islanders into 3 groups of 4. Placing two groups results in two possible outcomes. 

(1) If they balance, they are all real. That means one of the remaining is more or less.

Weigh 3 of the regular weighing guys against 3 of the remaining ones. If they balance, the last guy is more or less. 

If it is lighter, one of the ones on the seesaw must be less. Weigh 2 of the suspicious ones. If they balance, then the third guy is less. If they don't balance, the lighter one is the guy we are looking for.

If it is heavier, one of the ones on the seesaw must be more. Weigh 2 of the suspicious ones. If they balance, then the third guy is more. If they don't balance, the heavier one is the guy we are looking for.

(2) If at first they don't balance, the remaining 4 are real. Replace 3 of the ones on the "heavier" side with 3 of the ones on the "lighter" side and add 3 real guys on the "lighter" side. If the same side is still heavier, it means that either the one that wasn't replaced on the "heavier" side is heavier or the one not replaced on the "lighter" side is lighter. Choose either of them, and weigh them against a real guy.

Also, if they balance at the second use, then one of the three "lighter" ones were fake. Weigh two against each other, If they balance, then the third guy is fake. If they don't balance, the lighter one is the guy we are looking for.

My solution got deleted, so just to save my time, the answer is 175+6k.

1. Let the first term be a, and let the common difference be d. Then the four positive integers are a, a+d, a+2d, and a+3d. The sum of these four positive integers is 4a+6d=46, so 2a+3d=23. Solving for d, we find that =(23-2a)/3. The third term is a+2d=a+2(23-2a)/3=(46-a)/3. Thus, to maximize this expression, we minimize a, so a must be 1. Substituting 1 in, we find that the largest possible third term is 15.

2. 94 =F + 26D,  
27 =F + 93D, solve for F, D
F =120 - First term
D = -1 - Common difference

3.  As we begin, we note that no side of the triangle can exceed 29 units in length. Let’s start at the 20–20–20 triangle. We first count the number of triangles that have 2 sides of length less than 20. Let the third side be of length 20+k where 2≤k≤9. This we do by taking away k units in length from the first and the second sides, by subtracting at least one unit from each. For each k, We can do it in ⌊k/2⌋ ways. Thus, counting for each possible k, we have 1+1+2+2+3+3+4+4=201+1+2+2+3+3+4+4=20 cases.

Let’s now go back to the 20–20–20 triangle and count the number of cases where only one side has a length <20<20 units. We can take away 1 to 18 units from this one side and redistribute it to the other two such that no side exceeds 29 units in length. Let us take away k units from this one side, such that 1≤k≤9. In order for the triangles to be distinct, the redistribution can be done in ⌊k/2⌋+1⌊k/2⌋+1 ways. Then, for all possible values of k in this range, we have 1+2+2+3+3+4+4+5+5=29 cases. Now, let us consider the cases where 9
Finally, we add up all our cases:

1+20+29+25=751+20+29+25=75 distinct triangles. Also, subtract one because of the equilateral one.

4. I really can't think of another way other than listing.

The x-intercept refers to when y=0. 

Also, since the x-intercept is -5, then that means one of the legs of the right triangle is 5. 

Using the formula for the area of a triangle, \(\frac{5h}{2}=10 \rightarrow 5h=20 \rightarrow h=4\)

Now you have 2 points, (-5, 0) and (0, 4) (using (0, -4) also works i think)

Find the slope to be \(\boxed{\frac{4}{5}}\)

Work backwards. 

To get five on the last f(x), then x could be -3 for the first f(x). 

Or, on the f(x)=x+3, it would be impossible. 

So you need -3 after the first f(x).

To do that, you could start off with -6, get to -3 and get 5.

Or you could start off with x=1, get to -3, and get 5. 

There are two possible values for x: \(\boxed{\{-6, 1\}}\) .