Rom

\(y=\log_2\left(\dfrac{x}{x-1}\right) = \dfrac{1}{\ln(2)}\ln\left(\dfrac{x}{x-1}\right)\\ \dfrac{dy}{dx} = \dfrac{1}{\ln(2)}\cdot \dfrac{1}{\dfrac{x}{x-1}}\cdot \left(\dfrac{(x-1)-x}{(x-1)^2}\right)= \\ -\dfrac{1}{\ln(2)}\cdot \dfrac{x-1}{x}\cdot \dfrac{1}{(x-1)^2} = \\ -\dfrac{1}{\ln(2)}\dfrac{1}{x(x-1)} =\\ \dfrac{1}{\ln(2)}\dfrac{1}{x(1-x)}\)

\(\displaystyle \int_3^4 e^{5-x}~dx = \\ \left . -e^{5-x} \right |_3^4 = \\ e^2 - e\)

you want the max delta y which is (3-0)=3

and the min delta x which is (4-2)=2

So the greatest slope is 3/2

1) angle 2 and the 135deg angle are supplementary angles, their degree sum is 180 degrees.  Thus

2) angle 2 = 180deg - 135deg = 45deg

3) we know the angles in a triangle sum to 180 degrees.  The triangle shown has angle 1, angle 2=45deg, and a 90deg angle.

Thus angle 1 = 180-45-90 = 45deg

getting old or something, see answer #4

\(20 = 2^2 5 \text{ so we'll want to use prime factors }\\ 2^4 \cdot 3^1 \cdot 5^1 = 240\\ \text{This has }(4+1)(1+1)(1+1)=20 \text{ divisors}\)

I'm pretty sure for (2) we just find the smallest positive integer with 6 divisors and square it.

Proceeding as above

\(6=2\cdot 3 \text{ so choose}\\ (2^2\cdot 3)^2 = 144\)

huh... turns out the answer is 106

I'll have to look at this one.

incorrect answer deleted.  See answer #4

we put Alice and Andrew on opposite teams.

We then have \(\dbinom{6}{3}=20\) ways to select 3 players from the remaining six for one of the teams.

Once those 3 are selected the other 3 are automatically on the other team.

In your working you treated permutations of players within a team as important.  They aren't.