Rom

legs of a triangle a, b, c

the triangle inequality states

$|a|+|b| >|c| \\ \text{with strict inequality else we'd just have a line segment, not a triangle.}$

$|c| < |39|+|15| = 54 \Rightarrow c < 54 \\ \text{we also must have }\\ |c|+|15| > 39 \Rightarrow c > 24 \text{ thus}\\ 24 < c < 54$

$\text{If I understand what you're saying...}\\ 4\sqrt{abc}=5\\ 4\sqrt{abcd}=3\sqrt{10}$

$\dfrac{4\sqrt{abcd}}{4\sqrt{abd}}=\sqrt{d} = \dfrac{3\sqrt{10}}{5}\\ d = \left(\dfrac{3\sqrt{10}}{5}\right)^2 = \dfrac{9\cdot 10}{25} = \dfrac{18}{5}$

$\text{assuming a 365 day year}\\ nblink=3\dfrac{blink}{min}\cdot 60\dfrac{min}{hr}\cdot 14\dfrac{hr}{day}\cdot 365 \dfrac{day}{yr} = \\ 3\cdot 60 \cdot 14 \cdot 365 =919800 \dfrac{blink}{yr} \\ \text{so no, not quite a million blinks per year}$

I think the easiest way to do this is to parameterize the circle via a single variable and solve the minimization problem.

$\text{let }p \text{ be a point on the circle given by }(x-12)^2 + (y+5)^2 = 9 \\ \text{then p can be described as }\\ p=(3 \cos(\theta)+12,~3\sin(\theta)-5)$

$\text{The squared distance of this point to the origin is }\\ d^2 = (3 \cos(\theta)+12)^2 + (3\sin(\theta)-5)^2 =\\ 9 \sin ^2(\theta )-30 \sin (\theta )+9 \cos ^2(\theta )+72 \cos (\theta )+169 = \\ -30 \sin (\theta )+72 \cos (\theta )+178 = \\ 78 \cos(\theta+\arctan(5/12))+178$

$d^2 \text{will clearly be minimized when }\\ \theta + \arctan(5/12) = \pi\\ \theta = \pi - \arctan(5/12)\\ \text{at this angle }d^2 = -78 + 178 = 100\\ d=10$

The key here is to find x such that the determinant of F is zero.

$det(F) = 2\cdot 3 - (-4)\cdot x = 6+4x \\ \text{solving }det(F)=0 \text{ we get} \\ 6+4x = 0 \\ x = -\dfrac 3 2$

well let's multiply them

$\begin{pmatrix}4 &3\\1&1\end{pmatrix} \begin{pmatrix}1 &-3\\-1&4\end{pmatrix} =\begin{pmatrix}1&0\\0&1\end{pmatrix}$

so yes C and E are multiplicative inverses

There are a total of $\dbinom{6}{2}=15$ possible angles

But it turns out that many of the angles are congruent.

The 15 angles are made up of 1 + 2 + 3 + 4 + 5 copies of 5 different angles.

There are only 5 different possible angles

$\text{This is known as finding the partial fraction expansion}\\ \dfrac{x+7}{x^2-x-2} = \dfrac{x+y}{(x-2)(x+1)} = \dfrac{A}{x-2}+\dfrac{B}{x-1}$

$\dfrac{x+7}{(x-2)(x+1)} = \dfrac{A(x+1)+B(x-2)}{(x-2)(x+1)}\\ \forall x : x \neq 2,~-1 \\ x+7 = A(x+1) + B(x-2) \\ x = (A+B)x \Rightarrow \\ 1 = A+B\\ 7 = A -2B$

$B = 1-A\\ 7 = A - 2(1-A) = 3A-2\\ 3A = 9\\ A=3\\ B = 1-3 = -2\\ \dfrac{x+7}{(x-2)(x+1)}=\dfrac{3}{x-2}-\dfrac{2}{x+1}$

$\text{let }n_c \text{ be the number of cups of lemonade sold and }\\n_b \text{ be the number of lemon bars sold.}$

$\text{we have 3 constraints }\\ 2\cdot n_c + 1.5 \cdot n_b \geq 500 \\ 0.25 n_c + 0.2 n_b \leq 100\\ n_c \geq 150$

Oh oops, I missed that they called n_c = x and n_b = y, oh well you'll figure it out.

I don't see any solution method any better than just plugging in the points and checking.

so every pair but the first listed is acceptable.

The theorem being used here is that

$gcd(a^n -1,~a^m-1) = a^{gcd(m,n)}-1$

in this case $n=1001,~m=1012$

$1001 = 7 \cdot 11 \cdot 13 \\ 1012 = 2 \cdot 11 \cdot 23 \\ \text{so clearly }gcd(1001,~1012)=11\\ 2^{11}-1 = 2047 \text{ as guest correctly computed}$

The proof of this theorem can be found online