Rom

\(\text{surgery eh...}\\ x^2 + x + 3 = \left(x+\dfrac 1 2\right)^2 +\dfrac{11}{4} >0\\ 2x^2 + x - 6 = (2x-3 )(x+2)\\ 2x^2 + x - 6 \begin{cases}+ &x<2\\0&x=2\\- &2 < x < \dfrac 3 2\\ 0 &x=\dfrac 3 2\\+&\dfrac 3 2 < x\end{cases}\)

\(+ \div + \Rightarrow +\\ +\div - \Rightarrow -\\ \dfrac{x^2+x+3}{2x^2+x-6}\geq 0 \Rightarrow (-\infty,2]\cup \left[\dfrac 3 2,\infty\right)\)

all correct

#4

\(\text{assume we transfer }n \text{ birds at a time and it was assumed that }\\ \text{transferring }113 \text{ birds would take }t \text{ days. }\\ r \text{ is the common odd number of birds for the last trip}\\ 113 = n(t-1) + r\\ 131 = n(t+2)+r\)

\(\text{subtract 1 from 2}\\ 18=3n\\ n=6\)

\(131 = 6(t+2)+r\\ 119 = 6t + r\\ \text{note that we are told }r < n\\ \left \lfloor \dfrac{119}{6} \right \rfloor= 19\\ 119=19 \cdot 6 + r\\ 119=114+r\\ r=5 \)

2)

\(\text{we seek }d \text{ such that}\\ i d = \dfrac{18}{4},~j d = \dfrac{30}{4},~k d = \dfrac{11}{4}~ft,~i,j,k \in \mathbb{Z}\\ LCM(18,30,11) = 990\\ \text{so split up the longest side into }990 \text{ cubes}\\ d = \dfrac{30}{4} \times \dfrac{1}{990} = \dfrac{1}{132}~ft\\ \text{The prism will be split into }\\ \dfrac{\frac{18}{4}}{\frac{1}{132}}\times \dfrac{\frac{30}{4}}{\frac{1}{132}}\times \dfrac{\frac{11}{4}}{\frac{1}{132}}=\\ 595 \times 990 \times 363 = 213465780 \text{ cubes}\)

don't bump please, someone will get to it and bumping doesn't help.

How can a probability be 28 2/5 ?

0 <= Pr[anything] <= 1

\(P[\text{both same color}] = \\ P[\text{2 red}]+P[\text{2 blue}] = \\ \dfrac{5}{9}\dfrac{4}{8} + \dfrac{4}{9}\dfrac{3}{8} = \\ \dfrac{20 + 12}{72} = \\ \dfrac{4}{9}\)

#6) more diophantine equations

\(\text{Let the number of candies in the box = }N\\ N-3 \pmod{7}=0\\ N-6 \pmod{8} = 0\\ N+3 \pmod{5} = 0\)

\(N \pmod{7}=3,~N\pmod{8}=6\\ 7m+3 = 8n+6\\ m = \dfrac{8n+3}{7}\in \mathbb{Z}\\ 8n+3 \pmod{7}=0\\ n \pmod{7}=4\)

\(N=8(7k+4)+6 = 56k+38\\ 56k+38+3\pmod{5}=0\\ k + 1\pmod{5} = 0\\ k \pmod{5} = 4\\ N=56(5j+4)+38 = 280j+262\)

\(\text{we can't have negative candies so the smallest }N\\ \text{occurs when }j=0 \text{ i.e. when }N=262\)

#5)

\(N\pmod{9}=5\\ N\pmod{7}=3\\ N\pmod{5}=1\)

\(9m+5=7n+3\\ n=\dfrac{9m+2}{7} \in \mathbb{Z}\\ 9m+2 \pmod{7} = 0\\ 2m \pmod{7}=5\\ m \pmod{7} = 20 \pmod{7} = 6\\ m = 7k+6, ~k \in \mathbb{Z}\)

\(N=9(7k+6)+5 = 63k + 59\\ 63k+59 = 5j + 1\\ j = \dfrac{63k+58}{5}\in \mathbb{Z}\\ 63k+58 \pmod{5} = 0\\ 3k+3 \pmod{5} = 0\\ 3k \pmod{5} = 2\\ k \pmod{5} = 4\\ k = 5i + 4\)

\(N = 63(5i+4)+59\\ N = 315i + 311\\ i = 3 \text{ gets us the smallest 4 digit number}\\ N = 945 + 311 = 1256\)

\(\#3)\\ N = 3m + 2 = 4n+3\\ m = \dfrac{4n+1}{3} \in \mathbb{Z}\\ 4n+1 \pmod{3} = 0\\ n \pmod{3} = -1\\ n\pmod{3} = 2\\ n = 2+3k\\ N = 4n+3 = 4(2+3k)+3 = 12k+11\\ N \pmod{12} = 11\)

\(R=F_1+F_2 = (p-6)\hat{i} + (5-2p)\hat{j}\\ \text{This is parallel to the vector }\hat{i} - 3\hat{j}\\ \text{The easiest way to do this is just to equate the slopes}\\ \dfrac{5-2p}{p-6}=\dfrac{-3}{1} \\ 5-2p = 18-3p\\ p = 13\)

\(R = (7,-21) \text{ which is clearly parallel to }(1,-3)\)

\(\text{in two dimensions the ratio of the x and y coordinates}\\ \text{of parallel vectors will be equal}\\ \dfrac{q}{12} = \dfrac{-1}{9}\\ 9q = -12\\ q = -\dfrac 4 3 \)