\(\text{stellation $n$ has an $n \times n$ square of dots, and 4 identical triangular regions containing}\\ \dfrac{n(n-1)}{2} \text{ dots each}\\ dots_n = n^2 + 4\dfrac{n(n-1)}{2} = 3n^2 - 2n\\ \text{I leave you to plug 20 in}\)
\(\text{Each term is 4 less than the previous term}\\ s_n = 163-4(n-1) =167-4n\\ s_{23}=167-4(23-1)\\ \text{I leave you to finish}\)
The number of minutes it takes to be served is probably going to be exponentially distributed.
So No, Yes, No
original post has my two cents added
\(\text{For this problem I would choose }\\ x=f(t) = t^2,~y=g(t) = t\\ \text{Then }\textbf{dx}=(2t,1)dt\\ \textbf{F(x)$\cdot$ dx}= (t^4,t^2)\cdot (2t,1)dt =2 t^5+t^2~dt\)
I leave you to complete the final integral
\(\displaystyle \int_0^3 2t^5+t^2 ~dt\)
I think you got the answer Melody.
Mathematica returns
\(x = \dfrac{(-3 k+2) \pm \sqrt{k^2+4}}{2 k}\)
\(x \not \in \mathbb{R} \Rightarrow k^2 + 4 < 0\\ k^2 < -4\\ k \in i ((-\infty, 2)\cup (2, \infty))\)
which is a region on the imaginary axis.
\(\text{At a minimum }\dbinom{n}{1}=\dbinom{n}{n-1}=n \text{ both appear}\\ \text{So the minimum number is }2\)
deleted... had f(1)=5 not -5
I'm showing 2 as the first prime and 9973 as the 1229th prime.
10007 is the 1230th prime.
Oh, my bad... between 10 and 10000... blah
I get
(1,3,5,7) = (306,309,307,303)
as you did. Righto then.
deleted - see below
+6252 