Rom

I'm showing 2305

What class is this?  Is there supposed to be some clever way of figuring this out?

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\((0.4)n = 200\\ n = \dfrac{200}{0.4} = 500\)

\(\text{The problem is to show that }f(x) = f(x+q),~q \in \mathbb{Q}\\ r \in \mathbb{R}-\mathbb{Q} \Rightarrow q-r \in \mathbb{R}-\mathbb{Q}\\ \text{i.e. }q-r \text{ is irrational}\\ f(x) = f(x+q-r)\\ x+q-r \in \mathbb{R} \\ f(x) = f(x+q-r) = f(x+q-r + r) = f(x+q)\)

\(\text{so }f(x) = f(x+q)=f(x+r),~q \in \mathbb{Q},~r \in \mathbb{R}-\mathbb{Q} \\ \mathbb{Q} \cup \left( \mathbb{R}-\mathbb{Q} \right) = \mathbb{R}, \text{ thus}\\ f(x) = f(x + y),~\forall y \in \mathbb{R},~\text{ i.e. }f \text{ is constant}\)

I have to think more about this but I don't believe there are any beautiful colorings.

If the operation had been multiplication you could split the evens and odds up but odd plus odd = even.

\(\text{if the sign is similar to the banner then it's width must be}\\ w=\dfrac{17}{54}\cdot 36 = 17\cdot \dfrac 2 3 = \dfrac{34}{3}\\ \dfrac{A_{banner}}{A_{sign}} = \dfrac{54 \times 36}{17\times \frac{34}{3}} = \dfrac{2916}{289}\)

So distilling all this we have

5 earthlike planets

6 marslike planets

planets are distinct

earthlike planets use 2 units of colonists

marslike planets use 1 unit of colonists

There are 12 units of colonists to assign.

Probably easiest to work it looking at how many earthlike planets are colonized

There are only 6 marslike planets so there must be at least 3 earthlike planets colonized

There are only 5 earthlike planets so there can be at most 5 colonized

\(k ~Earth \Rightarrow 12-2k~Mars\\ N = \sum \limits_{k=3}^5 ~\left[\dbinom{5}{k}\right]\left[\dbinom{6}{12-2k}\right]\\ N=100\)

You want us to give you an answer to your test question?

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\(\text{There are a total of }\dbinom{4}{2}=6 \text{ pairs} \\ (3,6) \text{ is the only pair whose product is a multiple of }9\\ \text{So }p=\dfrac 1 6\)