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lisabon 2012 [21]
3 years ago
13

Paolo’s Pizza Pricing

Mathematics
1 answer:
omeli [17]3 years ago
4 0

The area of the pizza is πr² or in terms of the diameter,

A =  π(d/2)² = (π/4)d²

All our diameters are multiples of three, d=3k, so A=(9π/4)k²

The 9π/4 is just a constant, it won't affect the relative value.  

We can say little a is the amount of pizza in some units, such that

a = k²

We want to compare unit costs, cost per amount for the three pies, lower ratio means better value.   Let's call p the price of a pie in cents.

u = cents/values = p/k²

9 inch, k=3, p=1050, u=1050/3² = 116.666... cents per unit

12 inch, k=4, p=1500, u=1500/4² = 93.75  cents per unit

18 inch, k=6, p=1900, u=1900/6² = 52.777...  cents per unit

Clearly the 18 inch pizza is the best value.

Answer: 18-inch round pizza

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Determine whether the sequences converge.
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a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}

Notice that

\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}

So as n\to\infty you have a_n\to0. Clearly a_n must converge.

The second sequence requires a bit more work.

\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then a_n will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When n=2, you have

a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1

Assume a_k\ge a_{k-1}, i.e. that a_k=\sqrt{2a_{k-1}}\ge a_{k-1}. Then for n=k+1, you have

a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k

which suggests that for all n, you have a_n\ge a_{n-1}, so the sequence is increasing monotonically.

Next, based on the fact that both a_1=\sqrt2=2^{1/2} and a_2=2^{3/4}, a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}
a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}

and so on. We're getting an inkling that the explicit closed form for the sequence may be a_n=2^{(2^n-1)/2^n}, but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, a_1=2^{1/2}. Let's assume this is the case for n=k, i.e. that a_k. Now for n=k+1, we have

a_{k+1}=\sqrt{2a_k}

and so by induction, it follows that a_n for all n\ge1.

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ratelena [41]

Answer:

pyramid, cone, sphere, cube

Step-by-step explanation:

A symmetrical pattern is one that is arranged in such a way that there is a progressive repetition of a pattern. Each pattern formed is the same with respect to each other.

Eva could form series symmetrical patterns with the shapes he has, which depends on the arrangement of the shapes to form desired pattern.

One simple pattern to be formed is;

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pyramid, cone, sphere, cube

pyramid, cone, sphere, cube.

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<h3>Answer:</h3>

3)  likely

4)  1/2; equally likely

<h3>Step-by-step explanation:</h3>

3) You are being asked to translate a numerical value to a subjective statement. There are no hard-and-fast rules for this. Generally, the meanings of the terms you're asked to choose from are ...

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Answer:

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