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madreJ [45]
3 years ago
7

Suppose you pick 6 different numbers in [10]. Prove that 2 of the numbers are next to each other. (Hint: use the pigeonhole prin

ciple. What are the pigeons and what are the holes?)
Mathematics
1 answer:
Masja [62]3 years ago
4 0

Step-by-step explanation:

We are picking 6 numbers from the numbers 1,2,3,4,5,6,7,8,9,10. Since we care about numbers being next to each other, we might think of the 10 numbers as being distributed in 5 boxes (which you can think of as the holes):

|   1 2   |   3 4   |   5 6   |   7 8   |   9 10  |

So on the first box we have the numbers 1 and 2, on the second box we have the numbers 3 and 4, and so on. Since we are picking 6 numbers from those 10 numbers, that means we'll have to pick 6 boxes (and inside each box we pick a number), but we only have 5 available boxes, so by the pigeonhole principle, we'll have to pick 1 same box at least two times. Since on each picked box we'll need to pick a number, on this box which was picked two times, we will have to pick both of its numbers. And so those 2 numbers inside that box will be next to each other (meaning they're consecutive numbers).

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How to differentiate ?
Bas_tet [7]

Use the power, product, and chain rules:

y = x^2 (3x-1)^3

• product rule

\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d(x^2)}{\mathrm dx}\times(3x-1)^3 + x^2\times\dfrac{\mathrm d(3x-1)^3}{\mathrm dx}

• power rule for the first term, and power/chain rules for the second term:

\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(x-1)^2\times\dfrac{\mathrm d(3x-1)}{\mathrm dx}

• power rule

\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(3x-1)^2\times3

Now simplify.

\dfrac{\mathrm dy}{\mathrm dx} = 2x(3x-1)^3 + 9x^2(3x-1)^2 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2 \times (2(3x-1) + 9x) \\\\ \boxed{\dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2(15x-2)}

You could also use logarithmic differentiation, which involves taking logarithms of both sides and differentiating with the chain rule.

On the right side, the logarithm of a product can be expanded as a sum of logarithms. Then use other properties of logarithms to simplify

\ln(y) = \ln\left(x^2(3x-1)^3\right) \\\\ \ln(y) =  \ln\left(x^2\right) + \ln\left((3x-1)^3\right) \\\\ \ln(y) = 2\ln(x) + 3\ln(3x-1)

Differentiate both sides and you end up with the same derivative:

\dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac2x + \dfrac9{3x-1} \\\\ \dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \times x^2(3x-1)^3 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(15x-2)(3x-1)^2

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Step-by-step explanation:

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3 years ago
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AnnZ [28]

Answer:

Probability = \frac{243}{1024}

Step-by-step explanation:

Given

Sum = \{7, 11, 12\}

Rolls = 5

Required

Probability of not getting audited

If a pair of dice is rolled, the following are the observations of the sum

Outcomes = 36

Sum\ of\ 7 = 6

Sum\ of\ 11 = 2

Sum\ of\ 12 = 1

So, in a single roll; The probability of getting audited is:

p= \frac{1 + 2 + 6}{36}

p= \frac{9}{36}

p= \frac{1}{4}

The probability of not getting audited in a single roll is:

q = 1 - p --- Complement rule

q = 1 - \frac{1}{4}

Take LCM

q = \frac{4 - 1}{4}

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The probability of not getting audited in 5 rolls is:

Probability = q^5

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Alenkasestr [34]
<h3>Given</h3>

tan(x)²·sin(x) = tan(x)²

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Subtract the right side and factor. Then make use of the zero-product rule.

... tan(x)²·sin(x) -tan(x)² = 0

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This is an indeterminate form at x = π/2 and undefined at x = 3π/2. We can resolve the indeterminate form by using an identity for tan(x)²:

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Then our equation becomes

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Answer:

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Step-by-step explanation:

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