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sergejj [24]
3 years ago
11

If f(x) = ln(2), then limx--->2 (f(2)-f(x))/x-2

Mathematics
1 answer:
Blizzard [7]3 years ago
7 0

Answer:

  • as written, -2
  • with denominator parentheses, 0
  • with f(x)=ln(x) and denominator parentheses, -1/2

Step-by-step explanation:

The problem as stated asks for the limit as x approaches 2 of (0/x) -2.

As written, the limit is (0/2) -2 = -2.

<u>Explanation</u>: f(x) is a constant, so the numerator is 0. The ratio 0/x -2 is defined as -2 everywhere except x=0. So, the value at x=2 is 0/2 -2 = -2.

__

If you mean (f(2) -f(x))/(x -2), that limit is the limit of 0/(x-2) = 0 as x approaches 2.

<u>Explanation</u>: f(x) is a constant, so the numerator is 0. The ratio 0/(x-2) is zero everywhere except at x=2. The left limit and right limit are both 0 as x approaches 2. Since these limits agree, the limit is said to be 0.

__

If you mean f(x) = ln(x) and you want the limit of (f(2) -f(x))/(x -2), that value will be -1/2.

<u>Explanation</u>: The value of the ratio is 0/0 at x=2, so we can find the limit using L'Hôpital's rule. Differentiating numerator and denominator, we have ...

  lim = (-1/x)/(1)

The value is -1/2 at x=2.

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densk [106]

Answer:

P_{20} = 20 --- 20th percentile

P_{25} = 21.75  --- 25th percentile

P_{65} = 27.85   --- 65th percentile

P_{75} = 29.5   --- 75th percentile

Step-by-step explanation:

Given

27, 24, 21, 16, 30, 33, 28, and 24.

N = 8

First, arrange the data in ascending order:

Arranged data: 16, 21, 24, 24, 27, 28, 30, 33

Solving (a): The 20th percentile

This is calculated as:

P_{20} = 20 * \frac{N +1}{100}

P_{20} = 20 * \frac{8 +1}{100}

P_{20} = 20 * \frac{9}{100}

P_{20} = \frac{20 * 9}{100}

P_{20} = \frac{180}{100}

P_{20} = 1.8th\ item

This is then calculated as:

P_{20} = 1st\ Item +0.8(2nd\ Item - 1st\ Item)

P_{20} = 16 + 0.8*(21 - 16)

P_{20} = 16 + 0.8*5

P_{20} = 16 + 4

P_{20} = 20

Solving (b): The 25th percentile

This is calculated as:

P_{25} = 25 * \frac{N +1}{100}

P_{25} = 25 * \frac{8 +1}{100}

P_{25} = 25 * \frac{9}{100}

P_{25} = \frac{25 * 9}{100}

P_{25} = \frac{225}{100}

P_{25} = 2.25\ th

This is then calculated as:

P_{25} = 2nd\ item + 0.25(3rd\ item-2nd\ item)

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Solving (c): The 65th percentile

This is calculated as:

P_{65} = 65 * \frac{N +1}{100}

P_{65} = 65 * \frac{8 +1}{100}

P_{65} = 65 * \frac{9}{100}

P_{65} = \frac{65 * 9}{100}

P_{65} = \frac{585}{100}

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This is then calculated as:

P_{65} = 5th + 0.85(6th - 5th)

P_{65} = 27 + 0.85(28 - 27)

P_{65} = 27 + 0.85(1)

P_{65} = 27 + 0.85

P_{65} = 27.85

Solving (d): The 75th percentile

This is calculated as:

P_{75} = 75 * \frac{N +1}{100}

P_{75} = 75 * \frac{8 +1}{100}

P_{75} = 75 * \frac{9}{100}

P_{75} = \frac{75 * 9}{100}

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This is then calculated as:

P_{75} = 6th + 0.75(7th - 6th)

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Finding the numbers, which are equal to the sum of two odd number and it has to be single digit number.

Lets look into numbers which are odd and single digit.

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3 = 3+5, 3+7, 3+9

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Now, accumlating numbers which are fullfiling the criteria, however, making sure no number should get repeated.

∴ Numbers are: 4,6,8,10,12,14\ and\ 16

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Answer:

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We have been given that line m is parallel to line p, m\angle HEF=39^{o} and m\angle IGF=13^{o}.

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We can see that angle EFG is exterior angle of triangle GFJ. Since the measure of an exterior angle of a triangle equals to the sum of the opposite interior angles.

We can see that angle IGF and angle EJG are opposite interior angles of angle EFG.

m\angle EFG=m\angle IGF+m\angle EJG

Upon substituting our given values we will get,

m\angle EFG=13^{o}+39^{o}

m\angle EFG=52^{o}

Therefore, measure of angle EFG is 52 degrees.

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Answer:

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