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Tema [17]
3 years ago
9

What is the right-hand limit of as x approaches f(x) = x/x-5 as x approaches 5?

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

Answer:

+\infty

Step-by-step explanation:

The expression for the right hand limit of f(x) is:

f = \frac{x+\epsilon}{x + \epsilon - 5}

The expression can be simplified with some algebraic handling:

f = \frac{1}{\frac{x+\epsilon-5}{x+\epsilon} }

f = \frac{1}{1-\frac{5}{x+\epsilon} }

Let be x = 5, as \epsilon is approaching zero, f is positive and becomes bigger, diverging to +\infty.

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2 years ago
Select the correct answer.
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\sf{(x - 3)(x + 7)}

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3 0
3 years ago
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A number cube with the numbers 1 through 6 is rolled. What is the probability of the number cube showing a 4 or a number less th
Sonja [21]

Answer:

Step-by-step explanation:

The probability is

P

(

A

)

=

1

3

Explanation:

To calculate the probability you have to count the number of all possible results

|

Ω

|

and the number of results that fulfill the given condition

|

A

|

.

In this case

Ω

=

{

1

,

2

,

3

,

4

,

5

,

6

}

Thereare 6 possible result of a dice toss. So

|

Ω

|

=

6

The given condition is "the result is divisible by

3

", so

A

=

{

3

,

6

}

- those are the only numbers divisible by

3

, so

|

A

|

=

2

.

Finally to calculate the probability we have to divide

|

A

|

by

|

Ω

|

.

P

(

A

)

=

|

A

|

|

Ω

|

=

2

6

=

1

3

Note The probability is never larger than

1

, so if you get such result then there must be a mistake in calculations.

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