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Kipish [7]
2 years ago
9

For what value of constant c is the function k(x) continuous at x = 0 if k =

Mathematics
1 answer:
nlexa [21]2 years ago
5 0

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x)  is continuous, we take the limit of the parameter as follows:

\mathbf{ \lim_{x \to 0^-} k(x) =  \lim_{x \to 0^+} k(x) =  0 }

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= c }

Provided that:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }

Using l'Hospital's rule:

\mathbf{\implies  \lim_{x \to 0} \ \  \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}=  \lim_{x \to 0}   sec \ x  \ tan \ x = 0}

Therefore:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}=0 }

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

#SPJ1

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5 0
2 years ago
The vehicle preference of police officers and firefighters is given in the table.
Arisa [49]

Based on the information in the table, an example of independent events is  the: P(policer officer and chooses car).

<h3>What is an independent event?</h3>

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Based on the information provided, we can infer and logically deduce that an example of independent events is  the probability of being a policer officer and chooses a car because they aren't overlapping probabilities.

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8 0
1 year ago
Use sigma notation to represent the sum of the first seven terms of the following sequence -4,-6,-8.....
lina2011 [118]

Answer:Answer:

\sum\left {{7} \atop {1}} \right -n(3+n)

Step-by-step explanation:

Given the sequence -4,-6,-8..., in order to get sigma notation to represent the sum of the first seven terms of the sequence, we need to first calculate the sum of the first seven terms of the sequence as shown;

The sum of an arithmetic series is expressed as S_n = \frac{n}{2}[2a+(n-1)d]

n is the number of terms

a is the first term of the sequence

d is the common difference

Given parameters

n = 7, a = -4 and d = -6-(-4) = -8-(-6) = -2

Required

Sum of the first seven terms of the sequence

S_7 = \frac{7}{2}[2(-4)+(7-1)(-2)]\\\\S_7 =  \frac{7}{2}[-8+(6)(-2)]\\\\S_7 =  \frac{7}{2}[-8-12]\\\\\\S_7 = \frac{7}{2} * -20\\\\S_7 = -70

The sum of the nth term of the sequence will be;

S_n = \frac{n}{2}[2(-4)+(n-1)(-2)]\\\\S_n = \frac{n}{2}[-8+(-2n+2)]\\\\S_n = \frac{n}{2}[-6-2n]\\\\S_n =  \frac{-6n}{2} -  \frac{2n^2}{2}\\S_n = -3n-n^2\\\\S_n = -n(3+n)

The sigma notation will be expressed as \sum\left {{7} \atop {1}} \right -n(3+n). <em>The limit ranges from 1 to 7 since we are to  find  the sum of the first seven terms of the series.</em>

3 0
3 years ago
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