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beks73 [17]
2 years ago
13

This exercise refers to a standard deck of playing cards. Assume that 5 cards are randomly chosen from the deck.

Mathematics
1 answer:
icang [17]2 years ago
8 0
There is not enough information to answer this question
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2 years ago
Explain why f(x) = x^2+4x+3/x^2-x-2 is not continuous at x = -1.
liberstina [14]

Answer:

The value of x = -1 makes the denominator of the function equal to zero. That is why this value is not included in the domain of f(x)

Step-by-step explanation:

We have the following expression

f(x) = \frac{x^2+4x+3}{x^2-x-2}

Since the division between zero is not defined then the function f(x) can not include the values of x that make the denominator of the function zero.

Now we search that values of x make 0 the denominator factoring the polynomial x^2-x-2

We need two numbers that when adding them get as a result -1 and when multiplying those numbers, obtain -2 as a result.

These numbers are -2 and 1

Then the factors are:

(x-2) (x + 1)

We do the same with the numerator

x^2+4x+3

We need two numbers that when adding them get as a result 4 and when multiplying those numbers, obtain 3 as a result.

These numbers are 3 and 1

Then the factors are:

(x+3)(x + 1)

Therefore

f(x) = \frac{(x+3)(x+1)}{(x-2)(x+1)}

Note that \frac{(x+1)}{(x+1)}=1 only if x \neq -1

So since x = -1 is not included in the domain the function has a discontinuity in x = -1

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