If an equation is an identity, how many solutions does it have? zero one infinite

Mathematics

2 answers:

Galina-37 [17]3 years ago

6 0

<span>If it's an identity over an infinite field, then it has an infinite number of solutions.

</span>

Leokris [45]3 years ago

4 0

I Think Its C. Infinite

You might be interested in

What is the polynomial function of lowest degree with lead coefficient 1 and roots i, -2 and 2?

lutik1710 [3]

Answer:

Step-by-step explanation:

the polynomial function is; (x - i)(x-2)(x+2)

7 0

3 years ago

Read 2 more answers

Find the value of k such that lim-->4 (x^2+x-k)/(x-4) exists

7nadin3 [17]

Answer:

k=20

Step-by-step explanation:

when x approaches 4, the denominator x-4 approaches 0

if the denominator is 0, it means that this is invalid

if the function is a number over 0 when x=4, it represents a vertical asymptote, which means no limit

so the only way possible to let there be a limit is to let the function be 0/0 when we plug in x=4

so x^2 + x - k = 0 when x = 4

4^2 + 4 - k = 0 ==> 20 - k = 0 ==> k = 20

6 0

2 years ago

Select the correct answer.

olganol [36]

Answer: transportation

Step-by-step explanation:

7 0

2 years ago

Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the

kodGreya [7K]

Answer:

p=0.7103 (4-game series)
p=0.6480 (2-game series)

Step-by-step explanation:

Let X be the random variable equal the the first 4 straight wins. An overall win for the stronger team implies a negative binomial function with the parameters n=4, p=0.6:

$P(X=4)={{i-1}\choose {4-1}}0.6^40.4^{i-4},\ i=4,5,6,7$

#We find probabilities for the different values of i:

$P(X=4)={3\choose 3}0.6^4=0.1296\\\\P(X=5)={4\choose 3}0.6^40.4^1=0.2074\\\\P(X=6)={5\choose 3}0.6^40.4^2=0.2074\\\\P(X=4)={6\choose 3}0.6^40.4^3=0.1659$