Which expression is equivalent to 15n – 20?

Mathematics

2 answers:

crimeas [40]3 years ago

7 0

Answer:

5(3n-4)

Step-by-step explanation:

because(5*3n)-(5*4)=15n-20

Semenov [28]3 years ago

3 0

The only thing you can do with this expression is to factor a 5 out of the two terms: we have

$15n-20 = 5(3n-4)$

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Savatey [412]

Answer:

it wont let me write the graph

Step-by-step explanation:

7 0

3 years ago

Stephon makes the following statements: statement 1 lim x -3 f(x) exists and is equal to 1 statement 2 lim x 1 f(x) exists and i

Mandarinka [93]

Neither statement 1, nor statement 2 are correct

The given Stephon's statements are;

Statement 1; $\mathbf{\lim \limits _{x \to -3} f(x)} \ \mathbf{Exist} \ and \mathbf{\lim \limits _{x \to -3} f(x) = 1}$

Statement 2; $\mathbf{\lim \limits _{x \to 1} f(x)} \ \mathbf{Exist} \ and \mathbf{\lim \limits _{x \to 1} f(x) = 1}$

The analysis of the graph and reason for the answer

From the graphed line on the left of the y-axis, we have an open circle at x = -3, and an arrow at the other end pointing towards negative infinity, (-∞) which indicates that the domain is -∞ ≤ x < -3, therefore, at x = -3, f(x) does not exist, therefore, we can write the following statement

The limits of the domain and range of the graph includes;

$\mathbf{\lim \limits _{x \to -3} f(x)} = \mathbf{Does \ not \ exist}$

f(x) = Defined for -∞ ≤ x < -3

Similarly, from the graphed line on the right of the y-axis, we have an open circle at x = 1 and an arrow at the other end of the line f(x) = 4 pointing towards positive (+∞) infinity, which indicates the domain and the graph of the function is 1 < x ≤ ∞ , therefore, f(x) does not exist at x = 1, and we can write

$\mathbf{\lim \limits _{x \to 1} f(x)} = \mathbf{Does \ not \ exist}$