q(x)= x 2 −6x+9 x 2 −8x+15 q, left parenthesis, x, right parenthesis, equals, start fraction, x, squared, minus, 8, x, plus, 1
AURORKA [14]
According to the theory of <em>rational</em> functions, there are no <em>vertical</em> asymptotes at the <em>rational</em> function evaluated at x = 3.
<h3>What is the behavior of a functions close to one its vertical asymptotes?</h3>
Herein we know that the <em>rational</em> function is q(x) = (x² - 6 · x + 9) / (x² - 8 · x + 15), there are <em>vertical</em> asymptotes for values of x such that the denominator becomes zero. First, we factor both numerator and denominator of the equation to see <em>evitable</em> and <em>non-evitable</em> discontinuities:
q(x) = (x² - 6 · x + 9) / (x² - 8 · x + 15)
q(x) = [(x - 3)²] / [(x - 3) · (x - 5)]
q(x) = (x - 3) / (x - 5)
There are one <em>evitable</em> discontinuity and one <em>non-evitable</em> discontinuity. According to the theory of <em>rational</em> functions, there are no <em>vertical</em> asymptotes at the <em>rational</em> function evaluated at x = 3.
To learn more on rational functions: brainly.com/question/27914791
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Isolate z by itself:

To simplify variables with powers, subtract the smaller power from the bigger power:


The exponent will be
-3.
So, he deposited 17 dollars, ( 4 ) 5 dollar bills, and 6 twenty dollar bills.
5 + 5 + 5 + 5 = 20
20 x 6 = 120
120 + 20 = 140
140 + 17 = 157
He deposited a total of <u>157 dollars.</u>
hope this helps!
Answer:
We estimate to have 8.33 times the number 6 in 50 trials.
Step-by-step explanation:
Let us consider a success to get a 6. In this case, note that the probability of having a 6 in one spin is 1/6. We can consider the number of 6's in 50 spins to be a binomial random variable. Then, let X to be the number of trials we get a 6 out of 50 trials. Then, we have the following model.

We will estimate the number of times that she spins a 6 as the expected value of this random variable.
Recall that if we have X as a binomial random variable of n trials with a probability of success of p, then it's expected value is np.
Then , in this case, with n=50 and p=1/6 we expect to have
number of times of having a 6, which is 8.33.