Thet contradict each other, that's why both of them are incorrect.
<span>Suppose that a polynomial has four roots: s, t, u, and v. If the polynomial were evaluated at any of these values, it would have to be zero. Therefore, the polynomial can be written in this form.
p(x)(x - s)(x - t)(x - u)(x - v), where p(x) is some non-zero polynomial
This polynomial has a degree of at least 4. It therefore cannot be cubic.
Now prove Kelsey correct. We have already proved that there can be no more than three roots. To prove that a cubic polynomial with three roots is possible, all we have to do is offer a single example of that. This one will do.
(x - 1)(x - 2)(x - 3)
This is a cubic polynomial with three roots, and four or more roots are not possible for a cubic polynomial. Kelsey is correct.
Incidentally, if this is a roller coaster we are discussing, then a cubic polynomial is not such a good idea, either for a vertical curve or a horizontal curve. I hope this helps</span><span>
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Part A:
A component is one voter's vote. An outcome is a vote in favour of our candidate.
Since there are 100 voters, we can stimulate the component by using two randon digits from 00 - 99, where the digits 00 - 54 represents a vote for our candidate and the digits 55 - 99 represents a vote for the underdog.
Part B:
A trial is 100 votes. We can stimulate the trial by randomly picking 100 two-digits numbers from 00 - 99. Whoever gets the majority of the votes wins the trial.
Part C:
The response variable is whether the underdog wants to win or not. To calculate the experimental probability, divide the number of trials in which the simulated underdog wins by the total number of trials.
Answer:
y= 3/2x+40
Step-by-step explanation:
y= rise/run+ y-intercept
F(x)=x^2+3x+5
f(3+h)=(3+h)^2+3(3+h)+5
f(3+h)=9+6h+h^2+9+3h+5
f(3+h)=23+9h+h^2
The value of t that makes the factor e^(.032t) have the value of 2 can be found using logarithms.
2 = e^(0.032t)
ln(2) = ln(e^(0.032t)) = 0.032t
t = ln(2)/0.032 ≈ 21.66
It would take 21.66 years for the cost to double.
