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o-na [289]
3 years ago
9

according to the fundemental theorem of algebra, how many roots exist for the polynomial function? f(x) = (x^3-3x+1)^2

Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
3 0

Answer:

6

Step-by-step explanation:

First, we can expand the function to get its expanded form and to figure out what degree it is. For a polynomial function with one variable, the degree is the largest exponent value (once fully expanded/simplified) of the entire function that is connected to a variable. For example, x²+1 has a degree of 2, as 2 is the largest exponent value connected to a variable. Similarly, x³+2^5 has a degree of 2 as 5 is not an exponent value connected to a variable.

Expanding, we get

(x³-3x+1)²  = (x³-3x+1)(x³-3x+1)

= x^6 - 3x^4 +x³ - 3x^4 +9x²-3x + x³-3x+1

= x^6 - 6x^4 + 2x³ +9x²-6x + 1

In this function, the largest exponential value connected to the variable, x, is 6. Therefore, this is to the 6th degree. The fundamental theorem of algebra states that a polynomial of degree n has n roots, and as this is of degree 6, this has 6 roots

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In a random sample of 30 people who rode a roller coaster one day, the mean wait time is 46.7 minutes with a standard deviation
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Answer: C. (29,\ 37.8)

Step-by-step explanation:

The confidence interval for difference of two population mean is given by :-

\overline{x}_1-\overline{x}_2\pm z_{\alpha/2}\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}

Given : Level of significance : 1-\alpha:0.99

Then , significance level : \alpha: 1-0.99=0.01

Critical value : z_{\alpha/2}=z_{0.005}=2.576

n_1=30\ ;\ n_2=50\\\\\overline{x}_1=46.7\ ;\ \overline{x}_2=13.3\\\\s_1=9.2\ ;\ s_2=1.9

46.7-13.3\pm(2.576)\sqrt{\dfrac{9.2^2}{30}+\dfrac{1.9^2}{50}}\approx33.4\pm4.38=(29.02\ ,37.78)\approx(29,\ 37.8)

Hence,  a 99% confidence interval for the difference between the mean wait times of everyone who rode both rides (29,\ 37.8)

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A certain freely falling object, released from rest, requires 1.50 s to travel the last 30.0 m before it hit the ground. a.Find
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A.)
<span>s= 30m
u = ? ( initial velocity of the object )
a = 9.81 m/s^2 ( accn of free fall )
t = 1.5 s
s = ut + 1/2 at^2 
\[u = \frac{ S - 1/2 a t^2 }{ t }\]
\[u = \frac{ 30 - ( 0.5 \times 9.81 \times 1.5^2) }{ 1.5 } \]
\[u = 12.6 m/s\]
</span>
b.)
<span>s = ut + 1/2 a t^2
u = 0 ,
s = 1/2 a t^2
\[s = \frac{ 1 }{ 2 } \times a \times t ^{2}\]
\[s = \frac{ 1 }{ 2 } \times 9.81 \times \left( \frac{ 12.6 }{ 9.81 } \right)^{2}\]
\[s = 8.0917...\]
\[therfore total distance = 8.0917 + 30 = 38.0917.. = 38.1 m \] </span>
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