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3 years ago

What is the multiplicity of the root 2 in the equation x^3-8x^2+20x-16=0?

1 answer:

mote1985 [20]3 years ago

8 0

Factoring the cubic you get: <span>(x - 2)^2 (x-4) , so we know the multiplicity is the amount of times a root shows in an equation because the equation (x-2) has a exponent of 2 that means the root 2 has a multiplicity of 2.</span>
Answer: 2

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What is the vertex of the graph of y = 1/3 (x-9)2 + 5 Question 1 options: (9,5) (3,3) (3,5) (-9,5)

svlad2 [7]

ANSWER

The vertex of the graph of
$y = \frac{1}{3} {(x - 9)}^{2} + 5$
is

$(9,5)$

EXPLANATION

The vertex form of a parabola is given by

$y = a {(x - h)}^{2} + k$

where
$V(h,k)$
is the vertex of the parabola.

The function given to us is

$y = \frac{1}{3} {(x - 9)}^{2} + 5$
This is already in the vertex form.

When we compare this to the general vertex form, we have,

$a = \frac{1}{3}$

$h = 9$
and

$k = 5$

Therefore the vertex of the parabola is

$V(9,5)$

Hence the correct answer is option A.

7 0

3 years ago

Read 2 more answers

Least to greatest again? I can't visualize this one.. -11/5, -2.4, 1.6, 15/10, -2.25 Thanks

Tcecarenko [31]

Answer:

Hey again! Just remember about the number lines. If it's easier, you can use a calculator to divide the fractions to make them easier to visualize in decimal form.

The answer to this one is:

-2.4

-2.25

-11/5 (which is -2.2)

-15/10 (1.5)

-1.6

6 0

3 years ago

In analyzing the city and highway fuel efficiencies of many cars and trucks, the mean difference in fuel efficiencies for the 60

Olegator [25]

Answer:

$7.38-1.96\frac{2.51}{\sqrt{601}}=7.18$

$7.38+1.96\frac{2.51}{\sqrt{601}}=7.58$

For this case the confidence interval is given by :

$7.18 \leq \mu \leq 7.58$

Step-by-step explanation:

Data provided

$\bar X=7.38$ represent the sample mean for the fuel efficiencies

$\mu$ population mean

s=2.51 represent the sample standard deviation

n=601 represent the sample size

Confidence interval

The formula for the confidence interval of the true mean is given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom for this case is given by:

$df=n-1=601-1=600$

The Confidence level for this case is 0.95 or 95%, and the significance level $\alpha=0.05$ and $\alpha/2 =0.025$ and the critical value is given by $t_{\alpha/2}=1.96$