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Delicious77 [7]
3 years ago
11

Which polynomial function has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1? f(x) = (x – 2i)(x – 3i) f(x) =

(x + 2i)(x + 3i) f(x) = (x – 2)(x – 3)(x – 2i)(x – 3i) f(x) = (x + 2i)(x + 3i)(x – 2i)(x – 3i)
Mathematics
2 answers:
kykrilka [37]3 years ago
8 0

Answer:

P(x)=(x-2i)(x-3i)

Step-by-step explanation:

Build a Polynomial Knowing its Roots

If we know a polynomial has roots x1, x2, ..., xn, then it can be expressed as:

P(x)=a(x-x1)(x-x2)...(x-xn)

Where a is the leading coefficient.

Note the roots appear with their signs changed in the polynomial.

If the polynomial has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1, then:

P(x)=1(x-2i)(x-3i)

\mathbf{P(x)=(x-2i)(x-3i)}

jeyben [28]3 years ago
4 0

Answer:

a

Step-by-step explanation:

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Step-by-step explanation:

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6 0
3 years ago
Consider a system with one component that is subject to failure, and suppose that we have 115 copies of the component. Suppose f
castortr0y [4]

Answer:

Step-by-step explanation:

From the given information:

the mean (\mu) = 115 \times 20

= 2300

Standard deviation = 20 \times \sqrt{115}

Standard deviation (SD) = 214.4761

TO find:

a) P(x > 3500)= P(Z > \dfrac{3500-\mu}{214.4761})

P(x > 3500)= P(Z > \dfrac{3500-2300}{214.4761})

P(x > 3500)= P(Z > \dfrac{1200}{214.4761})

P(x > 3500)= P(Z >5.595)

From the Z-table, since 5.595 is > 3.999

P(x > 3500)=1-0.9999

P(x > 3500) = 0.0001

b)

Here, the replacement time for the mean (\mu) = \dfrac{0+0.5}{2}

= 0.25

Replacement time for the Standard deviation \sigma = \dfrac{0.5-0}{\sqrt{12}}

\sigma = 0.1443

For 115 component, the mean time = (115 × 20)+(114×0.25)

= 2300 + 28.5

= 2328.5

Standard deviation = \sqrt{(115\times 20^2) +(114\times (0.1443)^2)}

= \sqrt{(115\times 400) +(114\times 0.02082249}

= \sqrt{(46000) +2.37376386}

= \sqrt{(46000) +(2.37376386)}

= \sqrt{46002.374}

= 214.482

Now; the required probability:

P(x > 4125) = P(Z > \dfrac{4125- 2328.5}{214.482})

P(x > 4125) = P(Z > \dfrac{1796.5}{214.482})

P(x > 4125) = P(Z >8.376)

P(x > 4125) =1-  P(Z

From the Z-table, since 8.376 is > 3.999

P(x > 4125) = 1 - 0.9999

P(x > 4125) = 0.0001

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3 years ago
Choose the answer that validates that the rate of change is constant by showing that the ratios of the two quantities are propor
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Given:

The table of values is

Number of Students   :  7    14   21    28

Number of Textbooks : 35  70  105  140

To find:

The rate of change and showing that the ratios of the two quantities are proportional and equivalent to the unit rate.

Solution:

The ratio of number of textbooks to number of students are

\dfrac{35}{7}=5

\dfrac{70}{14}=5

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All the ratios of the two quantities are proportional and equivalent to the unit rate.

Let y be the number of textbooks and x be the number of students, then

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Hence the rate of change is constant that is 5.

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Step-by-step explanation:

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