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3 years ago

For which function defined by a polynomial are the zeros of the polynomial -4 and -6?

Mathematics

1 answer:

V125BC [204]3 years ago

4 0

If the zeros are x=-4 and -6, then the factors are:

(x+4)(x+6) which expanded is:

f(x)=x^2+10x+24

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Given that (x - 2) is a factor of this polynomial, use the Factor Theorem to find the value of a.

IceJOKER [234]

The factor theorem states that if (x-a) is a factor of P(x) the P(a) = 0 so we can write, for this polynomial:

2^(4) - 3(2)^3 + A(2)^2 - 6(2) + 14 = 0

16 - 24 + 4A - 12 + 14 = 0
4A = -16+24 + 12 - 14 = 6
A = 6/4 = 1.5

8 0

4 years ago

A type of long-range radio transmits data across the Atlantic Ocean. The number of errors in the transmission during any given a

Marianna [84]

Answer:

32.33% probability of having at least 3 erros in an hour.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

The mean number of errors is 2 per hour.

This means that $\mu = 2$

(a) What is the probability of having at least 3 errors in an hour?

Either you have 2 or less errors in an hour, or we have at least 3 errors. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 2) + P(X \geq 3) = 1$

We want $P(X \geq 3)$

$P(X \geq 3) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-2}*(2)^{0}}{(0)!} = 0.1353$

$P(X = 1) = \frac{e^{-2}*(2)^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*(2)^{2}}{(2)!} = 0.2707$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1353 + 0.2707 + 0.2707 = 0.6767$

$P(X \geq 3) = 1 - P(X \leq 2) = 1 - 0.6767 = 0.3233$

32.33% probability of having at least 3 erros in an hour.

6 0

3 years ago

Which of the statements below are true for linear functions? Select all that apply.

Sati [7]

Answer:

The general equation is y=mx+b

the graph contains a vertex

the slop of the graph is constant and can be defined as rise over run

8 0

3 years ago

Read 2 more answers

without using your calculator or computer graphing tool. match each function rule with its graph. In each case, write a sentence

valentinak56 [21]

Well I know that c. goes with I. just because I saw so many x^3 graphs.

For a. the parent function is x^2, so 0.5x^2 should look about the same as x^2, so the answer is VI.

4 0

3 years ago

Please help me to prove this! I need is no.(c). So, please help me do it.

zloy xaker [14]

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = 90° → A + B = 90° - C

→ C = 90° - (A + B)

Use the Double Angle Identity: cos 2A = 1 - 2 sin² A

→ sin² A = (1 - cos 2A)/2

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use the Product to Sum Identity: cos (A - B) - cos (A + B) = 2 sin A · sin B

Use the Cofunction Identities: cos (90° - A) = sin A

sin (90° - A) = cos A

Proof LHS → RHS:

LHS: sin² A + sin² B + sin² C

$\text{Double Angle:}\qquad \dfrac{1-\cos 2A}{2}+\dfrac{1-\cos 2B}{2}+\sin^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2-\cos 2A-\cos 2B\bigg)+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\sin^2 C$

$\text{Sum to Product:}\quad 1-\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\cos (A+B)\cdot \cos (A-B)+\sin^2 C$

Given: 1 - cos (90° - C) · cos (A - B) + sin² C

Cofunction: 1 - sin C · cos (A - B) + sin² C

Factor: 1 - sin C [cos (A - B) + sin C]

Given: 1 - sin C[cos (A - B) - sin (90° - (A + B))]

Cofunction: 1 - sin C[cos (A - B) - cos (A + B)]

Sum to Product: 1 - sin C [2 sin A · sin B]

= 1 - 2 sin A · sin B · sin C

LHS = RHS: 1 - 2 sin A · sin B · sin C = 1 - 2 sin A · sin B · sin C $\checkmark$

6 0

3 years ago