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3 years ago

Find the Polynomial with roots 3, -2, and 0. Urgent!!! Help

Mathematics

2 answers:

777dan777 [17]3 years ago

5 0

Answer:

Step-by-step explanation:

scoundrel [369]3 years ago

4 0

Answer:

f(x) = x³ - x² - 6x

Step-by-step explanation:

If a polynomial f(x) has roots x = a and x = b then the factors are

(x - a) and (x - b) and the polynomial is the product of the factors, that is

f(x) = (x - a)(x - b)

Here the roots are x = 3, x = - 2 and x = 0, thus

(x - 3), (x - (- 2)) and (x - 0) are the factors, that is

(x - 3), (x + 2) and x , hence the polynomial is

f(x) = x(x - 3)(x + 2) ← expand the pair of factors

= x(x² - x - 6) ← distribute parenthesis by x

= x³ - x² - 6x

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Nezavi [6.7K]

Answer:

−7≤a≤8

Step-by-step explanation:

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3 years ago

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3 years ago

Dude is making 12 pounds of nut mixture with macadamia nuts and almonds. Macadamia nuts cost $9 per pound and almonds cost $5.25

pshichka [43]

Number of pounds of macadamia nuts is 8 pounds and number of pounds of almonds is 4 pounds.

Step-by-step explanation:

Step 1:

Given total pounds of mixture = 12 pounds, cost of macadamia nuts per pound = $9, cost of almonds per pound = $5.25, total cost of mixture per pound = $7.75.

Let number of pounds of macadamia nuts be x and number of pounds of almonds be 12-x.

Step 2:

Form an equation using the above information.

⇒ 9x + 5.25 (12-x) = 12 × 7.75

⇒ 9x + 63 - 5.25x = 93

⇒ 9x - 5.25x = 30

⇒ 3.75x = 30

⇒ x = 8

Number of macadamia nuts is 8 pounds.

Step 3:

Calculate number pounds of almonds

⇒ Number of pounds of almonds = 12 - x = 4 pounds.

5 0

3 years ago

The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection.

This means that,

Probability that at least one of them needs replacement at the first inspection =

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$

5 0

3 years ago

If the graph of the function h is defined by = h(x)+x^2+9 is translated vertically downward by 9 units, it becomes the graph of

natulia [17]

The answer is g(x) = x².
Solution:
The graph of h(x) = x²+9 translated vertically downward by 9 units means that each point (x, h(x)) is shifted onto the point (x, h(x) - 9), that is,
(x, h(x)) → (x, h(x) - 9)
The translated graph that represents the function is defined by the expression for g(x):
g(x) = h(x) - 9 = x² + 9 - 9 = x²

h(x) = x²+9 → g(x) = x² shows that the graph of the equation g(x) = x² moves the graph of h(x) = x²+9 down nine units.

6 0

3 years ago