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3 years ago

Find a thrid degree polynomial equation with rational coeffcients that has roots -4 and 6 +i

1 answer:

4 0

For the coefficients to be rational, any complex roots must occur in conjugate pairs. So if $6+i$ is a root, then so must be $6-i$ .

Now such a third degree polynomial might be

$(x+4)(x-(6+i))(x-(6-i))=(x+4)(x^2-12x+37)=x^3-8x^2-11x+148$

The only variation to this would be multiplying throughout by some non-zero constant. This doesn't change the roots.

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can somebody please write what are the rules for multiplying integers and the rules for dividing integers in a short answer?

arsen [322]

Answer:

Negative + Negative = Positive

Negative + Positive = Negative

Positive + Negative = Negative

Positive + Positive = Negative

The rules are the same for division.

Step-by-step explanation:

4 0

3 years ago

When factoring the equation 9x2+9x+2 what two numbers will the "9x" term be broken into?

Andrew [12]

Answer:

it will be broken into 9*what ever x is

Hope This Helps!!!

6 0

2 years ago

Read 2 more answers

A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. (a) What must be true

valentina_108 [34]

Answer:

(a) The distribution of the sample mean ( $\bar x$ ) is N (68, 4.74²).

(b) The value of $P(\bar X$ is 0.7642.

Step-by-step explanation:

A random sample of size n = 10 is selected from a population.

Let the population be made up of the random variable X.

The mean and standard deviation of X are:

$\mu=68\\\sigma=15$

(a)

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. n = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

$\mu_{\bar x}=\mu=68$

And the standard deviation of the distribution of the sample mean is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74$

Thus, the distribution of the sample mean ( $\bar x$ ) is N (68, 4.74²).

(b)

Compute the value of $P(\bar X$ as follows:

$P(\bar X$

$=P(Z$

*Use a z-table for the probability.

Thus, the value of $P(\bar X$ is 0.7642.

(c)

Compute the value of $P(\bar X\geq 69.1)$ as follows:

Apply continuity correction as follows:

$P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)$

$=P(\bar X>69.6)$

$=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})$

$=P(Z>0.34)\\=1-P(Z$

Thus, the value of $P(\bar X\geq 69.1)$ is 0.3670.

7 0

3 years ago

WILL MARK BRAINLIEST! Shelly served the volleyball over the net 17 out of 25 times. What is her percentage of serving the ball o

Tju [1.3M]

Answer

68℅

Explanation

17/25 =

17 ÷ 25 =

0.68 =

0.68 × 100/100 =

0.68 × 100% =

(0.68 × 100)% =

68%;

Hope that this helps you and have a great day :)

8 0

3 years ago

Read 2 more answers

The arch beneath a bridge is semi-elliptical, a one-way roadway passes under the arch. The width of the roadway is 38 feet and

forsale [732]

Answer:

Only truck 1 can pass under the bridge.

Step-by-step explanation:

So, first of all, we must do a drawing of what the situation looks like (see attached picture).

Next, we can take the general equation of an ellipse that is centered at the origin, which is the following:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}$

where:

a= wider side of the ellipse

b= shorter side of the ellipse

in this case:

$a=\frac{38}{2}=19ft$

and

b=12ft

so we can go ahead and plug this data into the ellipse formula:

$\frac{x^2}{(19)^2}+\frac{y^2}{(12)^2}$

and we can simplify the equation, so we get:

$\frac{x^2}{361}+\frac{y^2}{144}$

So, we need to know if either truk will pass under the bridge, so we will match the center of the bridge with the center of each truck and see if the height of the bridge is enough for either to pass.

in order to do this let's solve the equation for y:

$\frac{y^{2}}{144}=1-\frac{x^{2}}{361}$

$y^{2}=144(1-\frac{x^{2}}{361})$

we can add everything inside parenthesis so we get:

$y^{2}=144(\frac{361-x^{2}}{361})$

and take the square root on both sides, so we get:

$y=\sqrt{144(\frac{361-x^{2}}{361})}$

and we can simplify this so we get:

$y=\frac{12}{19}\sqrt{361-x^{2}}$

and now we can evaluate this equation for x=4 (half the width of the trucks) so:

$y=\frac{12}{19}\sqrt{361-(8)^{2}}$

y=11.73ft

this means that for the trucks to pass under the bridge they must have a maximum height of 11.73ft, therefore only truck 1 is able to pass under the bridge since truck 2 is too high.

5 0

3 years ago