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3 years ago

Please show work I just need long division, I got synthetic division done .

Mathematics

2 answers:

romanna [79]3 years ago

6 0

Answer:

(10x^4-50x^3-800)/(x-6)

Step-by-step explanation:

Images with work attached

Maslowich3 years ago

3 0

Answer:

10 • (x^4 - 5x^3 - 80)

————————————————————

x - 6

Step-by-step explanation:

STEP 1:

Equation at the end of step 1

STEP 2:

Equation at the end of step 2

STEP 3:

10x4 - 50x3 - 800

Simplify —————————————————

x - 6

STEP 4:

Pulling out like terms

Pull out like factors :

10x^4 - 50x^3 - 800 = 10 • (x^4 - 5x^3 - 80)

Polynomial Roots Calculator :

Find roots (zeroes) of : F(x) = x^4 - 5x^3 - 80

Polynomial Roots Calculator is a set of methods aimed at finding values of x for which F(x)=0

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers x which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient

In this case, the Leading Coefficient is 1 and the Trailing Constant is -80.

The factor(s) are:

of the Leading Coefficient : 1

of the Trailing Constant : 1 ,2 ,4 ,5 ,8 ,10 ,16 ,20 ,40 ,80

P Q P/Q F(P/Q) Divisor

-1 1 -1.00 -74.00

-2 1 -2.00 -24.00

-4 1 -4.00 496.00

-5 1 -5.00 1170.00

-8 1 -8.00 6576.00

Polynomial Roots Calculator found no rational roots

Polynomial Long Division :

Dividing : x^4 - 5x^3 - 80

("Dividend")

By : x - 6 ("Divisor")

dividend x^4 - 5x^3 - 80

- divisor * x^3 x^4 - 6x^3

remainder x^3 - 80

- divisor * x^2 x^3 - 6x^2

remainder 6x^2 - 80

- divisor * 6x^1 6x^2 - 36x

remainder 36x - 80

- divisor * 36x^0 36x - 216

remainder 136

Quotient : x^3 + x^2 + 6x + 36

Remainder : 136

= 10 • (x^4 - 5x^3 - 80)

————————————————————

x - 6

Hope this helps, have a nice day/night! :D

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The average number of annual trips per family to amusement parks in the UnitedStates is Poisson distributed, with a mean of 0.6

IrinaK [193]

Answer:

a) 0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.

b) 0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.

c) 0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.

d) 0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.

e) 0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.

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 interval.

Poisson distributed, with a mean of 0.6 trips per year

This means that $\mu = 0.6n$ , in which n is the number of years.

a.The family did not make a trip to an amusement park last year.

This is P(X = 0) when n = 1, so $\mu = 0.6$ .

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

$P(X = 0) = \frac{e^{-0.6}*(0.6)^{0}}{(0)!} = 0.5488$

0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.

b.The family took exactly one trip to an amusement park last year.

This is P(X = 1) when n = 1, so $\mu = 0.6$ .

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

$P(X = 1) = \frac{e^{-0.6}*(0.6)^{1}}{(1)!} = 0.3293$

0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.

c.The family took two or more trips to amusement parks last year.

Either the family took less than two trips, or it took two or more trips. So

$P(X < 2) + P(X \geq 2) = 1$

We want

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1) = 0.5488 + 0.3293 = 0.8781$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.8781 = 0.1219$

0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.

d.The family took three or fewer trips to amusement parks over a three-year period.

Three years, so $\mu = 0.6(3) = 1.8$ .

This is

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = \frac{e^{-1.8}*(1.8)^{0}}{(0)!} = 0.1653$

$P(X = 1) = \frac{e^{-1.8}*(1.8)^{1}}{(1)!} = 0.2975$

$P(X = 2) = \frac{e^{-1.8}*(1.8)^{2}}{(2)!} = 0.2678$

$P(X = 3) = \frac{e^{-1.8}*(1.8)^{3}}{(3)!} = 0.1607$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1653 + 0.2975 + 0.2678 + 0.1607 = 0.8913$

0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.

e.The family took exactly four trips to amusement parks during a six-year period.

Six years, so $\mu = 0.6(6) = 3.6$ .

This is P(X = 4). So

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

$P(X = 4) = \frac{e^{-3.6}*(3.6)^{4}}{(4)!} = 0.1912$

0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.

4 0

3 years ago

Someone please help me ASAP

ivann1987 [24]

Answer:

the percentage share for BBC2 remained almost the same at about 11 % each year

if you look at the chart the BBC2 almost remains stable between 10 and 12 %

1980 ( between 39 and 51)

1985 ( between 37 and 49 ) and so on

( these numbers are not exactly the same , it is about or approximately)

7 0

4 years ago

Which point is a solution to the system of equations?<br> y = 3x +4<br> y =+2

Andru [333]

(-2/3, 2)

Solve for the first variable in one of the equations, then substitute the result into the other equation.

3 0

3 years ago

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Will mark Brainlest please answer. find the value of a,b. <br>,p,q from the equal order pairs

NARA [144]

Step-by-step explanation:

<h3>Question-1:</h3>

by order pair we obtain:

$\displaystyle \begin{cases} \displaystyle 3p = 2p - 1 \dots \dots i\\2q - p = 1 \dots \dots ii\end{cases}$

cancel 2p from the i equation to get a certain value of p:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q - p = 1 \end{cases}$

now substitute the value of p to the second equation:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q - ( - 1) = 1 \end{cases}$

simplify parentheses:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q + 1= 1 \end{cases}$

cancel 1 from both sides:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\2q = 0\end{cases}$

divide both sides by 2:

$\displaystyle \begin{cases} \displaystyle p = - 1 \\q = 0\end{cases}$

<h3>question-2:</h3>

by order pair we obtain:

$\displaystyle \begin{cases} \displaystyle 2x - y= 3 \dots \dots i\\3y= x + y \dots \dots ii\end{cases}$

cancel out y from the second equation:

$\displaystyle \begin{cases} \displaystyle 2x - y= 3 \dots \dots i\\ x = 2y \dots \dots ii\end{cases}$

substitute the value of x to the first equation:

$\displaystyle \begin{cases} \displaystyle 2.2y-y= 3 \\ x = 2y \end{cases}$

simplify:

$\displaystyle \begin{cases} \displaystyle 3y= 3 \\ x = 2y \end{cases}$

divide both sides by 3:

$\displaystyle \begin{cases} \displaystyle y= 1 \\ x = 2y \end{cases}$

substitute the value of y to the second equation which yields:

$\displaystyle \begin{cases} \displaystyle y= 1 \\ x = 2 \end{cases}$

<h3>Question-3:</h3>

by order pair we obtain;

$\displaystyle \begin{cases} \displaystyle 2p + q = 2 \dots \dots i\\3q + 2p = 3 \dots \dots ii\end{cases}$

rearrange:

$\displaystyle \begin{cases} \displaystyle 2p + q = 2 \\2p + 3q= 3 \end{cases}$

subtract and simplify

$\displaystyle \begin{array}{ccc} \displaystyle 2p + q = 2 \\2p + 3q= 3 \\ \hline - 2q = - 1 \\ q = \dfrac{1}{2} \end{array}$

substitute the value of q to the first equation:

$\displaystyle 2.p+ \frac{1}{2} = 2$

make q the subject of the equation:

$\displaystyle p = \frac{3}{4}$

hence,

$\displaystyle q = \frac{1}{2} \\ p = \frac{3}{4}$

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3 years ago

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What is the volume of a container with sides 1ft, 2ft, and 2.5 ft

almond37 [142]

Answer:

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Step-by-step explanation:

The formula for volume is length × width × height.

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