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2 years ago

Which polynomial function has a leading coefficient of 1 and roots (7 + i) and (5 – i) with multiplicity 1?

Mathematics

2 answers:

svlad2 [7]2 years ago

3 0

Answer:

C. $f(x)=(x-(7-i))(x-(5+i))(x-(7+i))(x-(5-i))$

Step-by-step explanation:

We want to find the equation of a polynomial the following properties;

i. Leading coefficient is 1

ii. roots (7 + i) and (5 – i) with multiplicity 1

Recall the complex conjugate properties of the roots of a polynomial.

According to this property, if

$a+bi$ is a root of a polynomial, then the complex conjugate, $a-bi$ is also a root.

This means that:

(7 - i) and (5 + i) with multiplicity 1 are also roots of this polynomial.

The complete set of roots are:

$x=(7+i),x=(7-i),x=(5-i),x=(5+i)$

Therefore the polynomial is:

$f(x)=(x-(7-i))(x-(5+i))(x-(7+i))(x-(5-i))$

The correct choice is C.

vovikov84 [41]2 years ago

3 0

Answer:

C. f(x)= (x-(7-i)) (x-(5+i))(x-(7+i))(x-(5-i)

Step-by-step explanation:

edge 2020

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You offer senior citizens a 20% discount on their tune-ups at your gas station. Assuming that an average of 50 senior citizens g

andreyandreev [35.5K]

Percentage of discount given to senior citizens for tune-ups at the gas station = 20%
Average number of senior citizens visiting the gas station per month = 50
Average price of tune-up before discount = $49.95
Total amount that would
have been collected before discount = 49.95 * 50
= 2497.50 dollars
Total amount of discount given = (20/100) * 2497.50
= 499.50 dollars
I hope that the procedure is clear enough for you to understand.

6 0

2 years ago

Ronnie wants to lower his utility bills to save money on his monthly expenses. He discovers that if he replaces his water heater

Soloha48 [4]

Alright, lets get started.

Last year, Ronnie's utility expenses were = 1935.67 $

After switching to electric water heater to solar water heater, his yearly utility bills get reduced.

This year, Ronnie's utility expenses are = 862.40 $

So the difference on utility bills Ronnie saved = $1935.67 - 862.40$

so, difference Ronnie saved = 1073.27 $

So, the percentage of amount Ronnie saved = $\frac{1073.27}{1935.67} *100$

Percentage of amount Ronnie saved = 55.45 %

Hence, 55.45 % Ronnie save on his utlility bills by switching to a solar water heater. : Answer

Hope it will help :)

5 0

2 years ago

Read 2 more answers

NOT is coded as LKF and FLY is coded as TNA, then which of the following word/s has/have correct coding? (1) TOP – FKJ (2) RUN –

AVprozaik [17]

Answer:

(1) and (3) are correct options

Step-by-step explanation:

NOT is coded as LKF
FLY is coded as TNA

Using alphabet, we can see that:

NOT = 14, 15, 20 ⇔ 13+1, 13+2, 13+7 ⇒ LKF = 12,11,6 = 13-1, 13-2, 13-7

Coding for each letter is 13 + x ⇒ 13 - x

(1) TOP ⇒ FKJ

TOP = 20,15,16 ⇒ 13-7, 13-2, 13-3 = 6,11,10 = FKJ
Correct

(2) RUN ⇒ IFM

RUN = 18,21,14 ⇒ 13-5,13-8,13-1 = 8,5,12 = HEL ≠ IFM
Incorrect

(3) MUG ⇒ MES

MUG = 13,21,7 ⇒ 13+0, 13-8, 13+6 = 13,5,19 = MES
Correct

(4) HOT ⇒ RKG

HOT = 8,15,20 ⇒ 13+5,13-2, 13-7 = 18,11,6 = RKF ≠ RKG
Incorrect

7 0

1 year ago

Jason grew from 36 inches to 40 inches in 1 year. By percent did his growth increase? Round your answer off to the nearest tenth

IrinaVladis [17]

Answer:

33%

Step-by-step explanation:

7 0

2 years ago

One of the industrial robots designed by a leading producer of servomechanisms has four major components. Components’ reliabilit

Ivahew [28]

Answer:

a) Reliability of the Robot = 0.7876

b1) Component 1: 0.8034

Component 2: 0.8270

Component 3: 0.8349

Component 4: 0.8664

b2) Component 4 should get the backup in order to achieve the highest reliability.

c) Component 4 should get the backup with a reliability of 0.92, to obtain the highest overall reliability i.e. 0.8681.

Step-by-step explanation:

Component Reliabilities:

Component 1 (R1) : 0.98

Component 2 (R2) : 0.95

Component 3 (R3) : 0.94

Component 4 (R4) : 0.90

a) Reliability of the robot can be calculated by considering the reliabilities of all the components which are used to design the robot.

Reliability of the Robot = R1 x R2 x R3 x R4

= 0.98 x 0.95 x 0.94 x 0.90

Reliability of the Robot = 0.787626 ≅ 0.7876

b1) Since only one backup can be added at a time and the reliability of that backup component is the same as the original one, we will consider the backups of each of the components one by one:

Reliability of the Robot with backup of component 1 can be computed by first finding out the chance of failure of the component along with its backup:

Chance of failure = 1 - reliability of component 1

= 1 - 0.98

= 0.02

Chance of failure of component 1 along with its backup = 0.02 x 0.02 = 0.0004

So, the reliability of component 1 and its backup (R1B) = 1 - 0.0004 = 0.9996

Reliability of the Robot = R1B x R2 x R3 x R4

= 0.9996 x 0.95 x 0.94 x 0.90

Reliability of the Robot = 0.8034

Similarly, to find out the reliability of component 2:

Chance of failure of component 2 = 1 - 0.95 = 0.05

Chance of failure of component 2 and its backup = 0.05 x 0.05 = 0.0025

Reliability of component 2 and its backup (R2B) = 1 - 0.0025 = 0.9975

Reliability of the Robot = R1 x R2B x R3 x R4

= 0.98 x 0.9975 x 0.94 x 0.90

Reliability of the Robot = 0.8270

Reliability of the Robot with backup of component 3 can be computed as:

Chance of failure of component 3 = 1 - 0.94 = 0.06

Chance of failure of component 3 and its backup = 0.06 x 0.06 = 0.0036

Reliability of component 3 and its backup (R3B) = 1 - 0.0036 = 0.9964

Reliability of the Robot = R1 x R2 x R3B x R4

= 0.98 x 0.95 x 0.9964 x 0.90

Reliability of the Robot = 0.8349

Reliability of the Robot with backup of component 4 can be computed as:

Chance of failure of component 4 = 1 - 0.90 = 0.10

Chance of failure of component 4 and its backup = 0.10 x 0.10 = 0.01

Reliability of component 4 and its backup (R4B) = 1 - 0.01 = 0.99

Reliability of the Robot = R1 x R2 x R3 x R4B

= 0.98 x 0.95 x 0.94 x 0.99

Reliability of the Robot = 0.8664

b2) According to the calculated values, the highest reliability can be achieved by adding a backup of component 4 with a value of 0.8664. So, Component 4 should get the backup in order to achieve the highest reliability.

c) 0.92 reliability means the chance of failure = 1 - 0.92 = 0.08

We know the chances of failure of each of the individual components. The chances of failure of the components along with the backup can be computed as:

Component 1 = 0.02 x 0.08 = 0.0016

Component 2 = 0.05 x 0.08 = 0.0040

Component 3 = 0.06 x 0.08 = 0.0048

Component 4 = 0.10 x 0.08 = 0.0080

So, the reliability for each of the component & its backup is:

Component 1 (R1BB) = 1 - 0.0016 = 0.9984

Component 2 (R2BB) = 1 - 0.0040 = 0.9960

Component 3 (R3BB) = 1 - 0.0048 = 0.9952

Component 4 (R4BB) = 1 - 0.0080 = 0.9920

The reliability of the robot with backups for each of the components can be computed as:

Reliability with Component 1 Backup = R1BB x R2 x R3 x R4

= 0.9984 x 0.95 x 0.94 x 0.90

Reliability with Component 1 Backup = 0.8024

Reliability with Component 2 Backup = R1 x R2BB x R3 x R4

= 0.98 x 0.9960 x 0.94 x 0.90

Reliability with Component 2 Backup = 0.8258

Reliability with Component 3 Backup = R1 x R2 x R3BB x R4

= 0.98 x 0.95 x 0.9952 x 0.90

Reliability with Component 3 Backup = 0.8339

Reliability with Component 4 Backup = R1 x R2 x R3 x R4BB

= 0.98 x 0.95 x 0.94 x 0.9920

Reliability with Component 4 Backup = 0.8681

Component 4 should get the backup with a reliability of 0.92, to obtain the highest overall reliability i.e. 0.8681.

4 0

2 years ago