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BabaBlast [244]
3 years ago
9

Three roots of the polynomial equation x4 - 4x3 - 2x2 + 12x + 9 = 0 are 3, -1 and -1. Explain why the fourth root must be a real

number. Find the fourth root.
Mathematics
1 answer:
LiRa [457]3 years ago
3 0
Descares Theorem says that the number of roots of a polynomial is equal to its highest exponent, and the roots could be real and/or imaginary (complex).

However is their is one complex root (a + i.b), it's conjugate is also
a root (a-i.b).

In the equation x⁴ - 4x³ - 2x² + 12x + 9 = 0, we have already 3 roots, then the 4rth cannot be a complex one since we don't have a fifth for its conjugate.

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2(x+3)^2+2(y)^2=32 what is the center point of the circle and how long is the radius
spayn [35]

Answer:

  • center (-3, 0)
  • radius 4

Step-by-step explanation:

The standard form equation of a circle is ...

  (x -h)^2 +(y -k)^2 = r^2 . . . . . . . . . center at (h, k), radius r

Divide your given equation by 2 to put it into standard form:

  (x +3)^2 +y^2 = 16

Comparing to the above, we see ...

  • h = -3
  • k = 0
  • r = √16 = 4

The center point of the circle is (-3, 0); the radius is 4 units long.

6 0
3 years ago
Evaluate the expression y=4. 5+y+8<br><br> A)72<br> B)21<br> C)17<br> D)13
Elden [556K]

Answer:

C)17

Step-by-step explanation:

5+y+8

let  y=4

5+4+8

9+8

17

3 0
3 years ago
Read 2 more answers
The expected number of typographical errors on a page of a certain magazine is .2. What is the probability that an article of 10
Pavel [41]

Answer:

a) The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

Step-by-step explanation:

Given : The expected number of typographical errors on a page of a certain magazine is 0.2.

To find : What is the probability that an article of 10 pages contains

(a) 0 and (b) 2 or more typographical errors?

Solution :

Applying Poisson distribution,

N\sim Pois(0.2)

P(N=r)=\frac{e^{-np}(np)^r}{r!}

where, n is the number of words in a page

and p is the probability of every word with typographical errors.

Here, n=10 and E(N)=np=0.2

a) The probability that an article of 10 pages contains 0 typographical errors.

Substitute r=0 in formula,

P(N=0)=\frac{e^{-0.2}(0.2)^0}{0!}

P(N=0)=\frac{e^{-0.2}}{1}

P(N=0)=e^{-0.2}

P(N=0)=0.8187

The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors.

Substitute r\geq 2 in formula,

P(N\geq 2)=1-P(N

P(N\geq 2)=1-[P(N=0)+P(N=1)]

P(N\geq 2)=1-[\frac{e^{-0.2}(0.2)^0}{0!}+\frac{e^{-0.2}(0.2)^1}{1!}]

P(N\geq 2)=1-[e^{-0.2}+e^{-0.2}(0.2)]

P(N\geq 2)=1-[0.8187+0.1637]

P(N\geq 2)=1-0.9825

P(N\geq 2)=0.0175

The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

6 0
3 years ago
1 inch = 2.54cm 1 mile = 1.6km a map has a scale of 1 inch represents 1 mile use the given conversions to show that 1cm on the m
lara31 [8.8K]
Note: 1 inch = 2.54 cm is an exact converion, while 1 mile=1.6km is only approximate.

Using conversion
1 inch=2.54 cm
1 mile = 1760*3*12 = 63360 in = 160934.4 cm 
If 1 inch = 1 mile, then 
2.54 cm : 1 mile = 1.609344 km
1 cm : 0.6336 km   (exactly), or
1 cm : 0.6 km (approximately)

Using both conversions:
1 inch = 2.54 cm : 1 mile = 1.6 km
=>
2.54 cm : 1.6 km
1 cm : 1.6/2.54=0.6299 km (approximately), or
1 cm : 0.6 km (approximately)
8 0
3 years ago
Please help! multiple choice! will give brainliest!
kipiarov [429]

Answer:

D

Step-by-step explanation:

8 0
3 years ago
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