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Tasya [4]
3 years ago
10

How many real and complex roots exists for a polynomial?

Mathematics
1 answer:
solmaris [256]3 years ago
7 0

Step-by-step explanation:

whenever a complex number is a root of a polynomial with real coefficients, its complex conjugate is also a root of that polynomial. as an example, we'll find the roots of the polynomial..

x^5 - x^4 + x^3 - x^2 - 12x + 12.

the fifth-degree polynomial does indeed have five roots; three real, and two complex.

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polet [3.4K]
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3 inches is 1/4 of a foot
Number line would be divided into 12. The first spot would be 1/12, the next would be 2/12 and so on. Make a mark at 1/12 for one inch and at 3/12 for 3 inches
3 0
3 years ago
Please help! math! 10 points was the max:( <br> will give brainliest!!
Nezavi [6.7K]

It's gonna be an irrational number because the length of the hypotenuse is 6.40312....... cm

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\sqrt{41}  = 6.403124237

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4 0
3 years ago
Read 2 more answers
How many different ways can you write a fraction that has a numerator of 2 as a sum of fractions
svetoff [14.1K]
Infinitely many ways! 
Suppose you have the fraction 2/d.

<span>Pick </span>any<span> pair of integers a and b where b ≠ 0.</span>
Then 2b-ad is and integer, as is bd so that (2b - ad)/bd is a fraction.

Consider the fractions a/b and (2b - ad)/bd

<span>Their sum is </span>
a/b + (2b-ad)/bd = ad/bd + (2b-ad)/bd = 2b/bd = 2/d - as required.

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5 0
3 years ago
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4vir4ik [10]
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Hope this helped!
8 0
3 years ago
M³×n=189 Find m and n
Flura [38]

Answer:

m=3, n = 7

Step-by-step explanation:

Assuming m and n are integers, the following are the possible values of m and n

Let m = 1, then m³ = 1 and n= 189

If m = 2 we get 8n = 189 and 189/8 is not an integer so m ≠ 2

If m = 3, we get 27n = 189 and n = 189/27 = 7

Any higher integer values of m will not result in n being an integer and I am assuming m > 1 so the correct answer is m=3, n = 7

3 0
1 year ago
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