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cupoosta [38]
2 years ago
13

2x^4 +4x^3 - 5x^2 - 5x +6 how many zeros are imaginary

Mathematics
1 answer:
cricket20 [7]2 years ago
3 0

Answer:

4 zeros

Step-by-step explanation:

The zeroes of

f

(

x

)

=

2

x

4

−

5

x

3

+

3

x

2

+

4

x

−

6

are

x

=

−

1

,

x

=

3

2

,

x

=

1

−

i

, and

x

=

1

+

i

.

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The smallest square number which can be divisible by 11 and 5​
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Answer:

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Step-by-step explanation:

let's look at it in this concept.

For a number to be divisible by 11 and 5. it must be a multiple of the LCM of 11 and 5.

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therefore the number is 55x, where x is a positive integer.

it is a said that the number is a perfect square

therefore the square root of 55x must be an integer.

\sqrt{55x}  =  \sqrt{5 \times 11 \times x}

the smallest value of x to make 55x a perfect square is....

x = 11 \times 5 = 55

Therefore the number is.... .

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B is the correct answer

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