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Anon25 [30]
3 years ago
15

How many and of which kind of roots does this equation have?

Mathematics
2 answers:
Kamila [148]3 years ago
5 0

Answer:

B

Step-by-step explanation:

The polynomial is of degree 5, thus will have 5 roots either real or complex or both.

Using Descartes' rule of signs.

Count the number of sign changes in f(x)

There are no sign changes thus there are no real positive roots.

Find f(- x) then count the number of sign changes.

f(- x) = - x^{5} + 3x^{4} - 8x³ + 14x² - 16x + 8

- / + / - / + / - / + ← 5 sign changes

Thus there are 5 negative real roots

Thus f(x) has 5 real roots → B

Marat540 [252]3 years ago
3 0

Answer:

mckpwmc dk

Step-by-step explanation:

dlkd mlk kdlw  

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stepan [7]

Answer:

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Step-by-step explanation:

The options that will be equivalent to T will have to be the options that have the same Cardinality as T. Cardinality refers to the number of elements in a set and in the set T, there are 4 elements being Tinkey-Winky, Laa-Laa,  Dipsy, Po so the Cardinality is 4.

The equivalent sets would therefore be sets with a cardinality of 4 as well and those are;

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