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amm1812
3 years ago
13

Interpreting the Multiplicity of a Zero:Determine the multiplicity of the roots of the function

Mathematics
2 answers:
olasank [31]3 years ago
8 0

For this case we have the following expression:

k (x) = x (x + 2) ^ 3 (x + 4) ^ 2 (x-5) ^ 4\\

The roots are:

x = -2\\\\x = -4\\\\x = 5\\\\x = 0\\

For example: For x = 0 we have

k (0) = 0 (0 + 2) ^ 3 (0 + 4) ^ 2 (0-5) ^ 4\\\\k (0) = 0 (2) ^ 3 (4) ^ 2 (-5) ^ 4\\\\k (0) = 0 (8) (16) (625)\\\\k (0) = 0\\

so it is shown that x = 0 is a root.

By definition, multiplicity represents the number of times a root is repeated in a polynomial, in turn it is given by the degree of the term that contains the root.

Thus:

The multiplicity of 0 is 1

The multiplicity of -2 is 3

The multiplicity of -4 is 2

The multiplicity of 5 is 4

Answer:

The multiplicity of 0 is 1

The multiplicity of -2 is 3

The multiplicity of -4 is 2

The multiplicity of 5 is 4

OlgaM077 [116]3 years ago
4 0

GGGGGGGGGGGGGGGGGGGGGGGGGGGOOOOOOOOOOOOOOOOOOTTTTTTTTTTTTTTTTTTTAAAAAAAAAAAAAAAAAAA snatch a lil bit of points rq

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Answer:

The given expression  (\frac{5}{6} a^{9}p^{5})  ^{3} = \frac{125}{216}  \times a^{(27) } \times p^{(15)}

Step-by-step explanation:

Here, the given expression is:  (\frac{5}{6} a^{9}p^{5})  ^{3}

Now, starting from the outer most bracket.

As we know :

(abc)^{n}   = (a)^{n} \times (b)^{n}  \times (c)^{n}

and (a^m)^{n} = a^{(m \times n)}

⇒ (\frac{5}{6} a^{9}p^{5})  ^{3} = (\frac{5}{6})^{3} \times (a^{9})^{3}   \times (p^{5}) ^{3}\\

=\frac{125}{216}  \times (a)^{(9\times3) } \times (p)^{(5 \times 3)}\\= \frac{125}{216}  \times a^{(27) } \times p^{(15)}

Hence, the given expression  (\frac{5}{6} a^{9}p^{5})  ^{3} = \frac{125}{216}  \times a^{(27) } \times p^{(15)}

7 0
3 years ago
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What quantum numbers specify these subshells 5s 2p 3d?
Ivan
5s subshell represent that n is equal to 5 and l is equal to 0.
2p subshell represent that n is equal to 2 and l is equal to 1.
3d subshell represent that n is equal to 3 and l is equal to 2.
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s means l = 0 
p means l = 1 
d means l = 2
f means l = 3 
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