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valina [46]
3 years ago
5

the polynomial y = x^{5} + x^{3} - x^{2} - 1 factors into (x^{2} + 1) (x^{3} - 1). what is the multiplicity of the root (x^{3} -

1)?
Mathematics
1 answer:
Anestetic [448]3 years ago
6 0

Answer:

The multiplicity of the root is 1

Step-by-step explanation:

Firstly, we need to understand what multiplicity is

By multiplicity, we simply refer to the number of times we have the certain root repeating itself

From the factorization, each of the given roots were only repeated once

That indicates that the given multiplicity is 1

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Oksanka [162]

Answer:

Option C

Step-by-step explanation:

When you put the write the scenario in an equation, it'll look like

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Factor these, and you shall get 25q^2+15q-15q-9. Simplify that into

25q^2-9

8 0
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Tpy6a [65]
The area of the polygon with vertices W (1, 1), X (4, 4), Y (7, 1), and S (4, −8) is 36.
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3 years ago
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hodyreva [135]
Answer C

When you multiply that together (c), it gives you p^2, -5p, -2p, and 10. Add the -5-2, and it gives you -7, or the original equation.
5 0
3 years ago
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TiliK225 [7]

Answer:

The answer is 9.

4 0
4 years ago
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What is the discontinuity and zero of the function f(x) = 3x^2 + x - 4 / x-1
7nadin3 [17]
Ans: Option (1) <span>Discontinuity at (1, 7), zero at ( negative four thirds , 0) 
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Explanation:
Given function:
f(x) =  \frac{3x^2 + x - 4}{x-1}

Now if we plug in the x = 1, we would have discontinuity as function goes to infinity.

Now for f(1):
f(x) = \frac{3x^2 + x - 4}{x-1} \\ f(x) = \frac{3x^2 + 4x - 3x - 4}{x-1} \\ f(x) = \frac{x(3x+4)-1(3x+4)}{x-1} \\ f(x) = \frac{(x-1)(3x+4)}{(x-1)} \\ f(x) = 3x+4 \\ now ~ insert ~ x=1: \\ f(1) = 3(1) + 4 = 7 \\ It~means~discontinuity~at~(1,7).\\ Now~let~us~find~zeros.\\ put~ f(x) = 3x+4 ~ equals~ to ~zero. \\ =\ \textgreater \ ~3x+4 = 0 \\ =\ \textgreater \  ~ x =  \frac{-4}{3} \\ Hence~zeros=(  \frac{-4}{3}, 0)

5 0
3 years ago
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