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vredina [299]
3 years ago
6

Which table shows a function that is decreasing only over the interval (–1, ∞)? A 2-column table with 6 rows. The first column i

s labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 1, negative 3, negative 5, negative 2, negative 1, 2. A 2-column table with 6 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 3, negative 5, negative 7, negative 6, 1, negative 1. A 2-column table with 6 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 4, negative 3, negative 1, 2, 1, negative 6. A 2-column table with 6 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 5, negative 1, 1, 0, negative 4, negative 8. Mark this and return
Mathematics
1 answer:
Lemur [1.5K]3 years ago
8 0

Answer:

B. f(x) ≤ 0 over the interval [0, 2]. D.f(x) > 0 over the interval (–2, 0). E.f(x) ≥ 0 over the interval [2, ).

Step-by-step explanation:

Those are the 3 answers. Just did it on edge.

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<span>S−L=−rL</span>
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<span><span><span>S−L</span>L</span>=−r</span>
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Find a polynomial function of degree 3 with 2, i, -i as zeros.
sergiy2304 [10]

Answer:

p(x)= x^3-2x^2+x-2

Step-by-step explanation:

Here we are given that a polynomial has zeros as 2 , i and -i . We need to find out the cubic polynomial . In general we know that if \alpha , \ \beta \ \& \ \gamma are the zeros of the cubic polynomial , then ,

\sf \longrightarrow p(x)= (x -\alpha )(x-\beta)(x-\gamma)

Here in place of the Greek letters , substitute 2,i and -i , we get ,

\sf\longrightarrow p(x)= (x -2 )(x-i)(x+i)

Now multiply (x-i) and (x+i ) using the identity (a+b)(a-b)=a² - b² , we have ,

\sf  \longrightarrow p(x)= (x-2)\{ x^2 - (i)^2\}

Simplify using i = √-1 ,

\sf \longrightarrow p(x)= (x-2)( x^2 + 1 )

Multiply by distribution ,

\sf \longrightarrow p(x)= x(x^2+1) -2(x^2+1)

Simplify by opening the brackets ,

\sf\longrightarrow p(x)= x^3+x-2x^2-2

Rearrange ,

\sf\longrightarrow \underline{\boxed{\blue{\sf p(x)= x^3-2x^2+x-2}}}

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