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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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A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 5 − x 2 . What are the dimensions
kherson [118]

Answer:

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola

y=5−x^2. What are the dimensions of such a rectangle with the greatest possible area?

Width =

Height =

Width =√10 and Height = \frac{10}{4}

Step-by-step explanation:

Let the coordinates of the vertices of the rectangle which lie on the given parabola y = 5 - x² ........ (1)

are (h,k) and (-h,k).

Hence, the area of the rectangle will be (h + h) × k

Therefore, A = h²k ..... (2).

Now, from equation (1) we can write k = 5 - h² ....... (3)

So, from equation (2), we can write

A =h^{2} [5-h^{2} ]=5h^{2} -h^{4}

For, A to be greatest ,

\frac{dA}{dh} =0 = 10h-4h^{3}

⇒ h[10-4h^{2} ]=0

⇒ h^{2} =\frac{10}{4} {Since, h≠ 0}

⇒ h = ±\frac{\sqrt{10} }{2}

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