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3 years ago

The areas of the sides of a rectangular box are

Mathematics

1 answer:

Rufina [12.5K]3 years ago

7 0

Answer:

<u>4900 In. Squared</u>

Step-by-step explanation:

The formula for calculating volume is Length X Base X Height

So as you can see, the way you would solve this problem si by doing

10 X 14 X 35 which would give you the final answer of

4900 Inches Squared or 4900 In.2

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Find the length of 1 side of square if the area is 49 sq.cm<br> method needed

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Answer:

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Step-by-step explanation:

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2 years ago

I don't know how to do this. Can someone explain how to please?

Natalija [7]

Answer:

f(5) = -2

Step-by-step explanation:

You look on the x-axis for 5 because f(5) means let x = 5. Then go down to the graph (the red zig-zaggy line) At the red line (you're still directly under the 5) look over to the y-axis to find the value. That's your answer. It is -2. The point (5, -2) is on the red graph.

f(5) = -2

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2 years ago

Please help quick will give brainliest!

Karo-lina-s [1.5K]

Answer:

The statement that correctly uses limits to determine the end behavior of f(x) is;

$\lim\limits_{x \to \pm \infty} \dfrac{7 \cdot x^2+ x + 1}{x^4 + 1}= \lim\limits_{x \to \pm \infty} \dfrac{7 }{x^2 }$ so the end behavior of the function is that as x → ±∞, f(x) → 0

Step-by-step explanation:

The given function is presented here as follows;

$f(x) = \dfrac{7 \cdot x^2+ x + 1}{x^4 + 1}$

The limit of the function is presented as follows;

$\lim\limits_{x \to \pm \infty} \dfrac{7 \cdot x^2+ x + 1}{x^4 + 1}$

Dividing the terms by x², we have;

$\lim\limits_{x \to \pm \infty} \dfrac{\dfrac{7 \cdot x^2}{x^2} + \dfrac{x}{x^2} + \dfrac{1}{x^2} }{\dfrac{x^4}{x^2} +\dfrac{1}{x^2} }= \lim\limits_{x \to \pm \infty} \dfrac{7 + \dfrac{1}{x} + \dfrac{1}{x^2} }{x^2 +\dfrac{1}{x^2} }$

As 'x' tends to ±∞, we have;

$\lim\limits_{x \to \pm \infty} \dfrac{7 + \dfrac{1}{x} + \dfrac{1}{x^2} }{x^2 +\dfrac{1}{x^2} } = \lim\limits_{x \to \pm \infty} \dfrac{7 + 0 + 0 }{x^2 +0 } = \lim\limits_{x \to \pm \infty} \dfrac{7 }{x^2 }$

However, we have that the end behavior of 7/x² as 'x' tends to ±∞ is 7/x² tends to 0;

Therefore, we have;

$f(x) \rightarrow 0 \ as \lim\limits_{x \to \pm \infty} \dfrac{7 }{x^2 }$

The statement that correctly uses limits to determine the end behavior of f(x) is therefor given as follows;

$\lim\limits_{x \to \pm \infty} \dfrac{7 \cdot x^2+ x + 1}{x^4 + 1}= \lim\limits_{x \to \pm \infty} \dfrac{7 }{x^2 }$ so the end behavior of the function is that as x → ±∞, f(x) → 0.

8 0

3 years ago