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Irina18 [472]
3 years ago
8

He distance between Miami and Naples is 107 miles. The distance between Miami and Jacksonville is about three times this distanc

e. Landon estimates the distance between Miami and Jacksonville is about 400 miles. Is Landon's estimate reasonable? Why or why not?
Mathematics
1 answer:
zmey [24]3 years ago
4 0

Answer:

Step-by-step explanation:

Distance between Miami and Naples = 107 miles

Distance between Miami and Jacksonville is about three times this distance.

Distance between Miami and Jacksonville = 3 * 107 miles

= 321 miles

Landon estimates the distance between Miami and Jacksonville is about 400 miles

The estimate is too high

He should have estimated 107 miles to 110 miles which gives a total of 330 miles

Estimated 400 miles compared to 330 miles is too high

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What is the value of the expression 3+2x (with an exponent of 3 on top of x) when x = 2?
Ivan

Answer:

A. 67

Step-by-step explanation:

3+2(2)^3

The first thing you would do is distribute the 2 and you would get

3+(4)^3

The next thing you would do is the exponent

3+(4*4*4) = 3+64

Then you would just add 3

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So the answer is A. 67.

3 0
3 years ago
Whats 20 x 1??????????
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any number times 1 is itself. 20 x 1 = 20.

5 0
3 years ago
Read 2 more answers
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
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