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iragen [17]
2 years ago
8

An exponential decay function represents a quantity that has a decreasing halving time true or false

Mathematics
1 answer:
timofeeve [1]2 years ago
4 0

Answer:

In mathematics, exponential decay describes the process of reducing an amount by a consistent percentage rate over a period of time. ... You can compare and contrast the differences between exponential growth and decay, but it's pretty straightforward: one increases the original amount and the other decreases it.

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Eli is moving to a new city. He packs his belongings in cube-shaped boxes that each have a volume of one cubic foot. He places t
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Answer:

jnnm mnn m

Step-by-step explanation:

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3 years ago
Solve (2 - x)^2 < 4/25
Semenov [28]

Answer:

8/5 < x< 12/5

Step-by-step explanation:

(2 - x)^2 < 4/25

Take the square root of each side

sqrt((2 - x)^2)  <±sqrt( 4/25)

Make two equations

2-x < 2/5    2-x > -2/5

Subtract 2 from each side

2-x-2 < 2/5 -2          2-x-2  > -2/5-2

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Multiply by -1, remembering to flip the inequality

x> 8/5                   x < 12/5

8/5 < x< 12/5

5 0
3 years ago
Read 2 more answers
A full moon occurs about every 30 days . If the last full moon occurred on a Friday, how many days will pass before a full moon
pentagon [3]
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3 years ago
The smallest number that has 1, 2, 3, 4, and 5 asfactors.
STALIN [3.7K]

Answer:

The answer is 60

8 0
2 years ago
Define fn : [0,1] --&gt; R by the
sasho [114]

Answer:

The sequence of functions \{x^{n}\}_{n\in \mathbb{N}} converges to the function

f(x)=\begin{cases}0&0\leq x.

Step-by-step explanation:

The limit \lim_{n\to \infty }c^{n} exists and converges to zero whenever \lvert c \rvert. But, if c=1 the sequence \{c^{n}\} is constant and all its terms are equal to 1, then converges to 1. Using this result, consider the sequence of functions \{f_{n}\} defined on the interval [0,1] by f_{n}(x)=x^{n}. Then, for all 0\leq x we have that \lim_{n\to \infty}x^{n}=0. Now, if x=1, then \lim_{n\to \infty }x^{n}=1. Therefore, the limit function of the sequence of functions is

f(x)=\begin{cases}0&0\leq x.

To show that the convergence is not uniform consider 0. For any n>1 choose x\in (0,1)  such that \varepsilon^{1/n}. Then

\varepsilon

This implies that the convergence is not uniform.

8 0
3 years ago
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