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Burka [1]
1 year ago
8

Consider the graph of the function f(x) = 25

Mathematics
1 answer:
trasher [3.6K]1 year ago
3 0

Considering it's horizontal asymptote, the statement describes a key feature of function g(x) = 2f(x) is given by:

Horizontal asymptote at y = 0.

<h3>What are the horizontal asymptotes of a function?</h3>

They are the limits of the function as x goes to negative and positive infinity, as long as these values are not infinity.

Researching this problem on the internet, the functions are given as follows:

  • f(x) = 2^x.
  • g(x) = 2f(x) = 2(2)^x

The limits are given as follows:

\lim_{x \rightarrow -\infty} g(x) = \lim_{x \rightarrow -\infty} 2(2)^x = \frac{2}{2^{\infty}} = 0

\lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} 2(2)^x = 2(2)^{\infty} = \infty

Hence, the correct statement is:

Horizontal asymptote at y = 0.

More can be learned about horizontal asymptotes at brainly.com/question/16948935

#SPJ1

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The ​half-life of a radioactive element is 130​ days, but your sample will not be useful to you after​ 80% of the radioactive nu
gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

\frac{dN}{dt}\propto -N

\Rightarrow \frac{dN}{dt} =-\lambda N

\Rightarrow \frac{dN}{N} =-\lambda dt

Integrating both sides

\int \frac{dN}{N} =\int\lambda dt

\Rightarrow ln|N|=-\lambda t+c

When t=0, N=N_0 = initial amount

\Rightarrow ln|N_0|=-\lambda .0+c

\Rightarrow c= ln|N_0|

\therefore ln|N|=-\lambda t+ln|N_0|

\Rightarrow ln|N|-ln|N_0|=-\lambda t

\Rightarrow ln|\frac{N}{N_0}|=-\lambda t.......(1)

                            \frac{N}{N_0}=e^{-\lambda t}.........(2)

Logarithm:

  • ln|\frac mn|= ln|m|-ln|n|
  • ln|ab|=ln|a|+ln|b|
  • ln|e^a|=a
  • ln|a|=b \Rightarrow a=e^b
  • ln|1|=0

130 days is the half-life of the given radioactive element.

For half life,

N=\frac12 N_0,  t=t_\frac12=130 days.

we plug all values in equation (1)

ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130

\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130

\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130

\rightarrow -ln|2|=-\lambda \times 130

\rightarrow \lambda= \frac{-ln|2|}{-130}

\rightarrow \lambda= \frac{ln|2|}{130}

We need to find the time when the sample remains 80% of its original.

N=\frac{80}{100}N_0

\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t

\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t

\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t

\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}

\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}

\Rightarrow t\approx 42

We can use the sample about 42 days.

7 0
3 years ago
What is the value of 7x 2squared + 2x when x = 3?
mash [69]

Answer:

Given expression has the value 69

Step-by-step explanation:

Given equation is:

=7x^2 + 2x

Now we have to put x = 3

So the equation will become:

=7(3)^2 + 2(3)

By simplifying:

As 3^{2} = 9

and 2*3 = 6

So the above equation will become:

=7(9 ) + 6\\= 63 + 6\\= 69

So the value of given expression is 69.

i hope it will help you!

6 0
3 years ago
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