The sac inside a plant cell where water wastes and food are stored is called

Vegetative reproduction is a type of asexual reproduction

weeeeeb [17]

Answer:

wait what was it

Explanation:

I know it isn't A. or B.

3 0

3 years ago

Read 2 more answers

Muscles like the diaphragm and intercostals contract and expand the chest cavity to lower pressure for breathing. This is an exa

Oxana [17]

Answer:

Environment.

Explanation:

The respiratory system interacts with the environment by contracting and expanding the diaphragm and chest in order to do breathing. When the muscles in the diaphragm contracts and the expansion of chest occurs, takes the air inside the body that is present in the outside environment for inhalation of oxygen while on the other hand, when the diaphragm expands and chest contract, the air inside the body is removed with the removal of carbondioxide gas.

4 0

3 years ago

Explain the need for a reference point when describing the direction of motion

Tems11 [23]

Answer:

It is important as it tells if the object is stationary or in motion

Explanation:

8 0

3 years ago

A fossil is discovered that has only 12.5% of the carbon-14 that it would have had originally (when the animal was alive). If th

mariarad [96]

Answer:

The fossil is 17,100 years old.

Explanation:

The decay equation:

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

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

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

Integrating both sides

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

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

When t=0, N= $N_0$ = initial amount

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

$\Rightarrow ln |N_0|=c$

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

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

$\Rightarrow ln |\frac {N}{N_0}|=-\lambda t$

$\Rightarrow \frac {N}{N_0}=e^{-\lambda t}$

$\Rightarrow N=N_0e^{-\lambda t}$

The decay equation is

$N=N_0e^{-\lambda t}$

Given that,

The half life of carbon - 14 is 5700 years.

For half life, $N=\frac{1}{2} N_0$

To find the value of $\lambda$ , we need to put the value of N and t in the decay equation.

$\frac12N_0=N_0e^{-\lambda \times 5700}$

$\Rightarrow \frac12=e^{-\lambda \times 5700}$ [ Divided $N_0$ both sides]

Taking ln both sides

$\Rightarrow ln| \frac12|=ln|e^{-\lambda \times 5700}|$

$\Rightarrow ln| \frac12|={-\lambda \times 5700}$

$\Rightarrow \lambda= \frac{ln| \frac12|}{-5700}$

$\Rightarrow \lambda= \frac{ln|1|-ln|2|}{-5700}$ [ $ln|\frac mn|= ln |m|-ln |n|$ ]

$\Rightarrow \lambda= \frac{ln|2|}{5700}$ [ln 1= 0]

The fossil has only 12.5% of the carbon carbon-14 that it would have had originally.

So, $N=\frac{12.5}{100} N_0$

Then,

$\frac{12.5}{100} N_0=N_0e^{-\frac{ln|2|}{5700}t$

$\Rightarrow \frac{12.5}{100} =e^{-\frac{ln|2|}{5700}t$

Taking ln both sides

$\Rightarrow ln|\frac{12.5}{100} |=ln|e^{-\frac{ln|2|}{5700}t}|$

$\Rightarrow ln|\frac{12.5}{100} |={-\frac{ln|2|}{5700}t}$

$\Rightarrow t=\frac{ ln|\frac{12.5}{100}|} {-\frac{ln|2|}{5700}}$

$\Rightarrow t=\frac{ ln|\frac{12.5}{100}|\times 5700} {-{ln|2|}}$

$\Rightarrow t=17,100$

The fossil is 17,100 years old.

7 0

4 years ago

how could variation within the fish population help individuals survive their encounter with the dolphins?

Mkey [24]

The right answer for the question that is being asked and shown above is that: "The <span>variation within the fish population help individuals survive their encounter with the dolphins is that dolphins will tend to be with them and will be served as their own food."</span>

3 0

3 years ago