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When a breeding group of animals is introduced into a restricted area such as a wildlife reserve, the population can be expected

jasenka [17]

Answer:

A. Initially, there were 12 deer.

B. N(10) corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. After 15 years, there will be 410 deer.

D. The deer population incresed by 30 specimens.

Step-by-step explanation:

$N=\frac{12.36}{0.03+0.55^t}$

The amount of deer that were initally in the reserve corresponds to the value of N when t=0

$N=\frac{12.36}{0.33+0.55^0}$

$N=\frac{12.36}{0.03+1} =\frac{12.36}{1.03} = 12$

A. Initially, there were 12 deer.

B. $N(10)=\frac{12.36}{0.03 + 0.55^t} =\frac{12.36}{0.03 + 0.0025}=\frac{12.36}{y}=380$

B. N(10) corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. $N(15)=\frac{12.36}{0.03+0.55^15}=\frac{12.36}{0.03 + 0.00013}=\frac{12.36}{0.03013}= 410$

C. After 15 years, there will be 410 deer.

D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:

ΔN=N(15)-N(10)

ΔN=410 deer - 380 deer

ΔN= 30 deer.

D. The deer population incresed by 30 specimens.

8 0

3 years ago

What is the area of the shaded segment?

Whitepunk [10]

Answer:

$6\pi-9\sqrt{3$

Step-by-step explanation:

The area of a pie-slice shape can be found with the equation $A=\pi nr^2/360$ where n is the angle of the shape and r is the radius of the circle. In this case, n=60 and r=6. Therefore, the total area of the pie-slice is $\pi (60)(6^2)/360=6 \pi$

Next, the triangle that makes up the unshaded part of the segment is an equilateral triangle because the angle is 60. The area of an equilateral triangle is $\sqrt{3}r^2/4$ where r is the length of one of its side (in this case, r=6). Plugging that in, we get an area of $9 \sqrt{3}$