Answer:
A. Initially, there were 12 deer.
B. <em>N(10)</em> 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:

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


A. Initially, there were 12 deer.
B. 
B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.
C. 
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.
I'm gonna gonna go with C
Answer:

Step-by-step explanation:
<em>y=mx+b</em>
<em>b is where the line touches the y-intercept</em>
<em>mx is the slope...rise over run (positive slope) OR fall over crawl (negative slope)</em>
<em />
b= 8
mx= -3/2
y= -3/2x+8
400
because you need to divide 36/0.09=400
Answer:
<em>The graph of </em>y + 1 = −3/5 (x − 4)<em> would be a straight line. The graph figure is attached below.</em>
Step-by-step explanation:
As the linear equation y + 1 = −3/5 (x − 4) is given.
Since y-y₁ = m (x - x₁) is the Point-slope form is the general form y-y₁=m(x-x₁) for linear equations.
Hence, from the linear equation we can determine the slop which is m = -3/5
Also, when we put x = 0 in the linear equation, we determine the y-intercept as follows:
y + 1 = −3/5 (x − 4)
y + 1 = -3/5(-4) ∵x = 0
y = 12/5 - 1
y = 7/5
y-intercept: 7/5
Hence,
Table for some points can be made for x and values as:
<h3><em>x y</em></h3>
0<em> </em><em> 7/5</em>
<em>1 4/5</em>
<em />
<em>The graph of </em>y + 1 = −3/5 (x − 4)<em> would be a straight line. The graph figure is attached below.</em>
<em />
<em>Keywords: graph, straight line</em>
<em>Learn more about graph of a straight line from brainly.com/question/11488685</em>
<em>
#learnwithBrainly</em>