Answer:
db / dt = kb
this becomes b(t) = Ce^(kt)
C = 100, the initial population
P(1) = 420 = 100 e^(1k)
4.2 = e^k
ln 4.2 = k
a) thus, b(t) = 100 e^(t ln 4.2)
b) b(3) = 100 e^(3 ln 4.2)
c) growth constant will still be ln 4.2 (constant percentage of populatioin)
d) 10000 = 100 e^(t ln 4.2)
100 = e^(t ln 4.2)
ln 100 = t ln 4.2
t = ln 100 / ln 4.2
Step-by-step explanation:
Step-by-step explanation:
slope=4-6/8-3=-2/5
chosing another point (x,y) we have
y-4/x-8= -2/5
y-4=-2/5(x-8)
y= -2/5x+16/5+4
Number ten is 60 73+36=108
The answer would probably be the last one cuz it’s true and it has more then one thing
Answer:
Rectangular area as a function of x : A(x) = 200*x + 2*x²
A(max) = 5000 m²
Dimensions:
x = 50 m
l = 100 m
Step-by-step explanation:
"x" is the length of the perpendicular side to the wall of the rectangular area to be fenced, and we call "l" the other side (parallel to the wall of the barn) then:
A(r) = x* l and the perimeter of the rectangular shape is
P = 2*x + 2*l but we won´t use any fencing material along the wll of the barn therefore
P = 2*x + l ⇒ 200 = 2*x + l ⇒ l = 200 - 2*x (1)
And the rectangular area as a function of x is:
A(x) = x * ( 200 - 2*x) ⇒ A(x) = 200*x + 2*x²
Taking derivatives on both sides of the equation we get:
A´(x) = 200 - 4*x ⇒ A´= 0
Then 200 - 4*x = 0 ⇒ 4*x = 200 ⇒ x = 50 m
We find the l value, plugging the value of x in equation (1)
l = 200 - 2*x ⇒ l = 200 - 2*50 ⇒ l = 100 m
A(max) = 100*50
A(max) = 5000 m²
in the x-axis. What is the function rule for 
?
