Answer:
Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.
How long will it take for this population to grow to a hundred rodents? To a thousand rodents?
Step-by-step explanation:
Use the initial condition when dp/dt = 1, p = 10 to get k;
Seperate the differential equation and solve for the constant C.
You have 100 rodents when:
You have 1000 rodents when:
Answer:
Step-by-step explanation:
2/
5
k−
3
/5
k+
1
/10
k
=
2
/5
k+
−3
/5
k+
1
/10
k
Combine Like Terms:
=
2
/5
k+
−3
/5
k+
1
/10
k
=(
2
/5
k+
−3
/5
k+
1
/10
k)
=
−1
/10
k
Answer:
=
−1
/10
k
Hello :
3x3 – 11x2 – 26x + 30 divided by x – 5 is :<span>3x2 + 4x – 6
because :
(x-5)(3x²+4x-6)=3x^3+4x²-6x-15x²-20x+30 =</span>3x3 – 11x2 – 26x + 30
Answer:
There are 16 Oak tree in the forest.
Step-by-step explanation:
Given: There are 4 Oak tree on every 10 pine trees.
There are total 24 more pine tree than Oak tree.
Using the ratio of trees to find the number of trees.
Lets assume there are total number of Oak tree be "x".
∴ Total number of Pine trees will be 
We know the ratio of Oak to Pine tree is 4:10 or 
⇒
Multiplying both side by 
⇒ 
Using distributive property of multiplication, distributing 4 with x and 24.
⇒ 
Multiplying both side by 10
⇒ 
subtracting both side by 4x
⇒ 
dividing both side by 6
⇒
∴ 
Hence, There are 16 Oak trees in the forest.
I don't see an inequality, but I can tell you that the sign of the inequality is flipped whenever you multiply or divide by a negative number.
