The rate of change will be given by:
v(h)=3√h=3h^0.5
The rate of change will be:
v'(h)=(3×0.5)x^-0.5=1.5x^-0.5
hence when h=2 meters, then the rate of change will be:
v'(h)=1.5×2^-0.5
v'(h)=1.061 m/s
Answer:
9;15
Step-by-step explanation:
Answer:
Population of mosquitoes in the area at any time t is:

Step-by-step explanation:
assume population at any time t = P(t)
population increases at a rate proportional to the current population:
⇒dP/dt ∝ P
----(1)
where k is constant rate at which population is doubled
solving (1)

---- (2)
initial population = 400,000
population is doubled every week
⇒P(1)=2P(0)
Using (2)


In presence of predators amount is decreased by 50,000 per day
Then amount decreased per week = 350,000
In this case (1) becomes
---(3)
solving (3) by calculating integrating factor

Multiplying I.F with all terms of (3)

Integrating w.r.to t




at t=0



So, population of mosquitoes in the area at any time t is

Answer: The slope is 3 and the y-intercept is -1
Step-by-step explanation: This is right bro. 100%
from the equation<span> x^2 - 6x + 7 = 0 </span>, b is equal to -6. The form should become y = (x-(b/2)^2) + c. (<span>b/2)^2 is equal to 9. Hence,
</span>
0 = (x - 3)^2 -9 + 7
<span>0 = (x - 3)^2 -2
</span>2 = <span>(x - 3)^2
x should be equal to -1.5858 and -4.4142</span>