Answer:
20 = initial population of the rabbits
1.014 = growth rate of the rabbits
the average rate of change from day 50 to day 100 is 0.8
Step-by-step explanation:
A population of rabbits in a lab, p(x), can be modeled by the function
p(x) = 20(1.014)^x
This model is exponential. Where 20 = initial population of the rabbits
1.014 = growth rate of the rabbits with 1.4% increase rate of the rabbits
To find the average rate of change from day 50 to day 100,
find the population p(50) and p(100). Subtract them and divide by 100 - 50 = 50.
p(50) = 20(1.014)50 = 40.08...
p(100) = 20(1.014)100 = 80.32...
(80.32 - 40.08) / (100 - 50) = 40.24/50 = 0.8048. which is approximately 0.8 to the nearest tenth.
The rate of change is 0.8.
This situation of the recipe can be represented as y=x+5.5 and it is expected that x is ≥ 0 and y is ≥ 5.5.
<h3>What is the equation for this situation?</h3>
It is known the total amount of flour (y) is equivalent to the total amount of whole wheat flour (5.5) added to the total white flour (x). Therefore, the equation is:
y = x + 5.5
The possible values are
x = It is expected x is equalthan 0 or greater to it since the minimum amount of white flour that can be added is 0.
y = It is expected y is equal than 5.5 or greater to it since even if no
white flour is added the minimum total is 5.5.
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Givens
Work week = 40 hours
Paid 7.50 an hour.
Sells 6300 worth of goods.
Equation
Take home pay = Sales * 5% + 7.50 * number of hours.
Sub and Solve
Take home pay = 6300 * 5% + 7.50 * 40
Take home pay = 6300 * 5/100 + 7.50 * 40
Take home pay = 315 + 300
Take home pay = 615 dollars.
Answer 615 <<<<
Answer:
28.27 cm/s
Step-by-step explanation:
Though Process:
- The punch glass (call it bowl to have a shape in mind) is in the shape of a hemisphere
- the radius
- Punch is being poured into the bowl
- The height at which the punch is increasing in the bowl is

- the exposed area is a circle, (since the bowl is a hemisphere)
- the radius of this circle can be written as

- what is being asked is the rate of change of the exposed area when the height
- the rate of change of exposed area can be written as
. - since the exposed area is changing with respect to the height of punch. We can use the chain rule:

- and since
the chain rule above can simplified to
-- we can call this Eq(1)
Solution:
the area of the exposed circle is

the rate of change of this area can be, (using chain rule)
we can call this Eq(2)
what we are really concerned about is how
changes as the punch is being poured into the bowl i.e 
So we need another formula: Using the property of hemispheres and pythagoras theorem, we can use:

and rearrage the formula so that a is the subject:

now we can derivate a with respect to h to get 

simplify

we can put this in Eq(1) in place of 
and since we know 

and now we use substitute this
. in Eq(2)

simplify,

This is the rate of change of area, this is being asked in the quesiton!
Finally, we can put our known values:

from the question


Answer:
30.2 is the answer (use distrubutyion in this problem)
Step-by-step explanation: