Answer:
B
Step-by-step explanation:
Hope this helps!
Can someone help me find the theta and radius please?
1 answer:
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Step-by-step explanation:
The unit rate would be how many eggs one chicken lays
To do this, you would create a ratio
30:12 = x:1
30 = 12x cross multiply
30/12 = 12x/12 isolate the variable
2.5 = x
The unit rate is 2.5:1 or :1
I hope this helps!!!
The graph of the given function is attached showing characteristics of the function.
The correct responses are;
- (a) F'(4) = -30 means that <u>on day 4, the number of fish in the bay is decreasing at a rate of 30 fish per day</u>.
- (b) Over the interval 0 ≤ t ≤ 8 the absolute minimum number of fish in the bay is;<u> 1,000</u>.
- (c) The values of <em>t</em> at which the rate of change is decreasing is; <u>3 ≤ t ≤ 8</u>
- (d) The rate of change of the number of pelicans flying near is; P' = <u>10·(3·c² - 24·c + 45)</u>
Reasons:
The given function is; f(t) = 10·(t³ - 12·t² + 45·t + 100)
Where;
t = Number of days
0 ≤ t ≤ 8
(a) The derivative of the given function is presented as follows;
F'(t) = 10 × (3·t² - 2×12·t + 45) = 10·(3·t² - 24·t + 45)
Therefore;
F'(4) = 10 × (3 × 4² - 24 × 4 + 45) = -30
Therefore F'(4) = -30 means that on day 4, the <u>number of fish in the bay is decreasing at 30 fish per day</u>.
(b) The absolute minimum is given as follows;
At a minimum or maximum value, F'(t) = 10·(3·t² - 24·t + 45) = 0
Which gives;
3·(t - 5)·(t - 3) = 0
t = 5, or t = 3
At t = 5, we have;
f(5) = 10 × (5³ - 12 × 5² + 45 × 5 + 100) = 1,500
At t = 3, we have;
f(3) = 10 × (3³ - 12 × 3² + 45 × 3 + 100) = 1,540
Therefore, f(5) = 1,500 is a local maximum
However, at x = 0, we have;
f(0) = 10 × (0³ - 12 × 0² + 45 × 0 + 100) = 1000
At x = 8, we have;
f(8) = 10 × (8³ - 12 × 8² + 45 × 8 + 100) = 2,040
Therefore, the absolute minimum is given at t = 8, where f(t) = <u>1,000</u>
(c) The values of <em>t</em> at which the rate of change in the number of fish in the
bay is decreasing is between the local maximum at t = 3, and the local
minimum at t = 5, which gives;
The rate of change of the number of fish is decreasing for values of t in the range;
- <u>3 ≤ t ≤ 5</u>
(d) P = 10·(t³ - 12·t² + 45·t + 100)
P' = 10·(3·t² - 24·t + 45)
At time t = c, we have;
- P' = 10·(3·c² - 24·c + 45)
<em>Based on a similar question online, we have;</em>
<em>(c) The interval over which the rate of change is decreasing</em>
<em>(d) The rate of change of the number of pelicans flying near the bay at t = c</em>
Learn more about differentiation of functions here:
