Can you help me in need help mad
1 answer:
You might be interested in
The correct answer is Maxine
Explanation:
One of the easiest ways for knowing if a fraction is greater than another is by converting fractions to decimal numbers. This implies dividing the numerator (top number) by the denominator (bottom number). In the case of fraction, the decimal number is 0.5 considering 1 divided into 2 is equal to 0.5. Now to know if other fractions are greater or smaller, this process needs to be repeated.
Gina:
Maxine:
Shari:
Vanessa:
According to this, the girl with a heigh increased greater than 1/2 inch is Maxine because 0.666 (Maxine heigh increase) is greater than 0.5 (1/2 inch).
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:
