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kykrilka [37]
4 years ago
13

20.65 in word form and expanded form

Mathematics
2 answers:
irinina [24]4 years ago
8 0
Twenty point sixty five
<span />
kiruha [24]4 years ago
3 0

Answer:

Word form: twenty and sixty-five hundredths

Expanded form: 20+.6+.05

Step-by-step explanation:

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The number of fish in a small bay is modeled by the function F defined by F(t)=10 (t3 â 12t2 + 45t +100), where t is measured in
RSB [31]

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:

brainly.com/question/1422315

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Solve the equation below for x: <br><br> logx+log(x-2)=log(3x)
Ugo [173]

Answer:

x=5

Step-by-step explanation:

Given equation is

\log\left(x\right)+\log(x-2)=\log(3x)

Apply formula: \log A+\log B=\log\left(AB\right)

\log\left(x(x-2)\right)=\log(3x)

\log\left(x^2-2x)\right)=\log(3x)

Since both sides have equal bases so we can drop the log

x^2-2x=3x

x^2-2x-3x=0

x^2-5x=0

x(x-5)=0

Which gives x=0 and x-5=0

or x=0 and x=5

Original problem \log\left(x\right)+\log(x-2)=\log(3x) is not defined at x=0 henc only x=5 is the final answer.

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Arrange the steps in the correct order to perform this subtraction problem.
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