Find the quadratic function that models the data as a function of x, the number of weeks. Use the model to estimate the number of fish at the lake on week 11.
Week : 0 1 2 3 4 5 6
Population : 495,483,481,489,507,535,573
Answers
Answer: b
f(x)=5x^2-17x+495
f(11)=913
Step-by-step explanation:
I did some research and found an exam review sheet that had this same problem (worked out). This was all that was written.
The quadratic function that models the data will be;
y = 5x² - 17x + 495
What is an expression?
Mathematical expression is defined as the collection of the numbers variables and functions by using operations like addition, subtraction, multiplication, and division.
Given that;
The model to estimate the number of fish at the lake on week 11 are;
Week = 0, 1, 2, 3, 4, 5, 6
Population = 495, 483, 481, 489, 507, 535, 573
Now,
Let the quadratic equation is;
y = ax² + bx + c
Substitute x = 0 and y = 495, we get;
495 = a × 0 + b × 0 + c
c = 495
Substitute x = 1 and y = 483, we get;
483 = a × 1 + b × 1 + c
483 = a + b + 495
a + b = 483 - 495
a + b = - 12 ... (i)
Substitute x = 2 and y = 481, we get;
481 = a × 4 + b × 2 + c
481 = 4a + 2b + 495
4a + 2b = 481 - 495
4a + 2b = - 14
2a + b = - 7 ... (ii)
Solve equation (i) and (ii), we get;
a = 5 and b = - 17
So, We get;
The quadratic equation is;
y = ax² + bx + c
y = 5x² - 17x + 495
Therefore, The quadratic function that models the data will be;
y = 5x² - 17x + 495
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Simplify the expression:
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collect like terms:
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The spread of a virus is modeled by V (t) = −t 3 + t 2 + 12t,
where V (t) is the number of people (in hundreds) with the virus and t is the number of weeks since the first case was observed.
(a) Sketch V (t).
(b) What is a reasonable domain of t for this problem?
(c) Find the average rate of infection from t = 0 to t = 2.
(d) Find the instantaneous rate of infection as a function of t using the limit definition of the derivative.
(e) Find V (2) and V ‘ (2). Write a sentence interpreting V (2) and V ‘ (2) in terms of the number of infected people. Make sure to include units.
(f) Sketch the tangent line to the graph you drew in a. at the point (2, V (2)). State the slope of the tangent line.
(g) Use V (2) and V ‘ (2) to estimate the value of V (2.1).
(h) Find the maximum number of people infected at the same time and when the maximum occurs. Determine the rate of infection at this time.
Answers
Functions can be used to model real life scenarios
The reasonable domain is [tex]\mathbf{[0,\infty)}[/tex].The average rate of change from t = 0 to 2 is 20 persons per weekThe instantaneous rate of change is [tex]\mathbf{V'(t) = -3t^2 + 2t + 12}[/tex].The slope of the tangent line at point (2,V(20) is 10 The rate of infection at the maximum point is 8.79 people per weekThe function is given as:
[tex]\mathbf{V(t) = -t^3 + t^2 + 12t}[/tex]
(a) Sketch V(t)
See attachment for the graph of [tex]\mathbf{V(t) = -t^3 + t^2 + 12t}[/tex]
(b) The reasonable domain
t represents the number of weeks.
This means that: t cannot be negative.
So, the reasonable domain is: [tex]\mathbf{[0,\infty)}[/tex]
(c) Average rate of change from t = 0 to 2
This is calculated as:
[tex]\mathbf{m = \frac{V(a) - V(b)}{a - b}}[/tex]
So, we have:
[tex]\mathbf{m = \frac{V(2) - V(0)}{2 - 0}}[/tex]
[tex]\mathbf{m = \frac{V(2) - V(0)}{2}}[/tex]
Calculate V(2) and V(0)
[tex]\mathbf{V(2) = (-2)^3 + (2)^2 + 12 \times 2 = 20}[/tex]
[tex]\mathbf{V(0) = (0)^3 + (0)^2 + 12 \times 0 = 0}[/tex]
So, we have:
[tex]\mathbf{m = \frac{20 - 0}{2}}[/tex]
[tex]\mathbf{m = \frac{20}{2}}[/tex]
[tex]\mathbf{m = 10}[/tex]
Hence, the average rate of change from t = 0 to 2 is 20
(d) The instantaneous rate of change using limits
[tex]\mathbf{V(t) = -t^3 + t^2 + 12t}[/tex]
The instantaneous rate of change is calculated as:
[tex]\mathbf{V'(t) = \lim_{h \to \infty} \frac{V(t + h) - V(t)}{h}}[/tex]
So, we have:
[tex]\mathbf{V(t + h) = (-(t + h))^3 + (t + h)^2 + 12(t + h)}[/tex]
[tex]\mathbf{V(t + h) = (-t - h)^3 + (t + h)^2 + 12(t + h)}[/tex]
Expand
[tex]\mathbf{V(t + h) = (-t)^3 +3(-t)^2(-h) +3(-t)(-h)^2 + (-h)^3 + t^2 + 2th+ h^2 + 12t + 12h}[/tex][tex]\mathbf{V(t + h) = -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h}[/tex]
Subtract V(t) from both sides
[tex]\mathbf{V(t + h) - V(t)= -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h - V(t)}[/tex]
Substitute [tex]\mathbf{V(t) = -t^3 + t^2 + 12t}[/tex]
[tex]\mathbf{V(t + h) - V(t)= -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h +t^3 - t^2 - 12t}[/tex]
Cancel out common terms
[tex]\mathbf{V(t + h) - V(t)= -3t^2h -3th^2 - h^3 + 2th+ h^2 + 12h}[/tex]
[tex]\mathbf{V'(t) = \lim_{h \to \infty} \frac{V(t + h) - V(t)}{h}}[/tex] becomes
[tex]\mathbf{V'(t) = \lim_{h \to \infty} \frac{ -3t^2h -3th^2 - h^3 + 2th+ h^2 + 12h}{h}}[/tex]
[tex]\mathbf{V'(t) = \lim_{h \to \infty} -3t^2 -3th - h^2 + 2t+ h + 12}[/tex]
Limit h to 0
[tex]\mathbf{V'(t) = -3t^2 -3t\times 0 - 0^2 + 2t+ 0 + 12}[/tex]
[tex]\mathbf{V'(t) = -3t^2 + 2t + 12}[/tex]
(e) V(2) and V'(2)
Substitute 2 for t in V(t) and V'(t)
So, we have:
[tex]\mathbf{V(2) = (-2)^3 + (2)^2 + 12 \times 2 = 20}[/tex]
[tex]\mathbf{V'(2) = -3 \times 2^2 + 2 \times 2 + 12 = 4}[/tex]
Interpretation
V(2) means that, 20 people were infected after 2 weeks of the virus spread
V'(2) means that, the rate of infection of the virus after 2 weeks is 4 people per week
(f) Sketch the tangent line at (2,V(2))
See attachment for the tangent line
The slope of this line is:
[tex]\mathbf{m = \frac{V(2)}{2}}[/tex]
[tex]\mathbf{m = \frac{20}{2}}[/tex]
[tex]\mathbf{m = 10}[/tex]
The slope of the tangent line is 10
(g) Estimate V(2.1)
The value of 2.1 is
[tex]\mathbf{V(2.1) = (-2.1)^3 + (2.1)^2 + 12 \times 2.1}[/tex]
[tex]\mathbf{V(2.1) = 20.35}[/tex]
(h) The maximum number of people infected at the same time
Using the graph, the maximum point on the graph is:
[tex]\mathbf{(t,V(t) = (2.361,20.745)}[/tex]
This means that:
The maximum number of people infected at the same time is approximately 21.
The rate of infection at this point is:
[tex]\mathbf{m = \frac{V(t)}{t}}[/tex]
[tex]\mathbf{m = \frac{20.745}{2.361}}[/tex]
[tex]\mathbf{m = 8.79}[/tex]
The rate of infection is 8.79 people per week
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Answer
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What is line?A simple line in Euclidean geometry. A line, according to Euclid, is the distance between two points and can stretch forever in either direction. While Euclid's initial definition is now regarded as a line segment, such an extension in both directions is now thought of as a line.
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Step-by-step explanation:
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The correct answer hope this helps and Brainliest?
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Answer:
Step-by-step explanation:
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Step-by-step explanation:
mark brainliest
Answer:
1.08 feet
Step-by-step explanation:
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1.08 feet