The fair market rental rate for Bloomington is $747 and for Kokomo is $654. The relative change from Bloomington to Kokomo is __
1 answer:
Answer:
<em>The relative change from Bloomington to Kokomo is -12.449...%</em>
Step-by-step explanation:
The fair market rental rate for Bloomington is $747 and for Kokomo is $654.
So, the amount of relative change from Bloomington to Kokomo will be: dollar.
<em>Formula for percentage of relative change = (Amount of relative change/ Initial amount)× 100%</em>
So, the percentage will be:
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Answer: It's a tie between f(x) and h(x). Both have the same max of y = 3
The highest point shown on the graph of f(x) is at (x,y) = (pi,3). The y value here is y = 3.
For h(x), the max occurs when cosine is at its largest: when cos(x) = 1.
So,
h(x) = 2*cos(x)+1
turns into
h(x) = 2*1+1
h(x) = 2+1
h(x) = 3
showing that h(x) maxes out at y = 3 as well
--------------------------------
Note: g(x) has all of its y values smaller than 0, so there's no way it can have a max y value larger than y = 3. See the attached image to see what this graph would look like if you plotted the 7 points. A parabola seems to form. Note how point D = (-3, -2) is the highest point for g(x). So the max for g(x) is y = -2
The highest point shown on the graph of f(x) is at (x,y) = (pi,3). The y value here is y = 3.
For h(x), the max occurs when cosine is at its largest: when cos(x) = 1.
So,
h(x) = 2*cos(x)+1
turns into
h(x) = 2*1+1
h(x) = 2+1
h(x) = 3
showing that h(x) maxes out at y = 3 as well
--------------------------------
Note: g(x) has all of its y values smaller than 0, so there's no way it can have a max y value larger than y = 3. See the attached image to see what this graph would look like if you plotted the 7 points. A parabola seems to form. Note how point D = (-3, -2) is the highest point for g(x). So the max for g(x) is y = -2
The number of months until the insect population reaches 40 thousand is 14.29 months and the limiting factor on the insect population as time progresses is 250 thousands.
Given that population P(t) (in thousands) of insects in t months after being transplanted is P(t)=(50(1+0.05t))/(2+0.01t).
(a) Firstly, we will find the number of months until the insect population reaches 40 thousand by equating the given population expression with 40, we get
P(t)=40
(50(1+0.05t))/(2+0.01t)=40
Cross multiply both sides, we get
50(1+0.05t)=40(2+0.01t)
Apply the distributive property a(b+c)=ab+ac, we get
50+2.5t=80+0.4t
Subtract 0.4t and 50 from both sides, we get
50+2.5t-0.4t-50=80+0.4t-0.4t-50
2.1t=30
Divide both sides with 2.1, we get
t=14.29 months
(b) Now, we will find the limiting factor on the insect population as time progresses by taking limit on both sides with t→∞, we get
(c) Further, we will sketch the graph of the function using the window 0≤t≤700 and 0≤p(t)≤700 as shown in the figure.
Hence, when the population P(t) (in thousands) of insects in t months after being transplanted by P(t)=(50(1+0.05t))/(2+0.01t) then the number of months until the insect population reaches 40 thousand 14.29 months and the limiting factor on the insect population is 250 thousand and the graph is shown in the figure.
Learn more about limiting factor from here brainly.com/question/18415071.
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