Answer: The answer is (A) [0,100].
Step-by-step explanation: Given in the question that a marine biologist Kelsey determined that the population of Blue Whales off the coast of Antarctica varies with time as shown in the given graph.
The population of Blue Whales (y) is plotted on the Y-axis and the time in months (x) is plotted on the X-axis.
We need to find the range of the population of Blue Whales off the coast of Antarctica from 0 to 10 months.
We can see from the attached graph that
when x = 0, then y = 0,
when x = 10, then y = 100.
That is, when time varies between 0 to 10 months, then the population of Blue Whales varies between 0 and 100.
Thus, the correct option is (A) [0,100].
E. 1/2
With trig functions, multiple x values correspond with the same y value.
Using an initial x value (the principle value), we can find other x values for the same y value, this is what we are are being asked to find in this question.
There are slightly different ways to find these x values (also known as solutions) for each of the basic trig functions.
The x values are in degrees for the basic trig functions.
For cosine, the rule is as follows:
360 - principle value;
this will give what I, personally, like to think of as a secondary principle value (this value is not actually recognised as a secondary principle value, I just like to think of it as such). Anyway, all other solutions can the be found by adding or substrating any integer multiple of 360 to/from the PV and 'secondary PV'.
For your question:
cos60 = 1/2
60 is the x value (PV)
so...
360 - 60 = 300 is the 'secondary PV'
Just to be clear, this means that if I were to find the cos300, I would get 1/2.
That is sufficient for explaining the answer to this particular question but if you wanted to find any other solution, you would just have to do either:
60 + or - n(360)
or...
300 + or - n(360),
where n = any integer
Step-by-step explanation:
i = interest 3% for 30 years
This is a simple dynamical system for whom the the solutions are given as
](https://tex.z-dn.net/?f=S%3DR%5B%5Cfrac%7B%28i%2B1%29%5En-1%7D%7Bi%7D%5D%28i%2B1%29)
putting values we get
S=2000[\frac{(1.03)^{30}-1}{0.03}](1.03)
= $98005.35
withdrawal of money takes place from one year after last payment
To determine the result we use the present value formula of an annuity date
we need to calculate R so putting the values and solving for R we get
R= $6542.2356
