. A biologist is comparing the growth of a population of flies per week to the number of flies an iguana will consume per week.
He has devised an equation to solve for which day (x) the iguana would be able to eat the entire population. The equation is 3x = 2x + 1. However, he has observed that the iguana cannot eat more than eight flies in one week.. . Explain to the biologist how he can solve this on a graph using a system of equations. Identify any possible constraints to the situation.
2 answers:
Answer:
The biologist can graph two functions, one for the flies population growth and one for the flies the iguana can eat, and see that they intersect at two points (0, 0) and (1, 3).
The constraints are:
0 < x ≤ 7
1 ≤ y ≤ 8
Therefore, the only solution to the equation can be (1, 3).
Explanation:
In order to solve the equation 3ˣ = 2x + 1, the biologist can break it into two parts.
The function for the popuation growth is: f(x) = 3ˣ
The function for the number of flies eaten is: g(x) = 2x + 1
Then, he can graph both functions (I used Geogebra, the graph is attached) and see where they intersect, which means the points where f(x) = g(x). Algebrically, it means find the soutions to the system:
From the graph we can see that the two functions intersect at (0, 0) and (1, 3). In order to see if these solutions are acceptable, we need to set the constraints.
We know that x represents the number of days in a week, therefore: 0 < x ≤ 7 (we cannot have 0 days, and they cannot be more than or equal to 7).
We also know that y represents the number of flies, therefore: 1 ≤ y ≤ 8 (we need to start with at least a population of 1 fly and we cannot have more than 8 flies eaten per week).
Due to these restraints, we have to reject the solution (0, 0) which would have no sense.
Hence, by graphing the two functions and setting the restraints, the biologist can find that (1, 3) solves his equation. This means that after 1 day the iguana would eat the entire population of 3 flies.
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Answer:
The answer is .
Step-by-step explanation:
First, it is important to recall that the group law is not commutative in general, so we cannot assume it here. In order to solve the exercise we need to remember the axioms of group, specially the existence of the inverse element, i.e., for each element there exist another element, denoted by
such that
, where
stands for the identity element of G.
So, given the equality we make a left multiplication by
and we obtain:
But, . Hence,
.
Now, in the equality we make a right multiplication by
, and we obtain
.
Recall that and
. Therefore,
.