Each Friday, the school prints 400 copies of the school newsletter. The equation c = 400w models the relationship between the nu
mber of weeks and the total number of copies of the newsletters printed. What is true of the graph of this scenario?
2 answers:
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Answer:
A.) Acceleration = 4.6 m/s^2
B.) Average speed = 47 m/s
Step-by-step explanation:
A.) From the graph, at t = 10s,
Velocity V = 46 m/s
Acceleration a = (change in V)/ time t
a = (46 - 0)/10
a = 46/10 = 4.6 m/s^2
B.) If we estimate the speed at 5 seconds intervals, then we have
Average speed = (36 + 46 + 49 + 50 + 50 +50 + 50) ÷ 7
Average speed = 331/7 = 47.29 m/s
Average speed = 47 m/s
Answer:
The maximum value is 1/27 and the minimum value is 0.
Step-by-step explanation:
Note that the given function is equal to then it means that it is positive i.e
.
Consider the function
We want that the gradient of this function is equal to zero. That is (the calculations in between are omitted)
Note that the last equation is our restriction. The restriction guarantees us that at least one of the variables is non-zero. We've got 3 options, either 1, 2 or none of them are zero.
If any of them is zero, we have that the value of the original function is 0. We just need to check that there exists a value for lambda.
Suppose that x is zero. Then, from the second and third equation we have that
. If lambda is not zero, then y =z. But, since
and lambda is not zero, this implies that x=y=z=0 which is not possible. This proofs that if one of the variables is 0, then lambda is zero. So, having one or two variables equal to zero are feasible solutions for the problem.
Suppose that only x is zero, then we have the solution set .
If both x,y are zero, then we have the solution set . We can find the different solution sets by choosing the variables that are set to zero.
NOw, suppose that none of the variables are zero.
From the first and second equation we have that
which implies
Also, from the first and third equation we have that
which implies
So, in this case, replacing this in the restriction we have , which gives as another solution set. On this set, we have
. Over this solution set, we have that the value of our function is