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Answer:
J
Step-by-step explanation:
If the side of the shape is increased by 3/2 then the total distance around the shape will be 3/2 bigger.
Example: If the pentagon has side length 2 then it becomes 2*3/2 = 6/2 = 3.
The original perimeter was 2+2+2+2+2 = 10.
The new perimeter is 3+3+3+3+3 = 15.
Did you notice that 
? J is correct.
Answer:
<em>The company needs to sell 40 desks to break even</em>
Step-by-step explanation:
<u>Application of Equations</u>
There is virtually no limit to the possible situations where equations can help to find the solution of specific problems related to areas like economy, where one could need to establish some important indicators about the business.
B. The fixed cost for Abstract Office Supplies to sell a new computer desk is $14,000. Each desk will cost $150 to produce. The cost function to produce X desks is
C(x)=150x+14,000
A. The revenue for each desk is estimated at $500, for X desks will be
R(x)=500x
C. The company will break even when the cost and the revenue are the same. We'll find how many desks need to be sold for that to happen. We equate
C(x)=R(x)
Or equivalently
150x+14,000=500x
Rearranging
500x-150x=14,000
350x=14,000
Solving for x
x=14,000/350= 40
The company needs to sell 40 desks to break even
Multiple both sides of the equation by V, and divide both sides by T.. You'll get PV/T=KT/V*(V/T)=K.
Answer:
When we have a function f(x), the domain of the function is the set of all the inputs that "work" (Not only in a mathematical way, the context is also important) with the function f(x)
In this case, we have a function M(p) = $2*p
This function represents the amount of money collected depending on the number of people who ride on the ferris whell.
Then p can be only a whole number (we can not have 1.5 people, only whole numbers of people).
And we also know that the maximum capacity of the ferris is 64 people.
Then:
p ≤ 64
And we also should add the restriction:
0 ≤ p ≤ 64
(Because p can't be smaller than zero)
Such that p should also be an integer, then, the domain is:
D: p ∈ Z, p ∈ {0, 1, 2, ..., 64}
