which figures do not have lateral area?
1 answer:
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Answer:
No lower than $208, and no higher than $956.8
Step-by-step explanation:
Notice that the revenue (income) function is represented by a quadratic expression (a parabola with branches pointing down), and the cost function is linear with negative slope.
When we plot them, we obtain the graphs shown in the attached image. Notice as well that there are two points of intersection for these two functions. The revenue exceeds (lays above) the cost in between the two points of intersection (marked in red and green).
For the company to make a profit, the revenue has to be larger than the cost, so we study the limiting values for that to happen, which is the points of intersection of these two curves.
The first one (pictured in red) is associated with coordinates (35, 33488) which tells us that is the production of 35 chairs with a revenue of $33,488. That is a cost per chair of $33,488 / 35 = $956.8
The second intersection (pictured in green) is associated with coordinates (125, 26000), which tells us that is the production of 125 chairs with a revenue of $26,000. That is a cost per chair of $26000 / 125 = $ 208
So these values give us the minimum and maximum that the company can charge to make a profit.
Answer:
The number of ways is equal to
Step-by-step explanation:
The multiplication principle states that If a first experiment can happen in n1 ways, then a second experiment can happen in n2 ways ... and finally a i-experiment can happen in ni ways therefore the total ways in which the whole experiment can occur are
n1 x n2 x ... x ni
Also, given n-elements in which we want to put them in a row, the total ways to do this are n! that is n-factorial.
For example : We want to put 4 different objects in a row.
The total ways to do this are ways.
Using the multiplication principle and the n-factorial number :
The number of ways to put all 40 in a row for a picture, with all 12 sophomores on the left,all 8 juniors in the middle, and all 20 seniors on the right are : The total ways to put all 12 sophomores in a row multiply by the ways to put the 8 juniors in a row and finally multiply by the total ways to put all 20 senior in a row ⇒