What is the difference between the area of the large triangle and the area of the small triangle?
2 answers:
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Micah was asked to add the following expressions:
First, he combined like terms in the numerator and kept the common denominator
First step is correct. He added the like terms in the numerator, because the denominators are same.
So he got ,
In the next step, he cannot cancel out x^2 from the top and bottom . Because x-4 and 3x+2 are added with x^2
If we have x^2 is multiplied with other terms at the top and bottom , then we can cancel out x^2.
So Micah added the expression incorrectly. Final answer is not correct.
By listing the samples in descending order we have - Sample B, Sample E, Sample A, Sample C, and Sample D.
Step-by-step explanation:
Step 1; First, we must convert all the temperatures into a common scale. The given temperatures are in three different units i.e. °C, K, and °F. Here we convert all the units into K.
To convert °C to K, we add 273.15 with the °C , 44 °C= 44 + 273.15 = 317.15 K.
To convert °F, we have the formula (112°F − 32) × () + 273.15, so
110 °F = (110 - 32) × () + 273.15 = 316.483 K,
112 °F = (112 - 32) × () + 273.15 = 317.594 K.
Step 2; By rewriting all the values on the same unit, we get;
Sample A = 317.15 K,
Sample B = 320 K,
Sample C = 316.483 K,
Sample D = 310 K,
Sample E = 317.594.
The descending order is as follows 320 K, 317.594K, 317.15 K, 316.483 K and 310 K which is Sample B, Sample E, Sample A, Sample C, and Sample D.
Answer:
(a) 5 senior, 4 apprentice
(b) 368 per shift
(c) 7.5 senior, 0 apprentice
Step-by-step explanation:
The problem can be described by two inequalities. On describes the limit on the number of elves in the shop; the other describes the limit on the total payroll. Let x and y represent the number of apprentice and senior elves, respectively. Then the inequalities for the first scenario are ...
x + y ≤ 9 . . . . . . . . . . . . total number of elves in the shop
5x +8y ≤ 480/8 . . . . . . candy canes per hour paid to elves
These two inequalities are graphed in the first attachment. They describe a solution space with vertices at ...
(x, y) = (0, 7.5), (4, 5), (9, 0)
__
(a) Santa wants to maximize the output of trucks, so wants to maximize the function t = 4x +6y.
At the vertices of the solution space, the values of this function are ...
t(0, 7.5) = 45
t(4, 5) = 46
t(9, 0) = 36
Output of trucks is maximized by a workforce of 4 apprentice elves and 5 senior elves.
__
(b) The above calculations show 46 trucks per hour can be made, so ...
46×8 = 368 . . . trucks in an 8-hour shift
__
(c) The new demands change the inequalities to ...
x + y ≤ 8 . . . . . . number of workers
7x +8y ≤ 60 . . . total wages (per hour)
The vertices of the feasible region for these condtions are ...
(x, y) = (0, 7.5), (4, 4), (8, 0)
From above, we know the truck output will be maximized at the vertex (x, y) = (0, 7.5). However, we know we cannot have 7.5 senior elves working in the shop. We can have 7 or 8 elves working.
If the workforce must remain constant, truck output is maximized by a workforce of 7 senior elves.
If the workforce can vary through the shift, truck output is maximized by adding one more senior elf in the shop for half a shift.
Santa should assign 7 senior elves for the entire shift, and 8 senior elves (one more) for half a shift.
_____
<em>Comment on apprentice elf wages</em>
At 5 candy canes for 4 trucks, apprentice elves produced trucks for a cost of 1.25 candy canes per truck. At 8 candy canes for 6 trucks, senior elves produced trucks for a cost of about 1.33 candy canes per truck. The reason for employing senior elves in the first scenario is that their productivity is 1.5 times that of apprentice elves while their cost per truck is about 1.07 times that of apprentice elves.
After the apprentice elves wages were increased, their cost per truck is 1.75 candy canes per truck, but their productivity hasn't changed. They have essentially priced themselves out of a job, because they are not competitive with senior elves.
