Answer:
b
Step-by-step explanation:
is the situation could the expression model.
The points you found are the vertices of the feasible region. I agree with the first three points you got. However, the last point should be (25/11, 35/11). This point is at the of the intersection of the two lines 8x-y = 15 and 3x+y = 10
So the four vertex points are:
(1,9)
(1,7)
(3,9)
(25/11, 35/11)
Plug each of those points, one at a time, into the objective function z = 7x+2y. The goal is to find the largest value of z
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Plug in (x,y) = (1,9)
z = 7x+2y
z = 7(1)+2(9)
z = 7+18
z = 25
We'll use this value later.
So let's call it A. Let A = 25
Plug in (x,y) = (1,7)
z = 7x+2y
z = 7(1)+2(7)
z = 7+14
z = 21
Call this value B = 21 so we can refer to it later
Plug in (x,y) = (3,9)
z = 7x+2y
z = 7(3)+2(9)
z = 21+18
z = 39
Let C = 39 so we can use it later
Finally, plug in (x,y) = (25/11, 35/11)
z = 7x+2y
z = 7(25/11)+2(35/11)
z = 175/11 + 70/11
z = 245/11
z = 22.2727 which is approximate
Let D = 22.2727
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In summary, we found
A = 25
B = 21
C = 39
D = 22.2727
The value C = 39 is the largest of the four results. This value corresponded to (x,y) = (3,9)
Therefore the max value of z is z = 39 and it happens when (x,y) = (3,9)
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Final Answer: 39
I hope this helps in answering your question! If you have any inquiries to the solution, feel free to leave a comment!
Answer:
10 minutes
Step-by-step explanation:
First off, you want to figure out how many photos it prints in one minute. 36 divided by 8 is 4.5. So you should add 4.5 to 36, which makes 40.5 in 9 minutes. Add another 4.5 and you get 45 which you should add another minute. That gives you 10 minutes.
Another way is to find how many per minute which would again be 4.5. 45 divided by 4.5 is 10. So once again, 10 minutes.
Answer:
See explanation
Step-by-step explanation:
In the figure below, segment CD is parallel to segment EF, DE is a transversal, then angles DIH and HGI are congruent as alternate interior angles when two parallel lines are cut by a transversal.
Consider triangles DIH and EGH. In these triangles,
as alternate interior angles;
as vertical angles;
because point H bisects segment DE (given).
Thus,
by AAS postulate
