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
First term (a) = -15
common difference(d)= -9-(-15)
= -9+15=6
So,
12th term(a12)= a +(n-1)d
= -15+(12-1)×6
= -15 + 11×6
= -15+66
= 51
Answer:
0.76 or 76%
Step-by-step explanation:
Please kindly check the attached files for details
Answer:
11 ft
Step-by-step explanation:
Given the two lengths of a triangle as
AB = 6ft
AC = 6ft
This is an isosceles triangle because only 2 sides are equal.
In an isosceles triangle, the sum of 2 (sides) lengths must be greater than the other length.
Therefore, let's assume the following:
i) AC + AB > BC
6 + 6 > BC
12 > BC (BC is less than 12)
BC < 12
ii) BC + AC > AB
BC + 6 > 6
BC > 6 - 6
BC > 0
Therefore the range of values for BC =
0 < BC < 12
Since BC must be bigger than one of the lengths and it must also be less than the sum of the 2 sides. The length of BC could be 11 because it is less than (6+6) 12 and greater than 6.
