Answer:
12x8=96, GCF is 4, LCM is 24
Step-by-step explanation:
Answer:
The origin is (0, 0) so let's say we have (0, 5) we find that this point is 5 units away from the origin. We find that by subtracting the origin's y coordinate from our (0, 5) point's y coordinate. This method can be used for the x-axis too.
Answer:
1
Step-by-step explanation:
Step 1: Simplify both sides of the equation.
x+24+13=2x−16
x+ 1 2 + 1 3 =2x+ −1 6
(x)+( 1 2 + 1 3 )=2x+ −1 6 (Combine Like Terms)
x+ 5 6 =2x+ −1 6
x+ 5 6 =2x+ −1 6
Step 2: Subtract 2x from both sides.
x+ 5 6 −2x=2x+ −1 6 −2x
−x+ 5 6 = −1 6
Step 3: Subtract 5/6 from both sides.
−x+ 5 6 − 5 6 = −1 6 − 5 6
−x=−1
Step 4: Divide both sides by -1.
−x −1 = −1 −1
x=1
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
