How would I show that on a number line?
1 answer:
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Answer:
1) x ≤ 2 or x ≥ 5
2) -6 < x < 2
Step-by-step explanation:
1) We have x^2 - 7x + 10, so let's factor this as if this were a regular equation:
x^2 - 7x + 10 = (x - 2)(x - 5)
So, we now have (x - 2)(x - 5) ≥ 0
Let's imagine this as a graph (see attachment). Notice that the only place that is above the number line is considered greater than 0, and that's when x ≤ 2 or x ≥ 5 (the shaded region).
2) Again, we have x^2 + 4x - 12, so factor this as if this were a regular equation:
x^2 + 4x - 12 = (x + 6)(x - 2)
So now we have (x + 6)(x - 2) < 0
Now imagine this as a graph again (see second attachment). Notice that the only place that is below 0 (< 0) is when -6 < x < 2 (the shaded region).
Hope this helps!
Answer:
The maximum is at (0,4)
Step-by-step explanation:
x≥0
y≥0
x+y ≤4
2x+y ≤6
We need to plot the four inequalities. The left side has no axis. The max is at 4. The minimum is at 0 for the shaded area
When we have a system of constraints, the minimum and maximum are at the vertices of the graph
There are 4 points
(0,0)
The intersection point of the 2 lines which is (2,2)
The intersection at the y axis (0,4)
The intersection at the x axis is (3,0)
Once we have these points we put them into the constraint equation
C = 30x+50y
(0,0)
C = 0+0 = 0 This would be the trivial point since nothing happens
(0,4)
C = 0+50(4) = 200
(3,0)
C = 30*3 = 90
(2,2)
C = 30(2) + 50(2) = 60+100 = 160
The maximum is at (0,4)
