For this equation, substitute y for 1/2x-5, since y=1/2x-5.
4x+2(1/2x-5)= 0
4x+x-10=0
Combine like terms.
5x-10= 0
Add 10 to both sides.
5x=10
Divide by 5.
x=2
Now, plug in x for y.
y=1/2(2)-5
y= 1-5
y=-4
(2,-4) <=== solution
We can check this by plugging it in.
4x+2y= 0
4(2)+2(-4)=0
8+2(-4)=0
8-8=0
0=0
This works.
I hope this helps!
~kaikers
Let the number of large bookcases be x and number of small bookcases be y, then
Maximise P = 80x + 50y;
subkect to:
6x + 2y ≤ 24
x, y ≥ 2
The corner points are (2, 2), (2, 6), (3.333, 2)
For (2, 2): P = 80(2) + 50(2) = 160 + 100 = 260
For (2, 6): P = 80(2) + 50(6) = 160 + 300 = 460
For (3.333, 2): P = 80(3.333) + 50(2) = 266.67 + 100 = 366.67
Therefore, for maximum profit, he should produce 2 large bookcases and 6 small bookcases.
The answer is "D" because u have to divided them up in to groups and d is the only one that add's even close to 32.
C 87 the median :::::::::::::