Step-by-step explanation:
you don't show us the choices.
anyway, the real graph must be similar to the one you are showing, as it goes to - and + infinity for x = 2.
because the denominator "x-2" will be 0 for x = 2.
but the horizontal limits are both y = +3 (and not 0).
because (3x-2)/(x-2) goes more and more to 3/1 the larger (or smaller in the - direction) x gets.
Step-by-step explanation:
let us give all the quantities in the problem variable names.
x= amount in utility stock
y = amount in electronics stock
c = amount in bond
“The total amount of $200,000 need not be fully invested at any one time.”
becomes
x + y + c ≤ 200, 000,
Also
“The amount invested in the stocks cannot be more than half the total amount invested”
a + b ≤1/2 (total amount invested),
=1/2(x + y + c).
(x+y-c)/2≤0
“The amount invested in the utility stock cannot exceed $40,000”
a ≤ 40, 000
“The amount invested in the bond must be at least $70,000”
c ≥ 70, 000
Putting this all together, our linear optimization problem is:
Maximize z = 1.09x + 1.04y + 1.05c
subject to
x+ y+ c ≤ 200, 000
x/2 +y/2 -c/2 ≤ 0
≤ 40, 000,
c ≥ 70, 000
a ≥ 0, b ≥ 0, c ≥ 0.
Answer is 150/x see photo for solution
The answer is -5 all you have to do is divide -15 by 3
Answer: 6
<u>Step-by-step explanation:</u>
Let x represent the number of plates bought for $6
Let y represent the Total number of plates bought = x + 1
<em>Remember that Mrs. Lim bought another plate for $18</em>
Total cost ÷ Total plates = Average Cost
6x + 18 = 8(x + 1)
6x + 18 = 8x + 8
18 = 2x + 8
10 = 2x
5 = x
y = x + 1
= 5 + 1
= 6
