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.
down 3 unit mean -3 outside root
right 2 unit mean -2 inside root,
so the answer is
540/9 = 60cm
408/10 = 40.8
60-40.8 = 19.2cm longer
Answer:
It's A.
Step-by-step explanation:
5x2 + 60x = 0
5x is the GCF so:
5x(x + 12 ) = 0
5x = 0, x + 12 = 0
x =0, x = -12.
Answer: x = -7/2
Step-by-step explanation:
10x+ 30 = 20 +8x+ 3
10x+ 30=23+8x
10x-8x = 23-30
2x=-7
( a helpful app to help with these type of problems is photomath)
