The distance covered by the hare and the tortoise in t seconds are 8t and 5t respectively. (Simple Speed-Distance-Time relation)
The tortoise gets a 510m headstart so at t=0 is 1490m.
The functions representing the distance of both of them from the finish line is,
F(x)=2000-8t,. for hare
G(x)=1490-5t,. for tortoise
Answer:
15.6 meters/minute
Step-by-step explanation:
Answer:
6
Step-by-step explanation:
Half of eight = four
3/4
Double three = six
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.
