Answer:
21
ctto JBarnum
Step-by-step explanation:
P+11=(1/2)(P-13)^2
P+11=(1/2)(P^2-26p+169)
2(P+11)=(1/2)(2)(P^2-26p+169)
2P+22=P^2-26p+169
0=P^2-28p+147 ( -7 and -21)
(P-7)(P-21)=0
P=7,21 (cant be 7 as 13 years from 7 is negative so peter is 21)
checking:
21+11=32
32=(1/2)(21-13)^2
32=(1/2)(8)^2
32=(1/2)64
32=32
correct
Let the number of Go-go scooter to be produced be x and that of Whip scooter y, then
Maximize: R = 3200x + 5000y
subject to:
x + y ≤ 120
3x + 6y ≤ 600
The corner points of the constraints are (0, 0), (0, 100), (40, 80), (120, 0)
For (0, 0): R = 0
For (0, 100): R = 3200(0) + 5000(100) = 500,000
For (40, 80): R = 3200(40) + 5000(80) = 528,000
For (120, 0): R = 3200(120) + 5000(0) = 384,000
Therefore, for maximum revenue, they should produce 40 Go-go scooters and 80 Whip scooters.
Answer:3040
Step-by-step explanation:16000/100=160x19=3040
Answer:
The loss on the sale of 50 type M cars is $8000, and the loss is decreasing at the rate of $500 per 50 type M car sold.
Step-by-step explanation:
Given
Required
Interpret
From the question, we understand that:
Number of cars
Profit on x cars
Using the above as a point of reference,
Rate of loss per x cars
So, the interpretations are:
A loss of 8000 on 50 cars
The loss rate is a reduction of 500 per 50 car
