Answer:
Wonka bars=3 and Everlasting Gobstoppers=24
Step-by-step explanation:
let the wonka bars be X
and everlasting gobstoppers be Y
the objective is to
maximize 1.3x+3.2y=P
subject to constraints
natural sugar
4x+2y=60------1
sucrose
x+3y=75---------2
x>0, y>0
solving 1 and 2 simultaneously we have
4x+2y=60----1
x+3y=75------2
multiply equation 2 by 4 and equation 1 by 1 to eliminate x we have
4x+2y=60
4x+12y=300
-0-10y=-240
10y=240
y=240/10
y=24
put y=24 in equation 2 we have'
x+3y=75
x+3(24)=75
x+72=75
x=75-72
x=3
put x=3 and y=24 in the objective function we have
maximize 1.3x+3.2y=P
1.3(3)+3.2(24)=P
3.9+76.8=P
80.7=P
P=$80.9
-7-w<10
Add 7 to each side
-w<17
Change signs and flip sign
w>17
Hope this helps :)
In order for the product of some numbers to end in a zero, the prime factors of ... So 1 x 2 x 3 x ... x 98 x 99 x 100 (AKA 100!) has 24 zeroes at the end
You find 8% of 55.35 which is 4.42
Then you find 15% of 55.35 which is 8.30
You add them together which equals 12.72
You then subtract 12.75 from 55.35 55.35-12.75=42.63
The original price was $42.63
Answer:
Neither
Step-by-step explanation:
