B: $195
C: $210
D: $225
Answer:
Option B.
Step-by-step explanation:
Let x be the number of cans of beans and y be the number of cans of corn.
Cafeteria’s budget allows it to purchase at most 60 cans of beans and 45 cans of corn.
1 can of beans feeds 5 students, and 1 can of corn feeds 6 students. Each student will have beans or corn, but not both, and there will be a maximum of 420 students at lunch.
Can of beans cost $2.00 and a can of corn cost $3.00.
Objective function, Z=2x+3y
The required linear programming problem is
Objective function, Z=2x+3y
Subject to the constraints
(Only 1st quadrant)
Draw the graph of these constraints as shown below.
The verities of common shaded region are (0,45), (30,45), (60,20), (60,0), (0,0).
Points Z=2x+3y
(0,0) 0
(0,45) 135
(30,45) 195
(60,20) 180
(60,0) 120
The maximum amount of money required to feed all of the students either beans or corn is $195.
Number of cans of beans = 30
Number of cans of corn = 45
Therefore, the correct option is B.
A: if a*b is odd than a and b are odd
B: if a and b are odd, than a*b is odd
C: a*b is odd
D: a and b are odd
-Show Schedule-
Chemistry-Every 10 minutes
Electricity-Every 20 minutes
Recycling-Every 6 minutes
Fossils-Every 45 minutes
The first showing for all shows is at 10:00 A.M.
The chemistry and recycling presentations coincide every 30 minutes starting from 10:00 A.M. They will occur at the same time at 10:30 A.M., 11:00 A.M., 11:30 A.M., etc., until 5:00 P.M.
To solve this problem, we need to find the common multiple of 10 and 6, which is the frequency in minutes of the chemistry and recycling presentations respectively. The least common multiple (LCM) of 10 and 6 is 30. So, the chemistry and recycling presentations will coincide every 30 minutes.
Starting at 10:00 A.M., they would coincide at 10:30 A.M., 11:00 A.M., 11:30 A.M., and continue in this pattern until 5:00 P.M.
#SPJ11
Answer:
All real numbers are solutions
Step-by-step explanation: