Let the school hire x buses and y vans.
A bus can hold 40 students and 3 teachers.
A van can hold 8 students and 1 teacher.
The number of students riding in buses and vans is at least 400, therefore
40x + 8y ≥ 400 (1)
The number of teachers riding in buses and vans is at most 36, therefore
3x + y ≤ 36 (2)
Write (1) and (2) as
y ≥ 50 - 5x (3)
y ≤ 36 - 3x (4)
The equality portion of the solution of (3) and (4) is
36 - 3x = 50 - 5x
2x = 14
x = 7 => y = 36 - 3*7 = 15
A graph of the inequalities indicates the acceptable solution in shaded color, as shown below.
The minimum cost of renting buses and vans is
7*$1200 + 15*$100 = $9900
Answer: The minimum cost is $9,900
Answer:
½log3 + ½logx
Step-by-step explanation:
½(log(3x))
½(log3 + logx)
½logx + ½log3
He had 40 pencils left after he gave away 8, so originally he had 40 + 8 pencils, which is 48.
Now, he bought 4 packages, which had a total of 48 pencils, so divide 48 by 4, which is 12. He had 12 pencils in each package.
To determine the solution arithmetically, first add 8 to 40, then divide 48 by 4.
To determine the solution algebraically, set up and solve the equation 40 = 4x - 8.
Each package contained 12 pencils.
Hope this helps
All you have to do here is find the lcm.
the time it will take them too complete the lap at the same time is 12 minutes.