I’m not sure probably just your internet!
Answer:
bus = 43
van 12
Step-by-step explanation:
This can be solved using simultaneous equations
Let v represent the number of students that a van carries
Let b represent the number of students that a bus carries
the following equations can be derived from the question
3v + 2b = 122 eqn 1
5v + 3b = 189 eqn 2
Multiply eqn 1 by 5 and eqn 2 by 3
15v + 10b = 610 eqn 3
15v + 9b = 567 eqn 4
Subtract equation 4 from 3
b = 43
Substitute for b in equation 1
3v + 2(43) = 122
solve for v
v = 12
They mowed total of 11/15 (5/15 + 2/5) already, so they still have 4/15.
Answer:
f(1) = 4; f(n) = 4 + d(n - 1), n > 0.
Step-by-step explanation:
This arithmetic sequence has a common difference of d with first term = 4.
f(1) = 4; f(n) = 4 + d(n - 1), n > 0.
Answer:
16 rides
Step-by-step explanation:
Option 1 . Admission fee = $10
Each ride = $0.50
Option 2 . Admission fee = $6
Each ride = $0.75
Let no. of rides be x
So, cost of ride according to option 1 = 0.50x
So, total cost after having x rides according to option 1 :
= 10+0.50x ---1
Cost of ride according to option 2 = 0.75x
So, total cost after having x rides according to option 2 :
= 6+0.75x --2
Now to find the beak even point i.e. having the same cost
Equate 1 and 2





Thus for 16 rides , the two options have the same cost .
Hence the break even point is 16 rides