Class A: 6v + 8b = 202
Class B: 12v + 16b = 284
Solve using the elimination method:
since 6v and 12v are perfect for elimination, multiply the class A equation by 2 so that the van variable cancels out:
12v + 16b = 404
12v + 10b = 284
Then subtract the bottom equation from the top:
6b = 120
b = 20
Now you know that each bus can hold 20 students.
Just plug this into one of the original equations to solve for vans:
6v + 8(20) = 202
6v + 160 = 202
6v = 42
v=7
So then you know that each van can hold 7 students.
Check:
12 (7) + 10 (20) = 284
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He switched the 5 and 6 into different places
6+g
Sum = addition
........
C; this is because the line of beat fit isn’t linear but curved in that one graph.
Answer:
The answer is:O f(1) = 10, f(n) = f(n − 1) + 2, for n > 1
Step-by-step explanation:
f1=10 because the first number is 10. then f(n)=f(n-1)+2 for N because it goes up 2 every time.
