Answer:
$34
Step-by-step explanation:
I'd work backwards.
For the third book, she paid all her remaining money. The problem says she paid "1/2 her leftover money + $3". This means that: (let m = money used to buy book 3)
m = 1/2m + 3
1/2m = 3
m = 6
For the second book: (let n = money before book 2)
n - m (money left after book 2) = 1/2n + 2
1/2n +2 is money used up for book, which is the same as n-m.
n = 1/2n + 2 + m
1/2n = 2 + m = 2 + 6
1/2n =8
n = 16
For the first book, she spent 1/2 her money + 1. If o = money before book 1 (or the whole allowance):
o - n = 1/2o + 1
o = 1/2o + n + 1 = 1/2o + 16 + 1
1/2o = 17
o = 34
Check!
Spent 17 (half 34) + 1 on book 1
16 left
Spent 8 (half 16) +2 on book 2
6 left
Spent 3 (half 6) + 3 on book 3
0 left
Use a system of equations.
(x=# of small dogs, y=# of big dogs)
x+y=9 (times by -5) -5x-5y=-45
5x+8y=51
Then add the equations together and the x's cancel.
3y=6
y=2
So he walked 2 big dogs and 7 little dogs.
Remark.
The problem is a bit indistinct. Where exactly are the two edges of the road? I'm going to say that they are the x intercepts, but that may not be true. Certainly it does not have to be true at all.
Graph.
A graph has been made for you. The maximum is marked for you. It is an approximation The actual height can be more accurately found.
Height
y = (-1/200)(x - 16)(x + 16)
y = (-1/200)*(x^2 - 256)
The maximum height for this graph only is when x = 0.Other graphs require completing the square.
y = (-1/200) * (-256)
y = 1.28 exactly. I thought the graph might be rounding the answer. It is not.
