1. If the total baggage fees were equal, you would have to equate the two situations. Suppose x is the weight above the limit on the flight home, while y is the weight above the limit on the flight back to college. Then, the equation would be:
20 + 8x = 26 + 5y
Now, we have two unknowns, but only one independent equation. This is unsolvable unless we make an assumption. Suppose the excess weight for both trips are the same such that x=y, then,
<em>20 + 8x = 26 + 5x</em>
2. Then, we solve for x.
8x - 5x = 26 - 20
3x = 6
x = 6/3 = <em>2 lb</em>
Thus, Elizabeth's bag was 2 lb over the weight limit.
3. The total baggage fee would then be:
20+8(2) = 26 + 5(2) = <em>$36</em>
I hope this helps you
y=2/3x+6
y-6=2/3x
x=2/3 (y-6)
f^-1 (x)= 2/3. (x-6)
Since the function is y = 3x, take any value of x and multiply it to get y.
If the y-coordinate is not 3 times the x-coordinate, it is not on the graph.
0 * 3 = 0, so (0, 0) is on the graph.
5 * 3 = 15, so (5, 15) is on the graph.
9 * 3 = 27, not 3, so (9, 3) is not on the graph.
7 * 3 = 21, so (7, 21) is on the graph.
You find the time and how fast it is going then you divide the distance by the time to find the unit rate. Hope it helped. :)