She is starting at 4000, so that is the y-intercept of the graph. She is descending at 40 feet per minute, so her rate of change or slope is -40. The graph would look like this:
Answer/Step-by-step explanation:
Total Distance = 3Ft
a) Determine the amplitude and period of a sinusoidal function that models the bobbing buoy.
Amplitude = 3/2 = 1.5 period = 10 second
No ph. Shift
b) Write an equation of a function that models the buoy with x=0 at its highest point.
y= 1.5 cos (
)
Answer:
See attachment
Step-by-step explanation:
The given system is
2x-y=4
x-y=-2
We write the system in slope-intercept form:
y=2x-4
y=x+2
The first equation has slope of 2 and y-intercept at (0,-4)
When y=0, x=2.
We plot (0,-4) and (2,0) to graph this first equation.
The second equation has a slope of 1 and a y-intercept at (0,2)
Also when y=0, x=-2.
We plot (0,2) and (-2,0) and draw a straight line through them.
The graph of the system is shown in the attachment.
<span>You can probably just work it out.
You need non-negative integer solutions to p+5n+10d+25q = 82.
If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80.
So this is the same as n + 2d + 5q ≤ 16
So now you simply have to "crank out" the cases.
Case q=0 [ n + 2d ≤ 16 ]
Case (q=0,d=0) → n = 0 through 16 [17 possibilities]
Case (q=0,d=1) → n = 0 through 14 [15 possibilities]
...
Case (q=0,d=7) → n = 0 through 2 [3 possibilities]
Case (q=0,d=8) → n = 0 [1 possibility]
Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81
Case q=1 [ n + 2d ≤ 11 ]
Case (q=1,d=0) → n = 0 through 11 [12]
Case (q=1,d=1) → n = 0 through 9 [10]
...
Case (q=1,d=5) → n = 0 through 1 [2]
Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42
Case q=2 [ n + 2 ≤ 6 ]
Case (q=2,d=0) → n = 0 through 6 [7]
Case (q=2,d=1) → n = 0 through 4 [5]
Case (q=2,d=2) → n = 0 through 2 [3]
Case (q=2,d=3) → n = 0 [1]
Total from case q=2: 1 + 3 + 5 + 7 = 16
Case q=3 [ n + 2d ≤ 1 ]
Here d must be 0, so there is only the case:
Case (q=3,d=0) → n = 0 through 1 [2]
So the case q=3 only has 2.
Grand total: 2 + 16 + 42 + 81 = 141 </span>