First, let's find the time where the ship arrives to point C
t=d/V, t= 45/30=1.5h
if the ship's departure time is at 1.5h, on the point C, let 's find its arrival to the point O, V=D/Dt, Dt=tfinal-tinitial
Dt=D/V= 20/30=0.6h=t final -t initial, implies t final=0.6+t initial=0.6+1.5=2.1h
so the ship will be on the radar at 2.1h,
Answer:
7. (-1, 10)
8. (7,-6)
9. C
10. One solution, Infinitely Many Solutions, No Solution
11. Adult tickets: 52 Student tickets: 68
Let us assume the two unknown numbers to be "x" and "y".
Then
x + y = 37
And
xy = - 210
x = - 210/y
Now let us put the value of "x" in the first equation, we get
x + y = 37
- (210/y) + y = 37
y^2 - 210 = 37y
y^2 - 37y - 210 = 0
y^2 - 42y + 5y - 210 = 0
y(y - 42) + 5(y - 42) = 0
(y - 42)(y + 5) = 0
So when
y - 42 = 0
y = 42
Then
x = -(210/42)
= - 5
Again when
y + 5 = 0
y = -5
Then
x = 42
So the two unknown numbers are 42 and -5. I hope the procedure is clear enough for you to understand.
Answer:
Step-by-step explanation:
The <em>change in y</em> is the difference in y-coordinates between the lower right point and the upper left point. Similarly, the <em>change in x</em> is the difference in the x-coordinates of those points.
