At noon, ship A is 180 km west of ship B. Ship A is sailing east at 40 km/h and ship B is sailing north at 35 km/h. How fast is
the distance between the ships changing at 4:00 PM?
1 answer:
Answer:
≈ 29 km/h
Step-by-step explanation:
If t is the number of hours since noon, then the distance between ship A and the origin is 180 − 40t, and the distance between ship B and the origin is 35t.
Using Pythagorean theorem, the distance between the ships is:
d² = (35t)² + (180 − 40t)²
d² = 1225t² + 32400 − 14400t + 1600t²
d² = 2825t² − 14400t + 32400
Taking derivative with respect to time:
2d dd/dt = 5650t − 14400
dd/dt = (2825t − 7200) / d
First, find d when t = 4.
d² = 2825 (4)² − 14400 (4) + 32400
d² = 20000
d = 100√2
Now find dd/dt.
dd/dt = (2825(4) − 7200) / (100√2)
dd/dt = 41/√2
dd/dt ≈ 29 km/h
You might be interested in
-12(-3) + 7
First, simplify 12 × (-3) to get -36. / Your problem should look like:
Second, add 36 + 7. / Your problem should look like: 43
Answer: 43
First, simplify 12 × (-3) to get -36. / Your problem should look like:
Second, add 36 + 7. / Your problem should look like: 43
Answer: 43
Answer:
The answer is "0.765 and 0.2353".
Step-by-step explanation:
Please find the complete question in the attached file.
In point a:
P(a substantive term only)
P(major health insurance only)
P(both)
P(renewal) =P(insurance and renewal term only)+P (substantial and renewable health insurance only)+P (both and renew)
In point b:
In reality, the probability of having both life and major medical insurance provided the policyholder would renew next year