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Montano1993 [528]
3 years ago
6

Philip drove a tractor-trailer rig onto a ferry boat. He left the rig in gear because of a problem with the breaks. Posted regul

ations prohibited the starting of engines before docking, but when the ferry was about 50 feet from the dock, Philip started his engine. That cause the tractor-trailer to jump forward and strike Herrick’s car, which in turn hit Patton’s car. Patton’s car, at the head of the line, crashed through the ferry’s barricades and plunged into the water. The car could not be recovered. What was the tort duty in this case? Where was the breach of the duty? What were the injuries? What was the proximate cause of the injury to Patton’s car?
Advanced Placement (AP)
1 answer:
gregori [183]3 years ago
4 0

Answer:

A) the tort duty in this case is for automoble owners to not start their engine before the ferry docks.

B) the breach of duty in this case is Philip starting his engine before the ferry docked.

C) the injury in this case is the destruction of the ferry's barricade and the loss of patton's car.

D) the proximate cause of injury to patton's car is as a result of Philip's negligence of the tort duty.

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Answer:

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Explanation:

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4 years ago
Which of the following people would be MOST likely to migrate?
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Which sum is prime? A) 1 + 3 B) 2 + 4 C) 5 +8 D) 7+9
puteri [66]

Answer:

C

Explanation:

In other option ,the result comes out to be even

in option A, its 4

in option B,its 6

in option C,its 13

in option D,its 16

4 0
3 years ago
I can’t figure this problem out
Natalka [10]

Answer:

Explanation:

Alright so the way to do this is to use properties of integrals to make our life easier.

So we have:

\int\limits^4_1 {(3f(x)+2)} \, dx

So lets break this up into two different integrals that represent the same area.

\int\limits^4_0 {f(x)} \, dx - \int\limits^1_0 {f(x)} \, dx = \int\limits^4_1 {f(x)} \, dx

Lets think about what is going on up there. The integral from four to zero gives us the area under the curve of f(x) from four to zero. If we subtract this from the integral from one to zero (the area under f from one to zero) we are left with the area under f from four to one! Hence:

\int\limits^4_1 {f(x)} \, dx

But since we have these values we can say that:

-3 - 2 = -5

Which means that \int\limits^4_1 {f(x)} \, dx = -5

So now we can evaluate \int\limits^4_1 {(3f(x)+2)} \, dx

Lets first break up our integrand into two integrals

\int\limits^4_1 {(3f(x)+2)} \, dx = 3\int\limits^4_1{f(x)} \, dx + 2\int\limits^4_1 {} \, dx

Now we can evaluate this:

We know that \int\limits^4_1 {f(x)} \, dx = -5

So:

3(-5)+2[x] where x is evaluated at 4 to 1 so

-15 + 2(3)

So we are left with -15 + 6 = -9

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3 years ago
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Answer:

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Explanation:

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