What is the equation of the parabola?

Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut

frozen [14]

Answer:

a) $\sqrt{144-288t+208t^2}$ b.) -12knots, 8 knots c) No e) $4\sqrt{13}$

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

a)

We know that at time t , the ship A has moved $12\dot t (n.m)$ and ship B has moved $8\dot t (n.m)$ . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

$\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}$

b)

We want to find $\frac{ds}{dt}$ for t=0 and t=1

$\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots$

c)

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = $s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0$

Since function $f(x)=199-288x+208x^2$ is quadratic, concave up and has no real roots, we know that $199-288x+208x^2>0$ for every t. So, the ships haven't seen each other.

d)

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

e)

Function ds/dt has a horizontal asympote in the first quadrant if

$\lim_{t \to \infty} \frac{ds}{dt}$

So, lets check this limit:

$\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}$

Notice that:

$4\sqrt{13}=\sqrt{12^2+5^2}$ =√(speed of ship A² + speed of ship B²)

5 0

3 years ago

Solve 3x + 9 = 24.<br> A. x = 5<br> B. x = 11<br> C. x = 12<br> D. x = 18

DIA [1.3K]

Answer:

A. x = 5

Step-by-step explanation:

3x + 9 = 24

→ Minus 9 from both sides

3x = 15

→ Divide both sides by 3

x = 5

4 0

1 year ago

PLEASE HELP ME WITH Question 5

kozerog [31]

There are many ways to solve number 5 as it is just an reduce/enlargement of equidistant fractions

The first method that came up to my mind is just dividing denominators.

For A) you can do 72 divided by 18 which would give you 4. Then you can divide the given numerator with the answer you got from dividing the denominators. This would be 8 divided by 4 which would give you two. This means that 2/18 is equal to 8/72. You can check by simplifying the fraction. Both 2/18 and 8/72 equal 9 when simplified, therefore correct.

You can use this method for b, either use a calculator because of the decimals or just remove the decimals and replace them later. Make sure when you use this method, you are always dividing the bigger number. So in B you would do 40.3 divided by 12.4

4 0

3 years ago

How to find the height of a tree in cm. 5m and 4m equal how many cm

erastova [34]

The answer is 9 cm

Hope that helps! :)

3 0

3 years ago

Read 2 more answers

Creat a table of values for the equation y=-x-3

nadya68 [22]

Answer:

x /\ y

1 /\ -4

2 /\ -5

3 /\ -6

4 /\ -7

... /\ ...

Step-by-step explanation:

4 0

3 years ago