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2 years ago

Helllllp plz!!!! I really need it!!!!

Mathematics

2 answers:

ahrayia [7]2 years ago

8 0

Answer:

The second one

Step-by-step explanation:

Its not the third one because it is an equilateral

its not the first one because it is not obtuse

The triangle is isosceles

obtuse means the angles are big

the angles are small so they are acute

I hope this helps you

atroni [7]2 years ago

7 0

Acute and equilateral, you can tell all the sides are equal based on the Tik marks

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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.

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:

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

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

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

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.

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

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

Evaluate b-(-1/8)+c where b=2 and c=-7/4

tatiyna

Answer:

3/8

Step-by-step explanation:

Evaluate b - -1/8 + c where b = 2 and c = -7/4:

b - (-1)/8 + c = 2 - (-1)/8 - 7/4

Put 2 + 1/8 - 7/4 over the common denominator 8. 2 + 1/8 - 7/4 = (8×2)/8 + 1/8 + (2 (-7))/8:

(8×2)/8 + 1/8 - (7×2)/8

8×2 = 16:

16/8 + 1/8 - (7×2)/8

2 (-7) = -14:

16/8 + 1/8 + (-14)/8

16/8 + 1/8 - 14/8 = (16 + 1 - 14)/8:

(16 + 1 - 14)/8

16 + 1 = 17:

(17 - 14)/8

| 1 | 7

- | 1 | 4

| 0 | 3:

Answer: 3/8

8 0

2 years ago

Umnica [9.8K]

The answer you are looking for is 12! have a great day!:)

6 0

3 years ago

Do you think the equation y=|x|+6 will be a function or not a function?

Lelechka [254]

It is a function, so yes ig

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3 years ago

A store sells five models of cameras for $190.00, $420.00, $270.00, $300.00, and $420.00. If the sales tax rate is 6%, what are

Juliette [100K]

mean= $339.20

median= $318.00

mode= $445.20

range= $243.80

3 0

2 years ago

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