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I am Lyosha [343]
3 years ago
13

A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectil

e is modeled using the equation h(t) = –16t2 + 48t + 190.
Mathematics
2 answers:
miskamm [114]3 years ago
6 0

Answer: C

Step-by-step explanation: took the test

Furkat [3]3 years ago
3 0

Answer:

t = 5.25 seconds

Step-by-step explanation:

Given that,

A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectile is modeled using the equation:

h(t)=-16t^2+48t+190

It is assumed to find the time when the projectile will hit the ground. When the projectile hit the ground, its height is equal to 0 such that,

-16t^2+48t+190=0

It forms a quadratic equation such that,

t=\dfrac{-b+\sqrt{b^{2}-4ac } }{2a},\dfrac{-b-\sqrt{b^{2}-4ac } }{2a}\\\\t=\dfrac{-48+\sqrt{(48)^{2}-4\times (-16)(190) } }{2\times (-16)},\dfrac{-48-\sqrt{(48)^{2}-4\times (-16)(190) } }{2\times (-16)}\\\\t=-2.25, 5.25

So, the projectile will hit the ground at t = 5.25 seconds.

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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.

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)

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LUCKY_DIMON [66]

Answer:

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Step-by-step explanation:

because it's 13 / 10 it means that there is a whole number , 13 - 10 equals 3 oh, so you divide 3 from 100

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Answer:

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Step-by-step explanation:

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84.15-1.8*8
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